Question 1

One method of diagnosing heteroscedasticity is to plot the residuals against the predicted values of y, then look for a change in the spread of the plotted values.

Accepted Answer

A) True 
 B)False

Question 2

The vertical spread of the data points about the regression line is measured by the y-intercept.

Accepted Answer

A) True 
 B)False

Question 3

If all the points in a scatter diagram lie on the least squares regression line, then the coefficient of correlation must be 1.0.

Accepted Answer

A) True 
 B)False

Question 4

The confidence interval estimate of the expected value of y will be narrower than the prediction interval for the same given value of x and confidence level. This is because there is less error in estimating a mean value as opposed to predicting an individual value.

Accepted Answer

A) True 
 B)False

Question 5

The confidence interval estimate of the expected value of y for a given value x, compared to the prediction interval of y for the same given value of x and confidence level, will be:

Accepted Answer

A) wider. 
B) the same. 
C) narrower. 
D) impossible to know.

Question 6

When the actual values y of a dependent variable and the corresponding predicted values $\hat { y }$ are the same, the standard error of estimate s$\varepsilon$ will be 0.0.

Accepted Answer

A) True 
 B)False

Question 7

In simple linear regression, the coefficient of correlation r and the least squares estimate b1 of the population slope $\beta$1:

Accepted Answer

A) must be equal. 
B) must have the same sign. 
C) are not related. 
D) None of these choices.

Question 8

Given the least squares regression line $\hat { y } = 2.48 - 1.63 x$ , and a coefficient of determination of 0.81, the coefficient of correlation is:

Accepted Answer

A) -0.66 
B) 0.81 
C) -0.90 
D) 0.90

Question 9

NARRBEGIN: Marc Anthony Concert
Marc Anthony Concert
At a recent Marc Anthony concert, a survey was conducted that asked a random sample of 20 people their age and how many concerts they have attended since the first of the year. The following data were collected:
 $$\begin{array}{l}
\begin{array} { | l | c c c c c c c c c c | } 
\hline \text { Age } & 62 & 57 & 40 & 49 & 67 & 54 & 43 & 65 & 54 & 41 \
\hline \text { Number af Concerts } & 6 & 5 & 4 & 3 & 5 & 5 & 2 & 6 & 3 & 1 \
\hline
\end{array}\

\begin{array} { | l | c c c c c c c c c c | } 
\hline \text { Age } & 44 & 48 & 55 & 60 & 59 & 63 & 69 & 40 & 38 & 52 \
\hline \text { Number of Concerts } & 3 & 2 & 4 & 5 & 4 & 5 & 4 & 2 & 1 & 3 \
\hline
\end{array}
\end{array}$$ An Excel output follows: $\begin{array} { l  } 
 \text {SUMMARY OUTPUT }& \text {DESCRIPTIVE STATISTICS }\
\begin{array}{lc}
\hline{\text { Reqression Statistics }} \
\hline \text { Multiple R } & 0.80203 \
\text { R Square } & 0.64326 \
\text { Adjusted R Square } & 0.62344 \
\text { Standard Error } & 0.93965 \
\text { Observations } & 20 \
\hline
\end{array}&\begin{array}{lclc}
\hline {\text { Age }} && {\text { Concerts }} \
\hline \text { Mean } & 53 & \text { Mean } & 3.65 \
\text { Standard Error } & 2.1849 & \text { Standard Error } & 0.3424 \
\text { Standard Deviation } & 9.7711 & \text { Standard Deviation } & 1.5313 \
\text { Sample Variance } & 95.4737 & \text { Samplevariance } & 2.3447 \
\text { Count } & 20 & \text { Court } & 20 \
\hline
\end{array}
\end{array}$
$\text { SPEARMAN RANK CORRELATION COEFFICIENT }=0.8306$
$\begin{array}{lccccc} 
& \text { df } & \text { SS } & \text { MS } & \text { F } & \text { Sign\&icance F } \
\hline \text { Regression } & 1 & 28.65711 & 28.65711 & 32.45653 & 2.1082 \mathrm{E}-05 \
\text { Residual } & 18 & 15.89289 & 0.88294 & & \
\text { Total } & 19 & 44.55 & & & \
\hline
\end{array}$

$\begin{array}{lcccccc} 
& \text { Coefficients} & \text { Standard Error } & \text { t Stat } & \text { Rvale } & \text { Lower 95\% } & \text { Upoer 95\% } \
\hline \text { Intercept } & -3.01152 & 1.18802 & -2.53491 & 0.02074 & -5.50746 & -0.5156 \
\text { Age } & 0.12569 & 0.02206 & 5.69706 & 0.00002 & 0.07934 & 0.1720 \
\hline
\end{array}$ NARREND
-{Marc Anthony Concert Narrative} Does it appear that the errors are normally distributed? Explain.

