Deck 25: Multiple and Logistic Regression

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Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below.

 <strong>Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below. ​   The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin: R1Length<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (BodyWeight<sub>i</sub>) + β<sub>2</sub> (Aviary0Wild1<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:  \begin{array}{l} \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 35.79 & 16.95 \\ \text { BodyWeight } & 2.826 & 1.030 \\ \text { Aviary0Wild1 } & -10.453 & 4.996 \end{array}\\ \mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \% \end{array}    \begin{array}{lrrr} \text { Analysis of Variance } & & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 2 & 879.79 & 439.90 \\ \text { Residual Error } & 22 & 2036.71 & 92.58 \\ \text { Total } & 24 & 2916.50 & \end{array}  Based on the software output provided, what is the equation for the multiple regression model?</strong> A)R1Length = 35.79 + 2.83 BodyWeight - 10.45 Aviary0Wild1 B)R1Length = 16.95 + 1.03 BodyWeight + 5.00 Aviary0Wild1 C)R1Length = 35.79 BodyWeight + 2.83 Aviary0Wild1 - 10.45 D)R1Length = 16.95 BodyWeight + 1.03 Aviary0Wild1 + 5.00 <div style=padding-top: 35px>  The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin:
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996S=9.62172RSq=30.2%\begin{array}{l}\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 35.79 & 16.95 \\\text { BodyWeight } & 2.826 & 1.030 \\\text { Aviary0Wild1 } & -10.453 & 4.996\end{array}\\\mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \%\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 2879.79439.90 Residual Error 222036.7192.58 Total 242916.50\begin{array}{lrrr}\text { Analysis of Variance } & & \\\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 2 & 879.79 & 439.90 \\\text { Residual Error } & 22 & 2036.71 & 92.58 \\\text { Total } & 24 & 2916.50 &\end{array} Based on the software output provided, what is the equation for the multiple regression model?

A)R1Length = 35.79 + 2.83 BodyWeight - 10.45 Aviary0Wild1
B)R1Length = 16.95 + 1.03 BodyWeight + 5.00 Aviary0Wild1
C)R1Length = 35.79 BodyWeight + 2.83 Aviary0Wild1 - 10.45
D)R1Length = 16.95 BodyWeight + 1.03 Aviary0Wild1 + 5.00
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Question
Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below.

 <strong>Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below. ​   The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin: R1Length<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (BodyWeight<sub>i</sub>) + β<sub>2</sub> (Aviary0Wild1<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:  \begin{array}{l} \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 35.79 & 16.95 \\ \text { BodyWeight } & 2.826 & 1.030 \\ \text { Aviary0Wild1 }-10.453 & 4.996 \end{array}\\ \mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \% \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 2 & 879.79 & 439.90 \\ \text { Residual Error } & 22 & 2036.71 & 92.58 \\ \text { Total } & 24 & 2916.50 & \end{array} \end{array}  Suppose we wish use the ANOVA F test to test the following hypotheses: H<sub>0</sub>: β<sub>1</sub> = β<sub>2</sub> = 0 H<sub>α</sub>: At least one of β<sub>1</sub> or β<sub>2</sub> is not 0 What is the value of the F statistic for these hypotheses?</strong> A)1.43 B)2.31 C)3.31 D)4.75 <div style=padding-top: 35px>  The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin:
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996S=9.62172RSq=30.2%\begin{array}{l}\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 35.79 & 16.95 \\\text { BodyWeight } & 2.826 & 1.030 \\\text { Aviary0Wild1 }-10.453 & 4.996\end{array}\\\mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \%\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 2879.79439.90 Residual Error 222036.7192.58 Total 242916.50\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 2 & 879.79 & 439.90 \\\text { Residual Error } & 22 & 2036.71 & 92.58 \\\text { Total } & 24 & 2916.50 &\end{array}\end{array} Suppose we wish use the ANOVA F test to test the following hypotheses:
H0: β1 = β2 = 0
Hα: At least one of β1 or β2 is not 0
What is the value of the F statistic for these hypotheses?

A)1.43
B)2.31
C)3.31
D)4.75
Question
Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below.

 <strong>Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below. ​   The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin: R1Length<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (BodyWeight<sub>i</sub>) + β<sub>2</sub> (Aviary0Wild1<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:  \begin{array}{l} \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 35.79 & 16.95 \\ \text { BodyWeight } & 2.826 & 1.030 \\ \text { Aviary0Wild1 }-10.453 & 4.996 \end{array}\\ \mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \% \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 2 & 879.79 & 439.90 \\ \text { Residual Error } & 22 & 2036.71 & 92.58 \\ \text { Total } & 24 & 2916.50 & \end{array} \end{array}  Suppose we wish use the ANOVA F test to test the following hypotheses: H<sub>0</sub>: β<sub>1</sub> = β<sub>2</sub> = 0 H<sub>α</sub>: At least one of β<sub>1</sub> or β<sub>2</sub> is not 0 What is the P-value of this test?</strong> A)Greater than 0.10 B)Between 0.05 and 0.10 C)Between 0.01 and 0.05 D)Less than 0.01 <div style=padding-top: 35px>  The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin:
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996S=9.62172RSq=30.2%\begin{array}{l}\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 35.79 & 16.95 \\\text { BodyWeight } & 2.826 & 1.030 \\\text { Aviary0Wild1 }-10.453 & 4.996\end{array}\\\mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \%\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 2879.79439.90 Residual Error 222036.7192.58 Total 242916.50\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 2 & 879.79 & 439.90 \\\text { Residual Error } & 22 & 2036.71 & 92.58 \\\text { Total } & 24 & 2916.50 &\end{array}\end{array} Suppose we wish use the ANOVA F test to test the following hypotheses:
H0: β1 = β2 = 0
Hα: At least one of β1 or β2 is not 0
What is the P-value of this test?

A)Greater than 0.10
B)Between 0.05 and 0.10
C)Between 0.01 and 0.05
D)Less than 0.01
Question
Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below.

 <strong>Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below. ​   The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin: R1Length<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (BodyWeight<sub>i</sub>) + β<sub>2</sub> (Aviary0Wild1<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:  \begin{array}{l} \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 35.79 & 16.95 \\ \text { BodyWeight } & 2.826 & 1.030 \\ \text { Aviary0Wild1 }-10.453 & 4.996 \end{array}\\ \mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \% \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 2 & 879.79 & 439.90 \\ \text { Residual Error } & 22 & 2036.71 & 92.58 \\ \text { Total } & 24 & 2916.50 & \end{array} \end{array}  What percent of variation in tail-feather length can be explained by the regression model?</strong> A)9% B)30% C)55% D)93% <div style=padding-top: 35px>  The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin:
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996S=9.62172RSq=30.2%\begin{array}{l}\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 35.79 & 16.95 \\\text { BodyWeight } & 2.826 & 1.030 \\\text { Aviary0Wild1 }-10.453 & 4.996\end{array}\\\mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \%\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 2879.79439.90 Residual Error 222036.7192.58 Total 242916.50\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 2 & 879.79 & 439.90 \\\text { Residual Error } & 22 & 2036.71 & 92.58 \\\text { Total } & 24 & 2916.50 &\end{array}\end{array} What percent of variation in tail-feather length can be explained by the regression model?

A)9%
B)30%
C)55%
D)93%
Question
Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below.

 <strong>Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below. ​   The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin: R1Length<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (BodyWeight<sub>i</sub>) + β<sub>2</sub> (Aviary0Wild1<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:  \begin{array}{l} \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 35.79 & 16.95 \\ \text { BodyWeight } & 2.826 & 1.030 \\ \text { Aviary0Wild1 } & -10.453 & 4.996 \end{array}\\ \mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \% \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 2 & 879.79 & 439.90 \\ \text { Residual Error } & 22 & 2036.71 & 92.58 \\ \text { Total } & 24 & 2916.50 & \end{array} \end{array}  What is a 95% confidence interval for 1, the coefficient of BodyWeight in this model?</strong> A)2.826 ± 1.030 B)2.826 ± 2.019 C)2.826 ± 2.126 D)2.826 ± 2.136 <div style=padding-top: 35px>  The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin:
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996S=9.62172RSq=30.2%\begin{array}{l}\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 35.79 & 16.95 \\\text { BodyWeight } & 2.826 & 1.030 \\\text { Aviary0Wild1 } & -10.453 & 4.996\end{array}\\\mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \%\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 2879.79439.90 Residual Error 222036.7192.58 Total 242916.50\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 2 & 879.79 & 439.90 \\\text { Residual Error } & 22 & 2036.71 & 92.58 \\\text { Total } & 24 & 2916.50 &\end{array}\end{array} What is a 95% confidence interval for 1, the coefficient of BodyWeight in this model?

A)2.826 ± 1.030
B)2.826 ± 2.019
C)2.826 ± 2.126
D)2.826 ± 2.136
Question
Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below.

 <strong>Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below. ​   The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin: R1Length<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (BodyWeight<sub>i</sub>) + β<sub>2</sub> (Aviary0Wild1<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:  \begin{array}{l} \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 35.79 & 16.95 \\ \text { BodyWeight } & 2.826 & 1.030 \\ \text { Aviary0Wild1 } & -10.453 & 4.996 \end{array}\\ \mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \% \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 2 & 879.79 & 439.90 \\ \text { Residual Error } & 22 & 2036.71 & 92.58 \\ \text { Total } & 24 & 2916.50 & \end{array} \end{array}  To determine whether the bird's origin affects tail-feather length in this model, we test the hypotheses H<sub>0</sub>: β<sub>2</sub> = 0, H<sub>α</sub>: β<sub>2</sub> ≠ 0 in this particular model. Using the estimates provided, what is the P-value for this test?</strong> A)Greater than 0.10 B)Between 0.05 and 0.10 C)Between 0.01 and 0.05 D)Less than 0.01 <div style=padding-top: 35px>  The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin:
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996S=9.62172RSq=30.2%\begin{array}{l}\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 35.79 & 16.95 \\\text { BodyWeight } & 2.826 & 1.030 \\\text { Aviary0Wild1 } & -10.453 & 4.996\end{array}\\\mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \%\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 2879.79439.90 Residual Error 222036.7192.58 Total 242916.50\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 2 & 879.79 & 439.90 \\\text { Residual Error } & 22 & 2036.71 & 92.58 \\\text { Total } & 24 & 2916.50 &\end{array}\end{array} To determine whether the bird's origin affects tail-feather length in this model, we test the hypotheses
H0: β2 = 0, Hα: β2 ≠ 0 in this particular model. Using the estimates provided, what is the P-value for this test?

A)Greater than 0.10
B)Between 0.05 and 0.10
C)Between 0.01 and 0.05
D)Less than 0.01
Question
Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below.

 <strong>Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below. ​   The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin: R1Length<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (BodyWeight<sub>i</sub>) + β<sub>2</sub> (Aviary0Wild1<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:  \begin{array}{l} \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 35.79 & 16.95 \\ \text { BodyWeight } & 2.826 & 1.030 \\ \text { Aviary0Wild1 } & -10.453 & 4.996 \end{array}\\ \mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \% \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 2 & 879.79 & 439.90 \\ \text { Residual Error } & 22 & 2036.71 & 92.58 \\ \text { Total } & 24 & 2916.50 & \end{array} \end{array}  Using a significance level of 0.05, what should you conclude about long-tailed finches, based on your calculations for a confidence interval for β<sub>1</sub> and for a test of hypothesis on β<sub>2</sub>?</strong> A)A bird's body weight and origin both significantly explain tail-feather length. B)A bird's body weight significantly helps explain tail-feather length, even after accounting for the bird's origin, and a bird's origin significantly helps explain tail-feather length, even after accounting for the bird's body weight. C)A bird's body weight significantly helps explain tail-feather length, even after accounting for the bird's origin, but a bird's origin does not significantly help explain tail-feather length after accounting for the bird's body weight. D)A bird's origin significantly helps explain tail-feather length even after accounting for the bird's body weight, but a bird's body weight does not significantly help explain tail-feather length after accounting for the bird's origin. <div style=padding-top: 35px>  The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin:
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996S=9.62172RSq=30.2%\begin{array}{l}\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 35.79 & 16.95 \\\text { BodyWeight } & 2.826 & 1.030 \\\text { Aviary0Wild1 } & -10.453 & 4.996\end{array}\\\mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \%\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 2879.79439.90 Residual Error 222036.7192.58 Total 242916.50\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 2 & 879.79 & 439.90 \\\text { Residual Error } & 22 & 2036.71 & 92.58 \\\text { Total } & 24 & 2916.50 &\end{array}\end{array} Using a significance level of 0.05, what should you conclude about long-tailed finches, based on your calculations for a confidence interval for β1 and for a test of hypothesis on β2?

A)A bird's body weight and origin both significantly explain tail-feather length.
B)A bird's body weight significantly helps explain tail-feather length, even after accounting for the bird's origin, and a bird's origin significantly helps explain tail-feather length, even after accounting for the bird's body weight.
C)A bird's body weight significantly helps explain tail-feather length, even after accounting for the bird's origin, but a bird's origin does not significantly help explain tail-feather length after accounting for the bird's body weight.
D)A bird's origin significantly helps explain tail-feather length even after accounting for the bird's body weight, but a bird's body weight does not significantly help explain tail-feather length after accounting for the bird's origin.
Question
Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below.

 <strong>Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below. ​   The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin: R1Length<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (BodyWeight<sub>i</sub>) + β<sub>2</sub> (Aviary0Wild1<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:  \begin{array}{l} \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 35.79 & 16.95 \\ \text { BodyWeight } & 2.826 & 1.030 \\ \text { Aviary0Wild1 }-10.453 & 4.996 \end{array}\\ \mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \% \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 2 & 879.79 & 439.90 \\ \text { Residual Error } & 22 & 2036.71 & 92.58 \\ \text { Total } & 24 & 2916.50 & \end{array} \end{array}  Given this regression model, what is an estimate of the average difference in tail-feather length between 18-g male finches either raised in an aviary or caught in the wild?</strong> A)2.8 g B)5.0 g C)9.6 g D)10.5 g <div style=padding-top: 35px>  The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin:
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996S=9.62172RSq=30.2%\begin{array}{l}\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 35.79 & 16.95 \\\text { BodyWeight } & 2.826 & 1.030 \\\text { Aviary0Wild1 }-10.453 & 4.996\end{array}\\\mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \%\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 2879.79439.90 Residual Error 222036.7192.58 Total 242916.50\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 2 & 879.79 & 439.90 \\\text { Residual Error } & 22 & 2036.71 & 92.58 \\\text { Total } & 24 & 2916.50 &\end{array}\end{array} Given this regression model, what is an estimate of the average difference in tail-feather length between 18-g male finches either raised in an aviary or caught in the wild?

A)2.8 g
B)5.0 g
C)9.6 g
D)10.5 g
Question
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.
μ
 <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot. μ   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall* } \text { stress } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% & \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array} \end{array}  Based on the software output provided, what is the equation for the multiple regression model?</strong> A)amplitude = 4.795 - 0.114 recall - 4.262 stress + 0.434 recall*stress B)amplitude = 1.329 + 0.107 recall + 2.205 stress + 0.168 recall*stress C)amplitude = 4.795 recall - 0.114 stress - 4.262 recall*stress + 0.434 D)amplitude = 1.329 recall + 0.107 stress + 2.205 recall*stress + 0.168 <div style=padding-top: 35px>  The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall*  stress 0.43370.1677 S=2.13946 R-Sq =22.8%\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall* } \text { stress } & 0.4337 & 0.1677 \\\mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% &\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 348.78816.263 Residual Error 36164.7834.577 Total 39213.571\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 3 & 48.788 & 16.263 \\\text { Residual Error } & 36 & 164.783 & 4.577 \\\text { Total } & 39 & 213.571 &\end{array}\end{array} Based on the software output provided, what is the equation for the multiple regression model?

A)amplitude = 4.795 - 0.114 recall - 4.262 stress + 0.434 recall*stress
B)amplitude = 1.329 + 0.107 recall + 2.205 stress + 0.168 recall*stress
C)amplitude = 4.795 recall - 0.114 stress - 4.262 recall*stress + 0.434
D)amplitude = 1.329 recall + 0.107 stress + 2.205 recall*stress + 0.168
Question
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.
μ  <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot. μ   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% & \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array} \end{array}  Suppose we wish use the ANOVA F test to test the following hypotheses: H<sub>0</sub>: β<sub>1</sub> = β<sub>2</sub> = β<sub>3</sub> = 0 H<sub>α</sub>: At least one of β<sub>1</sub> or β<sub>2</sub> or β<sub>3</sub> is not 0 What is the value of the F statistic for these hypotheses?</strong> A)2.14 B)3.55 C)4.58 D)5.42 <div style=padding-top: 35px>  The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall 0.43370.1677 S=2.13946 R-Sq =22.8%\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall } & 0.4337 & 0.1677 \\\mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% &\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 348.78816.263 Residual Error 36164.7834.577 Total 39213.571\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 3 & 48.788 & 16.263 \\\text { Residual Error } & 36 & 164.783 & 4.577 \\\text { Total } & 39 & 213.571 &\end{array}\end{array} Suppose we wish use the ANOVA F test to test the following hypotheses:
H0: β1 = β2 = β3 = 0
Hα: At least one of β1 or β2 or β3 is not 0
What is the value of the F statistic for these hypotheses?

A)2.14
B)3.55
C)4.58
D)5.42
Question
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.
μ  <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot. μ   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall }{ }^{\star} \text { stress } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% & \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array} \end{array}  Suppose we wish use the ANOVA F test to test the following hypotheses: H<sub>0</sub>: β<sub>1</sub> = β<sub>2</sub> = β<sub>3</sub> = 0 H<sub>α</sub>: At least one of β<sub>1</sub> or β<sub>2</sub> or β<sub>3</sub> is not 0 What is the P-value of the test?</strong> A)Greater than 0.10 B)Between 0.05 and 0.10 C)Between 0.01 and 0.05 D)Less than 0.01 <div style=padding-top: 35px>  The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall  stress 0.43370.1677 S=2.13946 R-Sq =22.8%\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall }{ }^{\star} \text { stress } & 0.4337 & 0.1677 \\\mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% &\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 348.78816.263 Residual Error 36164.7834.577 Total 39213.571\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 3 & 48.788 & 16.263 \\\text { Residual Error } & 36 & 164.783 & 4.577 \\\text { Total } & 39 & 213.571 &\end{array}\end{array} Suppose we wish use the ANOVA F test to test the following hypotheses:
H0: β1 = β2 = β3 = 0
Hα: At least one of β1 or β2 or β3 is not 0
What is the P-value of the test?

A)Greater than 0.10
B)Between 0.05 and 0.10
C)Between 0.01 and 0.05
D)Less than 0.01
Question
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.
μ  <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot. μ   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall } \text { stress } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \mathrm{R}-\mathrm{Sq}=22.8 \% & \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array} \end{array}  What is the percent of variation in brainwave relative amplitude that can be explained by the regression model?</strong> A)23% B)48% C)77% D)95% <div style=padding-top: 35px>  The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall  stress 0.43370.1677 S=2.13946RSq=22.8%\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall } \text { stress } & 0.4337 & 0.1677 \\\mathrm{~S}=2.13946 & \mathrm{R}-\mathrm{Sq}=22.8 \% &\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 348.78816.263 Residual Error 36164.7834.577 Total 39213.571\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 3 & 48.788 & 16.263 \\\text { Residual Error } & 36 & 164.783 & 4.577 \\\text { Total } & 39 & 213.571 &\end{array}\end{array} What is the percent of variation in brainwave relative amplitude that can be explained by the regression model?

A)23%
B)48%
C)77%
D)95%
Question
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.
μ  <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot. μ   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall }{ }^{*} \text { stress } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% & \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array} \end{array}  Here are the residual plots for this regression model:   Based on the study description, the scatterplot, and the residual plots, what should you conclude?</strong> A)The assumption of Normality is not met but the other assumptions are met. B)The assumption of constant variance is not met but the other assumptions are met. C)The assumption of independence is not met but the other assumptions are met. D)All of the assumptions for regression inference are met. <div style=padding-top: 35px>  The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall  stress 0.43370.1677 S=2.13946 R-Sq =22.8%\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall }{ }^{*} \text { stress } & 0.4337 & 0.1677 \\\mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% &\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 348.78816.263 Residual Error 36164.7834.577 Total 39213.571\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 3 & 48.788 & 16.263 \\\text { Residual Error } & 36 & 164.783 & 4.577 \\\text { Total } & 39 & 213.571 &\end{array}\end{array} Here are the residual plots for this regression model:
 <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot. μ   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall }{ }^{*} \text { stress } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% & \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array} \end{array}  Here are the residual plots for this regression model:   Based on the study description, the scatterplot, and the residual plots, what should you conclude?</strong> A)The assumption of Normality is not met but the other assumptions are met. B)The assumption of constant variance is not met but the other assumptions are met. C)The assumption of independence is not met but the other assumptions are met. D)All of the assumptions for regression inference are met. <div style=padding-top: 35px>  Based on the study description, the scatterplot, and the residual plots, what should you conclude?