Accepted Answer

The answer of NARRBEGIN: Marc Anthony Concert
Marc Anthony Concert
At a...

Question 10

In the least squares regression line $\tilde { y } = 3 - 2 x$ , the predicted value of y equals:

Accepted Answer

A) 1.0 when x = -1.0 
B) 2.0 when x = 1.0 
C) 2.0 when x =-1.0 
D) 1.0 when x = 1.0

Question 11

A regression analysis between sales (in $1,000) and advertising (in $1,000) resulted in the following least squares line: $\hat { y } = 80 + 5 x$ . This implies that:

Accepted Answer

A) as advertising increases by $1,000, sales increases by $5,000. 
B) as advertising increases by $1,000, sales increases by $80,000. 
C) as advertising increases by $5, sales increases by $80. 
D) None of these choices.

Question 12

The plot of residuals vs. predicted values should show no patterns if the conditions of a regression analysis are met.

Accepted Answer

A) True 
 B)False

Question 13

NARRBEGIN: Accidents & Rain
Accidents and Rain
A statistician investigating the relationship between the amount of rain (in inches) and the number of automobile accidents gathered data on accidents in her city for 10 randomly selected days throughout the year. The results are shown below. $$\begin{array} { | c | c | c | } 
\hline \text { Day } & \text { Rain } & \text { Number of Accidents } \
\hline 1 & 0.05 & 5 \
2 & 0.12 & 6 \
3 & 0.05 & 2 \
4 & 0.08 & 4 \
5 & 0.10 & \mathbf { 8 } \
6 & 0.35 & 14 \
7 & 0.15 & 7 \
8 & 0.30 & 13 \
9 & 0.10 & 7 \
10 & 0.20 & 10 \
\hline
\end{array}$$ NARREND
-{Accidents and Rain Narrative} What other variables might be associated with accidents, besides or along with rain?

Accepted Answer

The answer of NARRBEGIN: Accidents & Rain
Accidents and Rain
A statistician...

Question 14

The probability distribution of the error variable $\varepsilon$ is normal, with mean E($\varepsilon$) = 0, and standard deviation $\sigma$$\varepsilon$=1.

Accepted Answer

A) True 
 B)False

Question 15

In simple linear regression, most often we perform a two-tail test of the population slope $\beta$1 to determine whether there is sufficient evidence to infer that a linear relationship exists. The null hypothesis is stated as:

Accepted Answer

A) H0: $\beta$1 = 0 
B) H0: $\beta$1 = b1 
C) H0: $\beta$1 $\neq$ 0 
D) None of these choices.

Question 16

An outlier is an observation that is unusually small or unusually large.

Accepted Answer

A) True 
 B)False

Question 17

NARRBEGIN: Marc Anthony Concert
Marc Anthony Concert
At a recent Marc Anthony concert, a survey was conducted that asked a random sample of 20 people their age and how many concerts they have attended since the first of the year. The following data were collected:
 $$\begin{array}{l}
\begin{array} { | l | c c c c c c c c c c | } 
\hline \text { Age } & 62 & 57 & 40 & 49 & 67 & 54 & 43 & 65 & 54 & 41 \
\hline \text { Number af Concerts } & 6 & 5 & 4 & 3 & 5 & 5 & 2 & 6 & 3 & 1 \
\hline
\end{array}\

\begin{array} { | l | c c c c c c c c c c | } 
\hline \text { Age } & 44 & 48 & 55 & 60 & 59 & 63 & 69 & 40 & 38 & 52 \
\hline \text { Number of Concerts } & 3 & 2 & 4 & 5 & 4 & 5 & 4 & 2 & 1 & 3 \
\hline
\end{array}
\end{array}$$ An Excel output follows: $\begin{array} { l  } 
 \text {SUMMARY OUTPUT }& \text {DESCRIPTIVE STATISTICS }\
\begin{array}{lc}
\hline{\text { Reqression Statistics }} \
\hline \text { Multiple R } & 0.80203 \
\text { R Square } & 0.64326 \
\text { Adjusted R Square } & 0.62344 \
\text { Standard Error } & 0.93965 \
\text { Observations } & 20 \
\hline
\end{array}&\begin{array}{lclc}
\hline {\text { Age }} && {\text { Concerts }} \
\hline \text { Mean } & 53 & \text { Mean } & 3.65 \
\text { Standard Error } & 2.1849 & \text { Standard Error } & 0.3424 \
\text { Standard Deviation } & 9.7711 & \text { Standard Deviation } & 1.5313 \
\text { Sample Variance } & 95.4737 & \text { Samplevariance } & 2.3447 \
\text { Count } & 20 & \text { Court } & 20 \
\hline
\end{array}
\end{array}$
$\text { SPEARMAN RANK CORRELATION COEFFICIENT }=0.8306$
$\begin{array}{lccccc} 
& \text { df } & \text { SS } & \text { MS } & \text { F } & \text { Sign\&icance F } \
\hline \text { Regression } & 1 & 28.65711 & 28.65711 & 32.45653 & 2.1082 \mathrm{E}-05 \
\text { Residual } & 18 & 15.89289 & 0.88294 & & \
\text { Total } & 19 & 44.55 & & & \
\hline
\end{array}$