A)The assumption of Normality is not met but the other assumptions are met.
B)The assumption of constant variance is not met but the other assumptions are met.
C)The assumption of independence is not met but the other assumptions are met.
D)All of the assumptions for regression inference are met.
Question
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.
 <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{l} \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall }{ }^{*} \text { stress } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \mathrm{R}-\mathrm{Sq}=22.8 \% & \end{array}\\\\ \text { Analysis of Variance } \end{array}    \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array}  To determine whether the number of pictures recalled affects the brainwave relative amplitude in this model, we test the hypotheses H<sub>0</sub>: β<sub>1</sub> = 0, H<sub>a</sub>: β<sub>1</sub> ≠ 0 in this particular model. Using the estimates provided, what is the P-value for this test?</strong> A)Greater than 0.10 B)Between 0.05 and 0.10 C)Between 0.01 and 0.05 D)Less than 0.01 <div style=padding-top: 35px>  The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall  stress 0.43370.1677 S=2.13946RSq=22.8% Analysis of Variance \begin{array}{l}\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall }{ }^{*} \text { stress } & 0.4337 & 0.1677 \\\mathrm{~S}=2.13946 & \mathrm{R}-\mathrm{Sq}=22.8 \% &\end{array}\\\\\text { Analysis of Variance }\end{array}

 Source  DF  SS  MS  Regression 348.78816.263 Residual Error 36164.7834.577 Total 39213.571\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 3 & 48.788 & 16.263 \\\text { Residual Error } & 36 & 164.783 & 4.577 \\\text { Total } & 39 & 213.571 &\end{array} To determine whether the number of pictures recalled affects the brainwave relative amplitude in this model, we test the hypotheses H0: β1 = 0, Ha: β1 ≠ 0 in this particular model. Using the estimates provided, what is the P-value for this test?

A)Greater than 0.10
B)Between 0.05 and 0.10
C)Between 0.01 and 0.05
D)Less than 0.01
Question
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.
 <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall }{ }^{*} \text { stress } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% & \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array} \end{array}  What is a 95% confidence interval for β<sub>2</sub>, the coefficient of the indicator variable stress in this model?</strong> A)-4.26 ± 2.21 B)-4.26 ± 4.47 C)2.21 ± 1.93 D)2.21 ± 4.47 <div style=padding-top: 35px>  The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall  stress 0.43370.1677 S=2.13946 R-Sq =22.8%\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall }{ }^{*} \text { stress } & 0.4337 & 0.1677 \\\mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% &\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 348.78816.263 Residual Error 36164.7834.577 Total 39213.571\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 3 & 48.788 & 16.263 \\\text { Residual Error } & 36 & 164.783 & 4.577 \\\text { Total } & 39 & 213.571 &\end{array}\end{array} What is a 95% confidence interval for β2, the coefficient of the indicator variable stress in this model?

A)-4.26 ± 2.21
B)-4.26 ± 4.47
C)2.21 ± 1.93
D)2.21 ± 4.47
Question
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.
 <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall }{ }^{*} \text { stress } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% & \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array} \end{array}  The interaction term in this model is statistically significant at the 0.05 level. Given this and the scatterplot provided, what should you conclude?</strong> A)The effect of the number of pictures recalled on brainwave relative amplitude is substantially different when subjects are exposed to a stressful stimulus versus a neutral stimulus. B)The effect of the number of pictures recalled on brainwave relative amplitude is essentially the same when subjects are exposed to a stressful stimulus or a neutral stimulus. C)The number of pictures recalled does not affect brainwave relative amplitude at all. D)The type of stimulus received does not affect brainwave relative amplitude at all. <div style=padding-top: 35px>  The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall  stress 0.43370.1677 S=2.13946 R-Sq =22.8%\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall }{ }^{*} \text { stress } & 0.4337 & 0.1677 \\\mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% &\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 348.78816.263 Residual Error 36164.7834.577 Total 39213.571\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 3 & 48.788 & 16.263 \\\text { Residual Error } & 36 & 164.783 & 4.577 \\\text { Total } & 39 & 213.571 &\end{array}\end{array} The interaction term in this model is statistically significant at the 0.05 level. Given this and the scatterplot provided, what should you conclude?

A)The effect of the number of pictures recalled on brainwave relative amplitude is substantially different when subjects are exposed to a stressful stimulus versus a neutral stimulus.
B)The effect of the number of pictures recalled on brainwave relative amplitude is essentially the same when subjects are exposed to a stressful stimulus or a neutral stimulus.
C)The number of pictures recalled does not affect brainwave relative amplitude at all.
D)The type of stimulus received does not affect brainwave relative amplitude at all.
Question
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.
 <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall }^{*} \text { stress } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% & \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array} \end{array}  Based on the software output provided, what is the equation of the regression model when a stressful stimulus is received (indicator variable stress = 1)?</strong> A)amplitude = 0.533 - 0.114 recall B)amplitude = 0.533 + 0.320 recall C)amplitude = 4.795 - 0.114 recall D)amplitude = 4.795 + 0.320 recall <div style=padding-top: 35px>  The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall  stress 0.43370.1677 S=2.13946 R-Sq =22.8%\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall }^{*} \text { stress } & 0.4337 & 0.1677 \\\mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% &\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 348.78816.263 Residual Error 36164.7834.577 Total 39213.571\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 3 & 48.788 & 16.263 \\\text { Residual Error } & 36 & 164.783 & 4.577 \\\text { Total } & 39 & 213.571 &\end{array}\end{array} Based on the software output provided, what is the equation of the regression model when a stressful stimulus is received (indicator variable stress = 1)?

A)amplitude = 0.533 - 0.114 recall
B)amplitude = 0.533 + 0.320 recall
C)amplitude = 4.795 - 0.114 recall
D)amplitude = 4.795 + 0.320 recall
Question
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. Using software, we obtained the following output:
 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall* stress 0.43370.1677\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall* stress } & 0.4337 & 0.1677\end{array}

 <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. Using software, we obtained the following output:  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall* stress } & 0.4337 & 0.1677 \end{array}     Which of the following statements best describes the interval (2.465, 4.996)?</strong> A)A confidence interval for the mean brainwave relative amplitude of all adult men in a similar experiment exposed to a neutral stimulus and who remember 10 pictures B)A confidence interval for the mean brainwave relative amplitude of all adult men in a similar experiment exposed to a stressful stimulus and who remember 10 pictures C)A prediction interval for the brainwave relative amplitude of one adult man in a similar experiment exposed to a neutral stimulus and who remembers 10 pictures D)A prediction interval for the brainwave relative amplitude of one adult man in a similar experiment exposed to a stressful stimulus and who remembers 10 pictures <div style=padding-top: 35px>
Which of the following statements best describes the interval (2.465, 4.996)?

A)A confidence interval for the mean brainwave relative amplitude of all adult men in a similar experiment exposed to a neutral stimulus and who remember 10 pictures
B)A confidence interval for the mean brainwave relative amplitude of all adult men in a similar experiment exposed to a stressful stimulus and who remember 10 pictures
C)A prediction interval for the brainwave relative amplitude of one adult man in a similar experiment exposed to a neutral stimulus and who remembers 10 pictures
D)A prediction interval for the brainwave relative amplitude of one adult man in a similar experiment exposed to a stressful stimulus and who remembers 10 pictures
Question
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. Using software, we obtained the following output:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall*  stress 0.43370.1677\begin{array} { l r r } \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & - 0.1139 & 0.1069 \\ \text { stress } & - 4.262 & 2.205 \\ \text { recall* } \text { stress } & 0.4337 & 0.1677 \end{array}

 <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. Using software, we obtained the following output:   \begin{array} { l r r } \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & - 0.1139 & 0.1069 \\ \text { stress } & - 4.262 & 2.205 \\ \text { recall* } \text { stress } & 0.4337 & 0.1677 \end{array}     Which of the following statements best describes the interval (-0.804, 8.115)?</strong> A)A confidence interval for the mean brainwave relative amplitude of all adult men in a similar experiment exposed to a neutral stimulus and who remember 10 pictures B)A confidence interval for the mean brainwave relative amplitude of all adult men in a similar experiment exposed to a stressful stimulus and who remember 10 pictures C)A prediction interval for the brainwave relative amplitude of one adult man in a similar experiment exposed to a neutral stimulus and who remembers 10 pictures D)A prediction interval for the brainwave relative amplitude of one adult man in a similar experiment exposed to a stressful stimulus and who remembers 10 pictures <div style=padding-top: 35px>
Which of the following statements best describes the interval (-0.804, 8.115)?

A)A confidence interval for the mean brainwave relative amplitude of all adult men in a similar experiment exposed to a neutral stimulus and who remember 10 pictures
B)A confidence interval for the mean brainwave relative amplitude of all adult men in a similar experiment exposed to a stressful stimulus and who remember 10 pictures
C)A prediction interval for the brainwave relative amplitude of one adult man in a similar experiment exposed to a neutral stimulus and who remembers 10 pictures
D)A prediction interval for the brainwave relative amplitude of one adult man in a similar experiment exposed to a stressful stimulus and who remembers 10 pictures
Question
A study examined the effect of body weight and origin (aviary or wild) on the length of the R1 central tail feather in male finches. The following model is proposed for predicting the length of the R1 feather from the bird's weight (in grams) and origin (aviary = 0, wild = 1):
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to a sample of 25 male finches. Using software, we obtained the following output:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 35.79 & 16.95 \\\text { BodyWeight } & 2.826 & 1.030 \\\text { Aviary0Wild1 } & -10.453 & 4.996\end{array}

 Predicted Values for New Observations  NewObs  Fit  SE Fit 95%CI95% PI 176.224.32(67.26,85.17)(54.35,98.09)286.672.75(80.96,92.38)(65.91,107.43)\begin{array}{l}\text { Predicted Values for New Observations }\\\begin{array}{rrrcc}\text { NewObs } & \text { Fit } & \text { SE Fit } & 95 \% \mathrm{CI} & 95 \% \text { PI } \\1 & 76.22 & 4.32 & (67.26,85.17) & (54.35,98.09) \\2 & 86.67 & 2.75 & (80.96,92.38) & (65.91,107.43)\end{array}\end{array}

 Values of Predictors for New Observations  NewObs  BodyWeight  Aviary0Wild 118.01.0218.00.0\begin{array}{l}\text { Values of Predictors for New Observations }\\\begin{array}{crr}\text { NewObs } & \text { BodyWeight } & \text { Aviary0Wild } \\1 & 18.0 & 1.0 \\2 & 18.0 & 0.0\end{array}\end{array} What is the best description of the interval (67.26, 85.17)?

A)A confidence interval for the mean tail-feather length of all male finches that weigh 18 g and are raised in an aviary
B)A confidence interval for the mean tail-feather length of all male finches that weigh 18 g and are caught in the wild
C)A prediction interval for the tail-feather length of one male finch that weighs 18 g and was raised in an aviary
D)A prediction interval for the tail-feather length of one male finch that weighs 18 g and was caught in the wild
Question
A study examined the effect of body weight and origin (aviary or wild) on the length of the R1 central tail feather in male finches. The following model is proposed for predicting the length of the R1 feather from the bird's weight (in grams) and origin (aviary = 0, wild = 1):
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to a sample of 25 male finches. Using software, we obtained the following output:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996\begin{array} { l r r } \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 35.79 & 16.95 \\ \text { BodyWeight } & 2.826 & 1.030 \\ \text { Aviary0Wild1 } & - 10.453 & 4.996 \end{array}

Predicted Values for New Observations

 NewObs  Fit  SE Fit 95% CI 95% PI 176.224.32(67.26,85.17)(54.35,98.09)286.672.75(80.96,92.38)(65.91,107.43)\begin{array} { r r r c c } \text { NewObs } & \text { Fit } & \text { SE Fit } & 95 \% \text { CI } & 95 \% \text { PI } \\ 1 & 76.22 & 4.32 & ( 67.26,85.17 ) & ( 54.35,98.09 ) \\ 2 & 86.67 & 2.75 & ( 80.96,92.38 ) & ( 65.91,107.43 ) \end{array}

Values of Predictors for New Observations

 NewObs  BodyWeight  Aviary0Wild1 118.01.0218.00.0\begin{array} { c r r } \text { NewObs } & \text { BodyWeight } & \text { Aviary0Wild1 } \\ 1 & 18.0 & 1.0 \\ 2 & 18.0 & 0.0 \end{array} What is the best description of the interval (65.91, 107.43)?

A)A confidence interval for the mean tail-feather length of all male finches that weigh 18 g and are raised in an aviary
B)A confidence interval for the mean tail-feather length of all male finches that weigh 18 g and are caught in the wild
C)A prediction interval for the tail-feather length of one male finch that weighs 18 g and was raised in an aviary
D)A prediction interval for the tail-feather length of one male finch that weighs 18 g and was caught in the wild
Question
A study looked for a protein marker that might help predict active flare-ups in a particular chronic disease. A sample of 32 individuals with the disease had some blood taken to determine their disease status (active flare-up or inactive) and the relative abundance of the protein marker in their blood. We use logistic regression to model disease status as a function of relative abundance. Software gives the following information:
For Log Odds of Active/Inactive
 Term  Estimate  Std Error  Intercept 6.1862.416 Relative abundance 6.2292.651\begin{array} { c c c } \text { Term } & \text { Estimate } & \text { Std Error } \\\text { Intercept } & - 6.186 & 2.416 \\\text { Relative abundance } & 6.229 & 2.651\end{array} What is the probability p that a patient who has a protein marker relative abundance of 0.8 has an active flare-up?

A)0.231
B)0.300
C)0.511
D)0.770
Question
A study looked for a protein marker that might help predict active flare-ups in a particular chronic disease. A sample of 32 individuals with the disease had some blood taken to determine their disease status (active flare-up or inactive) and the relative abundance of the protein marker in their blood. We use logistic regression to model disease status as a function of relative abundance. Software gives the following information:
For Log Odds of Active/Inactive
 Term  Estimate  Std Error  Intercept 6.1862.416 Relative abundance 6.2292.651\begin{array} { c c c } \text { Term } & \text { Estimate } & \text { Std Error } \\\text { Intercept } & - 6.186 & 2.416 \\\text { Relative abundance } & 6.229 & 2.651\end{array} We want to know if the slope of the logistic regression model is significant. What is the value of the test statistic?

A)t = 2.35
B)t = 2.56
C)z = 2.35
D)z = 2.56
Question
A study looked for a protein marker that might help predict active flare-ups in a particular chronic disease. A sample of 32 individuals with the diseases had some blood taken to determine their disease status (active flare-up or inactive) and the relative abundance of the protein marker in their blood. We use logistic regression to model disease status as a function of relative abundance. Software gives the following information:
For Log Odds of Active/Inactive
 Term  Estimate  Std Error  Intercept 6.1862.416 tive abundance 6.2292.651\begin{array} { c c c } \text { Term } & \text { Estimate } & \text { Std Error } \\\text { Intercept } & - 6.186 & 2.416 \\\text { tive abundance } & 6.229 & 2.651\end{array} We want to know if the slope of the logistic regression model is significant. What is the P-value for the two-sided test?

A)Greater than 0.10
B)Between 0.05 and 0.10
C)Between 0.01 and 0.05
D)Less than 0.01
Question
A study looked for a protein marker that might help predict active flare-ups in a particular chronic disease. A sample of 32 individuals with the disease had some blood taken to determine their disease status (active flare-up or inactive) and the relative abundance of the protein marker in their blood. We use logistic regression to model disease status as a function of relative abundance. Software gives the following information:
For Log Odds of Active/Inactive
 Term  Estimate  Std Error  Intercept 6.1862.416 ive abundance 6.2292.651\begin{array} { c c c } \text { Term } & \text { Estimate } & \text { Std Error } \\\text { Intercept } & - 6.186 & 2.416 \\\text { ive abundance } & 6.229 & 2.651\end{array} What is a 95% confidence interval for the odds ratio eβ1?

A)(0.002, 507.2)
B)(0.096, 10.4)
C)(2.8, 91579.3)
D)(35.8, 7186.8)
Question
A case study enrolled case individuals with glioma (a type of brain cancer) and similar control individuals without glioma. Participants were asked whether they had been regular cell phone users for at least a year. We use logistic regression to model glioma status (1 = glioma, 0 = no glioma) as a function of regular cell phone use (1 = yes, 0 = no). Software gives the following information:
For log odds of glioma/no glioma
 Predictor  Coef  SE Coef  Constant 0.080740.04987 Regular use 0.025650.07275\begin{array} { l c c } \text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & - 0.08074 & 0.04987 \\\text { Regular use } & - 0.02565 & 0.07275\end{array} We want to know if the slope of the logistic regression model is significant. What is the value of the test statistic?

A)t = -1.62
B)t = -0.35
C)z = -1.62
D)z = -0.35
Question
A case study enrolled case individuals with glioma (a type of brain cancer) and similar control individuals without glioma. Participants were asked whether they had been regular cell phone users for at least a year. We use logistic regression to model glioma status (1 = glioma, 0 = no glioma) as a function of regular cell phone use (1 = yes, 0 = no). Software gives the following information:
For log odds of glioma/no glioma
 Predictor  Coef  SE Coef  Constant 0.080740.04987 Regular use 0.025650.07275\begin{array} { l c c } \text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & - 0.08074 & 0.04987 \\\text { Regular use } & - 0.02565 & 0.07275\end{array} We want to know if the slope of the logistic regression model is significant. What is the P-value for the two-sided test?

A)Greater than 0.10
B)Between 0.05 and 0.10
C)Between 0.01 and 0.05
D)Less than 0.01
Question
A case study enrolled case individuals with glioma (a type of brain cancer) and similar control individuals without glioma. Participants were asked whether they had been regular cell phone users for at least a year. We use logistic regression to model glioma status (1 = glioma, 0 = no glioma) as a function of regular cell phone use (1 = yes, 0 = no). Software gives the following information:
For log odds of glioma/no glioma
 Predictor  Coef  SE Coef  Constant 0.080740.04987 Regular use 0.025650.07275\begin{array} { l c c } \text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & - 0.08074 & 0.04987 \\\text { Regular use } & - 0.02565 & 0.07275\end{array} What is a 95% confidence interval for the odds ratio eβ1?

A)(-0.17, 0.12)
B)(0.12, 0.35)
C)(0.85, 1.12)
D)(0.97, 1.26)
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Deck 25: Multiple and Logistic Regression
1
Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below.

 <strong>Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below. ​   The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin: R1Length<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (BodyWeight<sub>i</sub>) + β<sub>2</sub> (Aviary0Wild1<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:  \begin{array}{l} \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 35.79 & 16.95 \\ \text { BodyWeight } & 2.826 & 1.030 \\ \text { Aviary0Wild1 } & -10.453 & 4.996 \end{array}\\ \mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \% \end{array}    \begin{array}{lrrr} \text { Analysis of Variance } & & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 2 & 879.79 & 439.90 \\ \text { Residual Error } & 22 & 2036.71 & 92.58 \\ \text { Total } & 24 & 2916.50 & \end{array}  Based on the software output provided, what is the equation for the multiple regression model?</strong> A)R1Length = 35.79 + 2.83 BodyWeight - 10.45 Aviary0Wild1 B)R1Length = 16.95 + 1.03 BodyWeight + 5.00 Aviary0Wild1 C)R1Length = 35.79 BodyWeight + 2.83 Aviary0Wild1 - 10.45 D)R1Length = 16.95 BodyWeight + 1.03 Aviary0Wild1 + 5.00  The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin:
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996S=9.62172RSq=30.2%\begin{array}{l}\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 35.79 & 16.95 \\\text { BodyWeight } & 2.826 & 1.030 \\\text { Aviary0Wild1 } & -10.453 & 4.996\end{array}\\\mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \%\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 2879.79439.90 Residual Error 222036.7192.58 Total 242916.50\begin{array}{lrrr}\text { Analysis of Variance } & & \\\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 2 & 879.79 & 439.90 \\\text { Residual Error } & 22 & 2036.71 & 92.58 \\\text { Total } & 24 & 2916.50 &\end{array} Based on the software output provided, what is the equation for the multiple regression model?