$\begin{array}{lcccccc} 
& \text { Coefficients} & \text { Standard Error } & \text { t Stat } & \text { Rvale } & \text { Lower 95\% } & \text { Upoer 95\% } \
\hline \text { Intercept } & -3.01152 & 1.18802 & -2.53491 & 0.02074 & -5.50746 & -0.5156 \
\text { Age } & 0.12569 & 0.02206 & 5.69706 & 0.00002 & 0.07934 & 0.1720 \
\hline
\end{array}$ NARREND
-{Marc Anthony Concert Narrative} Draw a histogram of the residuals.

Accepted Answer

The answer of NARRBEGIN: Marc Anthony Concert
Marc Anthony Concert
At a...

Question 18

NARRBEGIN: Oil Quality and Price
Oil Quality and Price
Quality of oil is measured in API gravity degrees--the higher the degrees API, the higher the quality. The table shown below is produced by an expert in the field who believes that there is a relationship between quality and price per barrel. $$\begin{array} { | c | c | } 
\hline \text { Oil degrees API } & \text { Price per barrel (in \$) } \
\hline 27.0 & 12.02 \
28.5 & 12.04 \
30.8 & 12.32 \
31.3 & 12.27 \
31.9 & 12.49 \
34.5 & 12.70 \
34.0 & 12.80 \
34.7 & 13.00 \
37.0 & 13.00 \
41.0 & 13.17 \
41.0 & 13.19 \
38.8 & 13.22 \
39.3 & 13.27 \
\hline
\end{array}$$ A partial Minitab output follows:
 $$\begin{array}{l}
\text { Dascriptive atafistics }\
\begin{array} { l l l l l } 
\text { Variable } & \mathrm { N } & \text { Mear } & \text { StDev } & \text { SE Mear } \
\text { Degrees } & 13 & 34.60 & 4.613 & 1.280 \
\text { Price } & 13 & 1270 & 0.757 & 0.127
\end{array}
\end{array}$$ 
$\begin{array}{lll} 
\text { Bavarifinces }\
& \text { Degrees } & \text { Price } \
\text { Degrees } & 21.281667 & \
\text { Price } & 2.026750 & 0208837
\end{array}$ $$\begin{array}{l}\text { Rederatian Antalyis }\
\begin{array} { l l l r } 
\text { predictor } & \text { Coef } & \text { StDev } & \text { T } & \text { P} \
\text { Constant } & 9.4349 & 0.2867 & 32.91&0.000 \
\text { Degrees } & 0.095235 &0.008220 & 11.59&0.000\
\end{array}\
S = 0 .1314 \quad R - S q = 92.46 \% \quad R - S q ( a d ) = 91.7 \%
\end{array}$$ $$\begin{array}{l}
\text { Analysis of Variance }\
\begin{array} { l l r r r } 
\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } &\text {P} \
\text { Regeression } & 1 & 2.3162 & 2.3162 & 134.24&0.000 \
\text { Residual Error } & 11 & 0.1898 & 0.0173 & \
\text { Total } & 12 & 2.5060 & &
\end{array}
\end{array}$$ NARREND
-{Oil Quality and Price Narrative} Does it appear that heteroscedasticity is a problem? Explain.

Accepted Answer

The answer of NARRBEGIN: Oil Quality and Price
Oil Quality and...