A)R1Length = 35.79 + 2.83 BodyWeight - 10.45 Aviary0Wild1
B)R1Length = 16.95 + 1.03 BodyWeight + 5.00 Aviary0Wild1
C)R1Length = 35.79 BodyWeight + 2.83 Aviary0Wild1 - 10.45
D)R1Length = 16.95 BodyWeight + 1.03 Aviary0Wild1 + 5.00
A
2
Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below.

 <strong>Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below. ​   The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin: R1Length<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (BodyWeight<sub>i</sub>) + β<sub>2</sub> (Aviary0Wild1<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:  \begin{array}{l} \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 35.79 & 16.95 \\ \text { BodyWeight } & 2.826 & 1.030 \\ \text { Aviary0Wild1 }-10.453 & 4.996 \end{array}\\ \mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \% \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 2 & 879.79 & 439.90 \\ \text { Residual Error } & 22 & 2036.71 & 92.58 \\ \text { Total } & 24 & 2916.50 & \end{array} \end{array}  Suppose we wish use the ANOVA F test to test the following hypotheses: H<sub>0</sub>: β<sub>1</sub> = β<sub>2</sub> = 0 H<sub>α</sub>: At least one of β<sub>1</sub> or β<sub>2</sub> is not 0 What is the value of the F statistic for these hypotheses?</strong> A)1.43 B)2.31 C)3.31 D)4.75  The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin:
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996S=9.62172RSq=30.2%\begin{array}{l}\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 35.79 & 16.95 \\\text { BodyWeight } & 2.826 & 1.030 \\\text { Aviary0Wild1 }-10.453 & 4.996\end{array}\\\mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \%\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 2879.79439.90 Residual Error 222036.7192.58 Total 242916.50\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 2 & 879.79 & 439.90 \\\text { Residual Error } & 22 & 2036.71 & 92.58 \\\text { Total } & 24 & 2916.50 &\end{array}\end{array} Suppose we wish use the ANOVA F test to test the following hypotheses:
H0: β1 = β2 = 0
Hα: At least one of β1 or β2 is not 0
What is the value of the F statistic for these hypotheses?

A)1.43
B)2.31
C)3.31
D)4.75
D
3
Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below.

 <strong>Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below. ​   The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin: R1Length<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (BodyWeight<sub>i</sub>) + β<sub>2</sub> (Aviary0Wild1<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:  \begin{array}{l} \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 35.79 & 16.95 \\ \text { BodyWeight } & 2.826 & 1.030 \\ \text { Aviary0Wild1 }-10.453 & 4.996 \end{array}\\ \mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \% \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 2 & 879.79 & 439.90 \\ \text { Residual Error } & 22 & 2036.71 & 92.58 \\ \text { Total } & 24 & 2916.50 & \end{array} \end{array}  Suppose we wish use the ANOVA F test to test the following hypotheses: H<sub>0</sub>: β<sub>1</sub> = β<sub>2</sub> = 0 H<sub>α</sub>: At least one of β<sub>1</sub> or β<sub>2</sub> is not 0 What is the P-value of this test?</strong> A)Greater than 0.10 B)Between 0.05 and 0.10 C)Between 0.01 and 0.05 D)Less than 0.01  The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin:
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996S=9.62172RSq=30.2%\begin{array}{l}\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 35.79 & 16.95 \\\text { BodyWeight } & 2.826 & 1.030 \\\text { Aviary0Wild1 }-10.453 & 4.996\end{array}\\\mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \%\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 2879.79439.90 Residual Error 222036.7192.58 Total 242916.50\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 2 & 879.79 & 439.90 \\\text { Residual Error } & 22 & 2036.71 & 92.58 \\\text { Total } & 24 & 2916.50 &\end{array}\end{array} Suppose we wish use the ANOVA F test to test the following hypotheses:
H0: β1 = β2 = 0
Hα: At least one of β1 or β2 is not 0
What is the P-value of this test?

A)Greater than 0.10
B)Between 0.05 and 0.10
C)Between 0.01 and 0.05
D)Less than 0.01
C
4
Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below.

 <strong>Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below. ​   The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin: R1Length<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (BodyWeight<sub>i</sub>) + β<sub>2</sub> (Aviary0Wild1<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:  \begin{array}{l} \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 35.79 & 16.95 \\ \text { BodyWeight } & 2.826 & 1.030 \\ \text { Aviary0Wild1 }-10.453 & 4.996 \end{array}\\ \mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \% \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 2 & 879.79 & 439.90 \\ \text { Residual Error } & 22 & 2036.71 & 92.58 \\ \text { Total } & 24 & 2916.50 & \end{array} \end{array}  What percent of variation in tail-feather length can be explained by the regression model?</strong> A)9% B)30% C)55% D)93%  The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin:
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996S=9.62172RSq=30.2%\begin{array}{l}\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 35.79 & 16.95 \\\text { BodyWeight } & 2.826 & 1.030 \\\text { Aviary0Wild1 }-10.453 & 4.996\end{array}\\\mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \%\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 2879.79439.90 Residual Error 222036.7192.58 Total 242916.50\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 2 & 879.79 & 439.90 \\\text { Residual Error } & 22 & 2036.71 & 92.58 \\\text { Total } & 24 & 2916.50 &\end{array}\end{array} What percent of variation in tail-feather length can be explained by the regression model?

A)9%
B)30%
C)55%
D)93%
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5
Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below.

 <strong>Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below. ​   The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin: R1Length<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (BodyWeight<sub>i</sub>) + β<sub>2</sub> (Aviary0Wild1<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:  \begin{array}{l} \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 35.79 & 16.95 \\ \text { BodyWeight } & 2.826 & 1.030 \\ \text { Aviary0Wild1 } & -10.453 & 4.996 \end{array}\\ \mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \% \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 2 & 879.79 & 439.90 \\ \text { Residual Error } & 22 & 2036.71 & 92.58 \\ \text { Total } & 24 & 2916.50 & \end{array} \end{array}  What is a 95% confidence interval for 1, the coefficient of BodyWeight in this model?</strong> A)2.826 ± 1.030 B)2.826 ± 2.019 C)2.826 ± 2.126 D)2.826 ± 2.136  The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin:
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996S=9.62172RSq=30.2%\begin{array}{l}\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 35.79 & 16.95 \\\text { BodyWeight } & 2.826 & 1.030 \\\text { Aviary0Wild1 } & -10.453 & 4.996\end{array}\\\mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \%\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 2879.79439.90 Residual Error 222036.7192.58 Total 242916.50\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 2 & 879.79 & 439.90 \\\text { Residual Error } & 22 & 2036.71 & 92.58 \\\text { Total } & 24 & 2916.50 &\end{array}\end{array} What is a 95% confidence interval for 1, the coefficient of BodyWeight in this model?

A)2.826 ± 1.030
B)2.826 ± 2.019
C)2.826 ± 2.126
D)2.826 ± 2.136
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6
Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below.

 <strong>Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below. ​   The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin: R1Length<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (BodyWeight<sub>i</sub>) + β<sub>2</sub> (Aviary0Wild1<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:  \begin{array}{l} \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 35.79 & 16.95 \\ \text { BodyWeight } & 2.826 & 1.030 \\ \text { Aviary0Wild1 } & -10.453 & 4.996 \end{array}\\ \mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \% \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 2 & 879.79 & 439.90 \\ \text { Residual Error } & 22 & 2036.71 & 92.58 \\ \text { Total } & 24 & 2916.50 & \end{array} \end{array}  To determine whether the bird's origin affects tail-feather length in this model, we test the hypotheses H<sub>0</sub>: β<sub>2</sub> = 0, H<sub>α</sub>: β<sub>2</sub> ≠ 0 in this particular model. Using the estimates provided, what is the P-value for this test?</strong> A)Greater than 0.10 B)Between 0.05 and 0.10 C)Between 0.01 and 0.05 D)Less than 0.01  The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin:
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996S=9.62172RSq=30.2%\begin{array}{l}\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 35.79 & 16.95 \\\text { BodyWeight } & 2.826 & 1.030 \\\text { Aviary0Wild1 } & -10.453 & 4.996\end{array}\\\mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \%\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 2879.79439.90 Residual Error 222036.7192.58 Total 242916.50\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 2 & 879.79 & 439.90 \\\text { Residual Error } & 22 & 2036.71 & 92.58 \\\text { Total } & 24 & 2916.50 &\end{array}\end{array} To determine whether the bird's origin affects tail-feather length in this model, we test the hypotheses
H0: β2 = 0, Hα: β2 ≠ 0 in this particular model. Using the estimates provided, what is the P-value for this test?

A)Greater than 0.10
B)Between 0.05 and 0.10
C)Between 0.01 and 0.05
D)Less than 0.01
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7
Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below.

 <strong>Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below. ​   The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin: R1Length<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (BodyWeight<sub>i</sub>) + β<sub>2</sub> (Aviary0Wild1<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:  \begin{array}{l} \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 35.79 & 16.95 \\ \text { BodyWeight } & 2.826 & 1.030 \\ \text { Aviary0Wild1 } & -10.453 & 4.996 \end{array}\\ \mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \% \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 2 & 879.79 & 439.90 \\ \text { Residual Error } & 22 & 2036.71 & 92.58 \\ \text { Total } & 24 & 2916.50 & \end{array} \end{array}  Using a significance level of 0.05, what should you conclude about long-tailed finches, based on your calculations for a confidence interval for β<sub>1</sub> and for a test of hypothesis on β<sub>2</sub>?</strong> A)A bird's body weight and origin both significantly explain tail-feather length. B)A bird's body weight significantly helps explain tail-feather length, even after accounting for the bird's origin, and a bird's origin significantly helps explain tail-feather length, even after accounting for the bird's body weight. C)A bird's body weight significantly helps explain tail-feather length, even after accounting for the bird's origin, but a bird's origin does not significantly help explain tail-feather length after accounting for the bird's body weight. D)A bird's origin significantly helps explain tail-feather length even after accounting for the bird's body weight, but a bird's body weight does not significantly help explain tail-feather length after accounting for the bird's origin.  The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin:
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996S=9.62172RSq=30.2%\begin{array}{l}\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 35.79 & 16.95 \\\text { BodyWeight } & 2.826 & 1.030 \\\text { Aviary0Wild1 } & -10.453 & 4.996\end{array}\\\mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \%\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 2879.79439.90 Residual Error 222036.7192.58 Total 242916.50\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 2 & 879.79 & 439.90 \\\text { Residual Error } & 22 & 2036.71 & 92.58 \\\text { Total } & 24 & 2916.50 &\end{array}\end{array} Using a significance level of 0.05, what should you conclude about long-tailed finches, based on your calculations for a confidence interval for β1 and for a test of hypothesis on β2?

A)A bird's body weight and origin both significantly explain tail-feather length.
B)A bird's body weight significantly helps explain tail-feather length, even after accounting for the bird's origin, and a bird's origin significantly helps explain tail-feather length, even after accounting for the bird's body weight.
C)A bird's body weight significantly helps explain tail-feather length, even after accounting for the bird's origin, but a bird's origin does not significantly help explain tail-feather length after accounting for the bird's body weight.
D)A bird's origin significantly helps explain tail-feather length even after accounting for the bird's body weight, but a bird's body weight does not significantly help explain tail-feather length after accounting for the bird's origin.
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8
Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below.

 <strong>Tail-feather length is a sexually dimorphic trait in long-tailed finches; that is, the trait differs substantially for males and for females of the same species. A bird's size and environment might also influence tail-feather length. Researchers studied the relationship between tail-feather length (measuring the R1 central tail feather) and weight in a sample of 20 male long-tailed finches raised in an aviary and 5 male long-tailed finches caught in the wild. The data are displayed in the scatterplot below. ​   The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin: R1Length<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (BodyWeight<sub>i</sub>) + β<sub>2</sub> (Aviary0Wild1<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:  \begin{array}{l} \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 35.79 & 16.95 \\ \text { BodyWeight } & 2.826 & 1.030 \\ \text { Aviary0Wild1 }-10.453 & 4.996 \end{array}\\ \mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \% \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 2 & 879.79 & 439.90 \\ \text { Residual Error } & 22 & 2036.71 & 92.58 \\ \text { Total } & 24 & 2916.50 & \end{array} \end{array}  Given this regression model, what is an estimate of the average difference in tail-feather length between 18-g male finches either raised in an aviary or caught in the wild?</strong> A)2.8 g B)5.0 g C)9.6 g D)10.5 g  The following model is proposed for predicting the length of the R1 feather from the bird's weight and origin:
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ . This model was fit to the sample of 25 male finches. The following results summarize the least-squares regression fit of this model:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996S=9.62172RSq=30.2%\begin{array}{l}\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 35.79 & 16.95 \\\text { BodyWeight } & 2.826 & 1.030 \\\text { Aviary0Wild1 }-10.453 & 4.996\end{array}\\\mathrm{S}=9.62172 \quad \mathrm{R}-\mathrm{Sq}=30.2 \%\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 2879.79439.90 Residual Error 222036.7192.58 Total 242916.50\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 2 & 879.79 & 439.90 \\\text { Residual Error } & 22 & 2036.71 & 92.58 \\\text { Total } & 24 & 2916.50 &\end{array}\end{array} Given this regression model, what is an estimate of the average difference in tail-feather length between 18-g male finches either raised in an aviary or caught in the wild?

A)2.8 g
B)5.0 g
C)9.6 g
D)10.5 g
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9
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.
μ
 <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot. μ   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall* } \text { stress } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% & \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array} \end{array}  Based on the software output provided, what is the equation for the multiple regression model?</strong> A)amplitude = 4.795 - 0.114 recall - 4.262 stress + 0.434 recall*stress B)amplitude = 1.329 + 0.107 recall + 2.205 stress + 0.168 recall*stress C)amplitude = 4.795 recall - 0.114 stress - 4.262 recall*stress + 0.434 D)amplitude = 1.329 recall + 0.107 stress + 2.205 recall*stress + 0.168  The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall*  stress 0.43370.1677 S=2.13946 R-Sq =22.8%\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall* } \text { stress } & 0.4337 & 0.1677 \\\mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% &\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 348.78816.263 Residual Error 36164.7834.577 Total 39213.571\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 3 & 48.788 & 16.263 \\\text { Residual Error } & 36 & 164.783 & 4.577 \\\text { Total } & 39 & 213.571 &\end{array}\end{array} Based on the software output provided, what is the equation for the multiple regression model?

A)amplitude = 4.795 - 0.114 recall - 4.262 stress + 0.434 recall*stress
B)amplitude = 1.329 + 0.107 recall + 2.205 stress + 0.168 recall*stress
C)amplitude = 4.795 recall - 0.114 stress - 4.262 recall*stress + 0.434
D)amplitude = 1.329 recall + 0.107 stress + 2.205 recall*stress + 0.168
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10
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.
μ  <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot. μ   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% & \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array} \end{array}  Suppose we wish use the ANOVA F test to test the following hypotheses: H<sub>0</sub>: β<sub>1</sub> = β<sub>2</sub> = β<sub>3</sub> = 0 H<sub>α</sub>: At least one of β<sub>1</sub> or β<sub>2</sub> or β<sub>3</sub> is not 0 What is the value of the F statistic for these hypotheses?</strong> A)2.14 B)3.55 C)4.58 D)5.42  The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall 0.43370.1677 S=2.13946 R-Sq =22.8%\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall } & 0.4337 & 0.1677 \\\mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% &\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 348.78816.263 Residual Error 36164.7834.577 Total 39213.571\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 3 & 48.788 & 16.263 \\\text { Residual Error } & 36 & 164.783 & 4.577 \\\text { Total } & 39 & 213.571 &\end{array}\end{array} Suppose we wish use the ANOVA F test to test the following hypotheses:
H0: β1 = β2 = β3 = 0
Hα: At least one of β1 or β2 or β3 is not 0
What is the value of the F statistic for these hypotheses?

A)2.14
B)3.55
C)4.58
D)5.42
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11
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.
μ  <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot. μ   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall }{ }^{\star} \text { stress } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% & \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array} \end{array}  Suppose we wish use the ANOVA F test to test the following hypotheses: H<sub>0</sub>: β<sub>1</sub> = β<sub>2</sub> = β<sub>3</sub> = 0 H<sub>α</sub>: At least one of β<sub>1</sub> or β<sub>2</sub> or β<sub>3</sub> is not 0 What is the P-value of the test?</strong> A)Greater than 0.10 B)Between 0.05 and 0.10 C)Between 0.01 and 0.05 D)Less than 0.01  The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall  stress 0.43370.1677 S=2.13946 R-Sq =22.8%\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall }{ }^{\star} \text { stress } & 0.4337 & 0.1677 \\\mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% &\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 348.78816.263 Residual Error 36164.7834.577 Total 39213.571\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 3 & 48.788 & 16.263 \\\text { Residual Error } & 36 & 164.783 & 4.577 \\\text { Total } & 39 & 213.571 &\end{array}\end{array} Suppose we wish use the ANOVA F test to test the following hypotheses:
H0: β1 = β2 = β3 = 0
Hα: At least one of β1 or β2 or β3 is not 0
What is the P-value of the test?

A)Greater than 0.10
B)Between 0.05 and 0.10
C)Between 0.01 and 0.05
D)Less than 0.01
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12
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.
μ  <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot. μ   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall } \text { stress } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \mathrm{R}-\mathrm{Sq}=22.8 \% & \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array} \end{array}  What is the percent of variation in brainwave relative amplitude that can be explained by the regression model?</strong> A)23% B)48% C)77% D)95%  The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall  stress 0.43370.1677 S=2.13946RSq=22.8%\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall } \text { stress } & 0.4337 & 0.1677 \\\mathrm{~S}=2.13946 & \mathrm{R}-\mathrm{Sq}=22.8 \% &\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 348.78816.263 Residual Error 36164.7834.577 Total 39213.571\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 3 & 48.788 & 16.263 \\\text { Residual Error } & 36 & 164.783 & 4.577 \\\text { Total } & 39 & 213.571 &\end{array}\end{array} What is the percent of variation in brainwave relative amplitude that can be explained by the regression model?

A)23%
B)48%
C)77%
D)95%
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Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.
μ  <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot. μ   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall }{ }^{*} \text { stress } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% & \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array} \end{array}  Here are the residual plots for this regression model:   Based on the study description, the scatterplot, and the residual plots, what should you conclude?</strong> A)The assumption of Normality is not met but the other assumptions are met. B)The assumption of constant variance is not met but the other assumptions are met. C)The assumption of independence is not met but the other assumptions are met. D)All of the assumptions for regression inference are met.  The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall  stress 0.43370.1677 S=2.13946 R-Sq =22.8%\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall }{ }^{*} \text { stress } & 0.4337 & 0.1677 \\\mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% &\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 348.78816.263 Residual Error 36164.7834.577 Total 39213.571\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 3 & 48.788 & 16.263 \\\text { Residual Error } & 36 & 164.783 & 4.577 \\\text { Total } & 39 & 213.571 &\end{array}\end{array} Here are the residual plots for this regression model:
 <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot. μ   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall }{ }^{*} \text { stress } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% & \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array} \end{array}  Here are the residual plots for this regression model:   Based on the study description, the scatterplot, and the residual plots, what should you conclude?</strong> A)The assumption of Normality is not met but the other assumptions are met. B)The assumption of constant variance is not met but the other assumptions are met. C)The assumption of independence is not met but the other assumptions are met. D)All of the assumptions for regression inference are met.  Based on the study description, the scatterplot, and the residual plots, what should you conclude?

A)The assumption of Normality is not met but the other assumptions are met.
B)The assumption of constant variance is not met but the other assumptions are met.
C)The assumption of independence is not met but the other assumptions are met.
D)All of the assumptions for regression inference are met.
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Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.
 <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{l} \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall }{ }^{*} \text { stress } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \mathrm{R}-\mathrm{Sq}=22.8 \% & \end{array}\\\\ \text { Analysis of Variance } \end{array}    \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array}  To determine whether the number of pictures recalled affects the brainwave relative amplitude in this model, we test the hypotheses H<sub>0</sub>: β<sub>1</sub> = 0, H<sub>a</sub>: β<sub>1</sub> ≠ 0 in this particular model. Using the estimates provided, what is the P-value for this test?</strong> A)Greater than 0.10 B)Between 0.05 and 0.10 C)Between 0.01 and 0.05 D)Less than 0.01  The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall  stress 0.43370.1677 S=2.13946RSq=22.8% Analysis of Variance \begin{array}{l}\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall }{ }^{*} \text { stress } & 0.4337 & 0.1677 \\\mathrm{~S}=2.13946 & \mathrm{R}-\mathrm{Sq}=22.8 \% &\end{array}\\\\\text { Analysis of Variance }\end{array}

 Source  DF  SS  MS  Regression 348.78816.263 Residual Error 36164.7834.577 Total 39213.571\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 3 & 48.788 & 16.263 \\\text { Residual Error } & 36 & 164.783 & 4.577 \\\text { Total } & 39 & 213.571 &\end{array} To determine whether the number of pictures recalled affects the brainwave relative amplitude in this model, we test the hypotheses H0: β1 = 0, Ha: β1 ≠ 0 in this particular model. Using the estimates provided, what is the P-value for this test?

A)Greater than 0.10
B)Between 0.05 and 0.10
C)Between 0.01 and 0.05
D)Less than 0.01
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15
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.
 <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall }{ }^{*} \text { stress } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% & \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array} \end{array}  What is a 95% confidence interval for β<sub>2</sub>, the coefficient of the indicator variable stress in this model?</strong> A)-4.26 ± 2.21 B)-4.26 ± 4.47 C)2.21 ± 1.93 D)2.21 ± 4.47  The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall  stress 0.43370.1677 S=2.13946 R-Sq =22.8%\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall }{ }^{*} \text { stress } & 0.4337 & 0.1677 \\\mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% &\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 348.78816.263 Residual Error 36164.7834.577 Total 39213.571\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 3 & 48.788 & 16.263 \\\text { Residual Error } & 36 & 164.783 & 4.577 \\\text { Total } & 39 & 213.571 &\end{array}\end{array} What is a 95% confidence interval for β2, the coefficient of the indicator variable stress in this model?

A)-4.26 ± 2.21
B)-4.26 ± 4.47
C)2.21 ± 1.93
D)2.21 ± 4.47
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16
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.
 <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall }{ }^{*} \text { stress } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% & \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array} \end{array}  The interaction term in this model is statistically significant at the 0.05 level. Given this and the scatterplot provided, what should you conclude?</strong> A)The effect of the number of pictures recalled on brainwave relative amplitude is substantially different when subjects are exposed to a stressful stimulus versus a neutral stimulus. B)The effect of the number of pictures recalled on brainwave relative amplitude is essentially the same when subjects are exposed to a stressful stimulus or a neutral stimulus. C)The number of pictures recalled does not affect brainwave relative amplitude at all. D)The type of stimulus received does not affect brainwave relative amplitude at all.  The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall  stress 0.43370.1677 S=2.13946 R-Sq =22.8%\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall }{ }^{*} \text { stress } & 0.4337 & 0.1677 \\\mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% &\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 348.78816.263 Residual Error 36164.7834.577 Total 39213.571\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 3 & 48.788 & 16.263 \\\text { Residual Error } & 36 & 164.783 & 4.577 \\\text { Total } & 39 & 213.571 &\end{array}\end{array} The interaction term in this model is statistically significant at the 0.05 level. Given this and the scatterplot provided, what should you conclude?

A)The effect of the number of pictures recalled on brainwave relative amplitude is substantially different when subjects are exposed to a stressful stimulus versus a neutral stimulus.
B)The effect of the number of pictures recalled on brainwave relative amplitude is essentially the same when subjects are exposed to a stressful stimulus or a neutral stimulus.
C)The number of pictures recalled does not affect brainwave relative amplitude at all.
D)The type of stimulus received does not affect brainwave relative amplitude at all.
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17
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.
 <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The data are displayed in the following scatterplot.   The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model: ​  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall }^{*} \text { stress } & 0.4337 & 0.1677 \\ \mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% & \end{array}    \begin{array}{l} \text { Analysis of Variance }\\ \begin{array}{lrrr} \text { Source } & \text { DF } & \text { SS } & \text { MS } \\ \text { Regression } & 3 & 48.788 & 16.263 \\ \text { Residual Error } & 36 & 164.783 & 4.577 \\ \text { Total } & 39 & 213.571 & \end{array} \end{array}  Based on the software output provided, what is the equation of the regression model when a stressful stimulus is received (indicator variable stress = 1)?</strong> A)amplitude = 0.533 - 0.114 recall B)amplitude = 0.533 + 0.320 recall C)amplitude = 4.795 - 0.114 recall D)amplitude = 4.795 + 0.320 recall  The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. The following results summarize the least-squares regression fit of this model:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall  stress 0.43370.1677 S=2.13946 R-Sq =22.8%\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall }^{*} \text { stress } & 0.4337 & 0.1677 \\\mathrm{~S}=2.13946 & \text { R-Sq }=22.8 \% &\end{array}

 Analysis of Variance  Source  DF  SS  MS  Regression 348.78816.263 Residual Error 36164.7834.577 Total 39213.571\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } \\\text { Regression } & 3 & 48.788 & 16.263 \\\text { Residual Error } & 36 & 164.783 & 4.577 \\\text { Total } & 39 & 213.571 &\end{array}\end{array} Based on the software output provided, what is the equation of the regression model when a stressful stimulus is received (indicator variable stress = 1)?

A)amplitude = 0.533 - 0.114 recall
B)amplitude = 0.533 + 0.320 recall
C)amplitude = 4.795 - 0.114 recall
D)amplitude = 4.795 + 0.320 recall
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18
Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. Using software, we obtained the following output:
 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall* stress 0.43370.1677\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 4.795 & 1.329 \\\text { recall } & -0.1139 & 0.1069 \\\text { stress } & -4.262 & 2.205 \\\text { recall* stress } & 0.4337 & 0.1677\end{array}

 <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. Using software, we obtained the following output:  \begin{array}{lrr} \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & -0.1139 & 0.1069 \\ \text { stress } & -4.262 & 2.205 \\ \text { recall* stress } & 0.4337 & 0.1677 \end{array}     Which of the following statements best describes the interval (2.465, 4.996)?</strong> A)A confidence interval for the mean brainwave relative amplitude of all adult men in a similar experiment exposed to a neutral stimulus and who remember 10 pictures B)A confidence interval for the mean brainwave relative amplitude of all adult men in a similar experiment exposed to a stressful stimulus and who remember 10 pictures C)A prediction interval for the brainwave relative amplitude of one adult man in a similar experiment exposed to a neutral stimulus and who remembers 10 pictures D)A prediction interval for the brainwave relative amplitude of one adult man in a similar experiment exposed to a stressful stimulus and who remembers 10 pictures
Which of the following statements best describes the interval (2.465, 4.996)?

A)A confidence interval for the mean brainwave relative amplitude of all adult men in a similar experiment exposed to a neutral stimulus and who remember 10 pictures
B)A confidence interval for the mean brainwave relative amplitude of all adult men in a similar experiment exposed to a stressful stimulus and who remember 10 pictures
C)A prediction interval for the brainwave relative amplitude of one adult man in a similar experiment exposed to a neutral stimulus and who remembers 10 pictures
D)A prediction interval for the brainwave relative amplitude of one adult man in a similar experiment exposed to a stressful stimulus and who remembers 10 pictures
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Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The following model is proposed for predicting the brainwave relative amplitude ("amplitude") from the number of images recalled ("recall"), the indicator variable reflecting the nature of the stimulus ("stress"), and an interaction term ("recall*stress"):
Amplitudei = β0 + β1 (recalli) + β2 (stressi) + β3 (recall*stressi) + εi
Where the deviations εi are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. Using software, we obtained the following output:

 Predictor  Coef  SE Coef  Constant 4.7951.329 recall 0.11390.1069 stress 4.2622.205 recall*  stress 0.43370.1677\begin{array} { l r r } \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & - 0.1139 & 0.1069 \\ \text { stress } & - 4.262 & 2.205 \\ \text { recall* } \text { stress } & 0.4337 & 0.1677 \end{array}

 <strong>Researchers investigated the effects of acute stress on emotional picture processing and recall by randomly assigning 40 adult males to receive either a stressful cold stimulus or a neutral warm stimulus before viewing pictures. The researchers computed the relative brainwave amplitude (in microvolts, μV) for each subject viewing unpleasant versus neutral pictures. They also recorded how many pictures the subjects were able to recall 24 hours later. The following model is proposed for predicting the brainwave relative amplitude (amplitude) from the number of images recalled (recall), the indicator variable reflecting the nature of the stimulus (stress), and an interaction term (recall*stress): Amplitude<sub>i</sub> = β<sub>0</sub> + β<sub>1</sub> (recall<sub>i</sub>) + β<sub>2</sub> (stress<sub>i</sub>) + β<sub>3</sub> (recall*stress<sub>i</sub>) + ε<sub>i</sub> Where the deviations ε<sub>i</sub> are assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to the sample of 40 adult males. Using software, we obtained the following output:   \begin{array} { l r r } \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 4.795 & 1.329 \\ \text { recall } & - 0.1139 & 0.1069 \\ \text { stress } & - 4.262 & 2.205 \\ \text { recall* } \text { stress } & 0.4337 & 0.1677 \end{array}     Which of the following statements best describes the interval (-0.804, 8.115)?</strong> A)A confidence interval for the mean brainwave relative amplitude of all adult men in a similar experiment exposed to a neutral stimulus and who remember 10 pictures B)A confidence interval for the mean brainwave relative amplitude of all adult men in a similar experiment exposed to a stressful stimulus and who remember 10 pictures C)A prediction interval for the brainwave relative amplitude of one adult man in a similar experiment exposed to a neutral stimulus and who remembers 10 pictures D)A prediction interval for the brainwave relative amplitude of one adult man in a similar experiment exposed to a stressful stimulus and who remembers 10 pictures
Which of the following statements best describes the interval (-0.804, 8.115)?

A)A confidence interval for the mean brainwave relative amplitude of all adult men in a similar experiment exposed to a neutral stimulus and who remember 10 pictures
B)A confidence interval for the mean brainwave relative amplitude of all adult men in a similar experiment exposed to a stressful stimulus and who remember 10 pictures
C)A prediction interval for the brainwave relative amplitude of one adult man in a similar experiment exposed to a neutral stimulus and who remembers 10 pictures
D)A prediction interval for the brainwave relative amplitude of one adult man in a similar experiment exposed to a stressful stimulus and who remembers 10 pictures
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20
A study examined the effect of body weight and origin (aviary or wild) on the length of the R1 central tail feather in male finches. The following model is proposed for predicting the length of the R1 feather from the bird's weight (in grams) and origin (aviary = 0, wild = 1):
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to a sample of 25 male finches. Using software, we obtained the following output:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996\begin{array}{lrr}\text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & 35.79 & 16.95 \\\text { BodyWeight } & 2.826 & 1.030 \\\text { Aviary0Wild1 } & -10.453 & 4.996\end{array}

 Predicted Values for New Observations  NewObs  Fit  SE Fit 95%CI95% PI 176.224.32(67.26,85.17)(54.35,98.09)286.672.75(80.96,92.38)(65.91,107.43)\begin{array}{l}\text { Predicted Values for New Observations }\\\begin{array}{rrrcc}\text { NewObs } & \text { Fit } & \text { SE Fit } & 95 \% \mathrm{CI} & 95 \% \text { PI } \\1 & 76.22 & 4.32 & (67.26,85.17) & (54.35,98.09) \\2 & 86.67 & 2.75 & (80.96,92.38) & (65.91,107.43)\end{array}\end{array}

 Values of Predictors for New Observations  NewObs  BodyWeight  Aviary0Wild 118.01.0218.00.0\begin{array}{l}\text { Values of Predictors for New Observations }\\\begin{array}{crr}\text { NewObs } & \text { BodyWeight } & \text { Aviary0Wild } \\1 & 18.0 & 1.0 \\2 & 18.0 & 0.0\end{array}\end{array} What is the best description of the interval (67.26, 85.17)?

A)A confidence interval for the mean tail-feather length of all male finches that weigh 18 g and are raised in an aviary
B)A confidence interval for the mean tail-feather length of all male finches that weigh 18 g and are caught in the wild
C)A prediction interval for the tail-feather length of one male finch that weighs 18 g and was raised in an aviary
D)A prediction interval for the tail-feather length of one male finch that weighs 18 g and was caught in the wild
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A study examined the effect of body weight and origin (aviary or wild) on the length of the R1 central tail feather in male finches. The following model is proposed for predicting the length of the R1 feather from the bird's weight (in grams) and origin (aviary = 0, wild = 1):
R1Lengthi = β0 + β1 (BodyWeighti) + β2 (Aviary0Wild1i) + εi
Where the deviations εi were assumed to be independent and Normally distributed with mean 0 and standard deviation σ. This model was fit to a sample of 25 male finches. Using software, we obtained the following output:
 Predictor  Coef  SE Coef  Constant 35.7916.95 BodyWeight 2.8261.030 Aviary0Wild1 10.4534.996\begin{array} { l r r } \text { Predictor } & \text { Coef } & \text { SE Coef } \\ \text { Constant } & 35.79 & 16.95 \\ \text { BodyWeight } & 2.826 & 1.030 \\ \text { Aviary0Wild1 } & - 10.453 & 4.996 \end{array}

Predicted Values for New Observations

 NewObs  Fit  SE Fit 95% CI 95% PI 176.224.32(67.26,85.17)(54.35,98.09)286.672.75(80.96,92.38)(65.91,107.43)\begin{array} { r r r c c } \text { NewObs } & \text { Fit } & \text { SE Fit } & 95 \% \text { CI } & 95 \% \text { PI } \\ 1 & 76.22 & 4.32 & ( 67.26,85.17 ) & ( 54.35,98.09 ) \\ 2 & 86.67 & 2.75 & ( 80.96,92.38 ) & ( 65.91,107.43 ) \end{array}

Values of Predictors for New Observations

 NewObs  BodyWeight  Aviary0Wild1 118.01.0218.00.0\begin{array} { c r r } \text { NewObs } & \text { BodyWeight } & \text { Aviary0Wild1 } \\ 1 & 18.0 & 1.0 \\ 2 & 18.0 & 0.0 \end{array} What is the best description of the interval (65.91, 107.43)?

A)A confidence interval for the mean tail-feather length of all male finches that weigh 18 g and are raised in an aviary
B)A confidence interval for the mean tail-feather length of all male finches that weigh 18 g and are caught in the wild
C)A prediction interval for the tail-feather length of one male finch that weighs 18 g and was raised in an aviary
D)A prediction interval for the tail-feather length of one male finch that weighs 18 g and was caught in the wild
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22
A study looked for a protein marker that might help predict active flare-ups in a particular chronic disease. A sample of 32 individuals with the disease had some blood taken to determine their disease status (active flare-up or inactive) and the relative abundance of the protein marker in their blood. We use logistic regression to model disease status as a function of relative abundance. Software gives the following information:
For Log Odds of Active/Inactive
 Term  Estimate  Std Error  Intercept 6.1862.416 Relative abundance 6.2292.651\begin{array} { c c c } \text { Term } & \text { Estimate } & \text { Std Error } \\\text { Intercept } & - 6.186 & 2.416 \\\text { Relative abundance } & 6.229 & 2.651\end{array} What is the probability p that a patient who has a protein marker relative abundance of 0.8 has an active flare-up?

A)0.231
B)0.300
C)0.511
D)0.770
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A study looked for a protein marker that might help predict active flare-ups in a particular chronic disease. A sample of 32 individuals with the disease had some blood taken to determine their disease status (active flare-up or inactive) and the relative abundance of the protein marker in their blood. We use logistic regression to model disease status as a function of relative abundance. Software gives the following information:
For Log Odds of Active/Inactive
 Term  Estimate  Std Error  Intercept 6.1862.416 Relative abundance 6.2292.651\begin{array} { c c c } \text { Term } & \text { Estimate } & \text { Std Error } \\\text { Intercept } & - 6.186 & 2.416 \\\text { Relative abundance } & 6.229 & 2.651\end{array} We want to know if the slope of the logistic regression model is significant. What is the value of the test statistic?

A)t = 2.35
B)t = 2.56
C)z = 2.35
D)z = 2.56
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24
A study looked for a protein marker that might help predict active flare-ups in a particular chronic disease. A sample of 32 individuals with the diseases had some blood taken to determine their disease status (active flare-up or inactive) and the relative abundance of the protein marker in their blood. We use logistic regression to model disease status as a function of relative abundance. Software gives the following information:
For Log Odds of Active/Inactive
 Term  Estimate  Std Error  Intercept 6.1862.416 tive abundance 6.2292.651\begin{array} { c c c } \text { Term } & \text { Estimate } & \text { Std Error } \\\text { Intercept } & - 6.186 & 2.416 \\\text { tive abundance } & 6.229 & 2.651\end{array} We want to know if the slope of the logistic regression model is significant. What is the P-value for the two-sided test?

A)Greater than 0.10
B)Between 0.05 and 0.10
C)Between 0.01 and 0.05
D)Less than 0.01
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25
A study looked for a protein marker that might help predict active flare-ups in a particular chronic disease. A sample of 32 individuals with the disease had some blood taken to determine their disease status (active flare-up or inactive) and the relative abundance of the protein marker in their blood. We use logistic regression to model disease status as a function of relative abundance. Software gives the following information:
For Log Odds of Active/Inactive
 Term  Estimate  Std Error  Intercept 6.1862.416 ive abundance 6.2292.651\begin{array} { c c c } \text { Term } & \text { Estimate } & \text { Std Error } \\\text { Intercept } & - 6.186 & 2.416 \\\text { ive abundance } & 6.229 & 2.651\end{array} What is a 95% confidence interval for the odds ratio eβ1?

A)(0.002, 507.2)
B)(0.096, 10.4)
C)(2.8, 91579.3)
D)(35.8, 7186.8)
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26
A case study enrolled case individuals with glioma (a type of brain cancer) and similar control individuals without glioma. Participants were asked whether they had been regular cell phone users for at least a year. We use logistic regression to model glioma status (1 = glioma, 0 = no glioma) as a function of regular cell phone use (1 = yes, 0 = no). Software gives the following information:
For log odds of glioma/no glioma
 Predictor  Coef  SE Coef  Constant 0.080740.04987 Regular use 0.025650.07275\begin{array} { l c c } \text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & - 0.08074 & 0.04987 \\\text { Regular use } & - 0.02565 & 0.07275\end{array} We want to know if the slope of the logistic regression model is significant. What is the value of the test statistic?

A)t = -1.62
B)t = -0.35
C)z = -1.62
D)z = -0.35
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27
A case study enrolled case individuals with glioma (a type of brain cancer) and similar control individuals without glioma. Participants were asked whether they had been regular cell phone users for at least a year. We use logistic regression to model glioma status (1 = glioma, 0 = no glioma) as a function of regular cell phone use (1 = yes, 0 = no). Software gives the following information:
For log odds of glioma/no glioma
 Predictor  Coef  SE Coef  Constant 0.080740.04987 Regular use 0.025650.07275\begin{array} { l c c } \text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & - 0.08074 & 0.04987 \\\text { Regular use } & - 0.02565 & 0.07275\end{array} We want to know if the slope of the logistic regression model is significant. What is the P-value for the two-sided test?

A)Greater than 0.10
B)Between 0.05 and 0.10
C)Between 0.01 and 0.05
D)Less than 0.01
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Unlock for access to all 28 flashcards in this deck.
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28
A case study enrolled case individuals with glioma (a type of brain cancer) and similar control individuals without glioma. Participants were asked whether they had been regular cell phone users for at least a year. We use logistic regression to model glioma status (1 = glioma, 0 = no glioma) as a function of regular cell phone use (1 = yes, 0 = no). Software gives the following information:
For log odds of glioma/no glioma
 Predictor  Coef  SE Coef  Constant 0.080740.04987 Regular use 0.025650.07275\begin{array} { l c c } \text { Predictor } & \text { Coef } & \text { SE Coef } \\\text { Constant } & - 0.08074 & 0.04987 \\\text { Regular use } & - 0.02565 & 0.07275\end{array} What is a 95% confidence interval for the odds ratio eβ1?

A)(-0.17, 0.12)
B)(0.12, 0.35)
C)(0.85, 1.12)
D)(0.97, 1.26)
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Unlock for access to all 28 flashcards in this deck.