Question 19

NARRBEGIN: Grateful Dead Concert
Grateful Dead Concert
At a recent Grateful Dead concert, a survey was conducted that asked a random sample of 20 people their age and how many concerts they have attended since the first of the year. It is suspected that older concert goers tend to go to more of his concerts in one year than younger concert goers. The data and analysis are shown below. $$\begin{array}{l}
\begin{array} { | l | c c c c c c c c c c | } 
\hline \text { Age } & 62 & 57 & 40 & 49 & 67 & 54 & 43 & 65 & 54 & 41 \
\hline \text { Number af Concerts } & 6 & 5 & 4 & 3 & 5 & 5 & 2 & 6 & 3 & 1 \
\hline
\end{array}\\
\begin{array} { | l | c c c c c c c c c c | } 
\hline \text { Age } & 44 & 48 & 55 & 60 & 59 & 63 & 69 & 40 & 38 & 52 \
\hline \text { Number of Concerts } & 3 & 2 & 4 & 5 & 4 & 5 & 4 & 2 & 1 & 3 \
\hline
\end{array}
\end{array}$$ An Excel output follows: SUMMARY OUTPUT
$\begin{array}{lc}
\hline {\text { Regression Statistics }} \
\hline \text { Multiple R } & 0.80203 \
\text { R Square } & 0.64326 \
\text { Adjusted R Square } & 0.62344 \
\text { Standard Error } & 0.93965 \
\text { Observations } & 20 \
\hline
\end{array}$

$\text { DESCRIPTIVE STATISTICS }$
$\begin{array}{lclc}
\hline{\text { AGe }}& & {\text { Concerts }} \
\hline \text { Mean } & 53 & \text { Mean } & 3.65 \
\text { Standard Error } & 2.1849 & \text { Standard Error } & 0.3424 \
\text { Standard Deviation } & 9.7711 & \text { Standard Deviation } & 1.5313 \
\text { Sample Variance } & 95.4737 & \text { Sample Variance } & 2.3447 \
\text { Count } & 20 & \text { Court } & 20 \
\hline
\end{array}$

$\text { SPEARMAN RANK CORRELATION COEFFICIENT }=0.8306$
$\begin{array}{l}
\text { ANOVA }\
\begin{array}{lccccc}
\hline & df & S S & M S & F & \text { Signficance F } \
\hline \text { Regression } & 1 & 28.65711 & 28.65711 & 32.45653 & 2.1082 E-05 \
\text { Residual } & 18 & 15.89289 & 0.88294 & & \
\text { Total } & 19 & 44.55 & & & \
\hline
\end{array}
\end{array}$
$\begin{array}{lcccccc}
\hline & \text { Coefficients Standard Error } & \text { tStat } & \text { p-value } & \text { Lower 95\% } & \text { Upper 95\% } \
\hline \text { Intercept } & -3.01152 & 1.18802 & -2.53491 & 0.02074 & -5.50746 & -0.5156 \
\text { Age } & 0.12569 & 0.02206 & 5.69706 & 0.00002 & 0.07934 & 0.1720 \
\hline
\end{array}$ NARREND
-{Grateful Dead Concert Narrative} Determine the coefficient of determination and discuss what its value tells you about the two variables.

Accepted Answer

The answer of NARRBEGIN: Grateful Dead Concert
Grateful Dead Concert
At a...

Question 20

NARRBEGIN: Truck Speed & Gas Mileage
Truck Speed and Gas Mileage
An economist wanted to analyze the relationship between the speed of a truck (x) and its gas mileage (y). As an experiment a truck is operated at several different speeds and for each speed the gas mileage is measured. These data are shown below. $$\begin{array} { | l | l l l l l l l | } 
\hline \text { Speed } & 25 & 35 & 45 & 50 & 60 & 65 & 70 \
\hline \text { Gas Mileage } & 40 & 39 & 37 & 33 & 30 & 27 & 25 \
\hline
\end{array}$$ NARREND
-{Truck Speed and Gas Mileage Narrative} Does this data provide sufficient evidence at the 5% significance level to infer that a linear relationship exists between speed and gas mileage?

Accepted Answer

The answer of NARRBEGIN: Truck Speed & Gas Mileage
Truck Speed...

Exam 16: Simple Linear Regression and Correlation

One method of diagnosing heteroscedasticity is to plot the residuals against the predicted values of y, then look for a change in the spread of the plotted values.

The vertical spread of the data points about the regression line is measured by the y-intercept.

If all the points in a scatter diagram lie on the least squares regression line, then the coefficient of correlation must be 1.0.

The confidence interval estimate of the expected value of y will be narrower than the prediction interval for the same given value of x and confidence level. This is because there is less error in estimating a mean value as opposed to predicting an individual value.

The confidence interval estimate of the expected value of y for a given value x, compared to the prediction interval of y for the same given value of x and confidence level, will be:

When the actual values y of a dependent variable and the corresponding predicted values $\hat { y }$ are the same, the standard error of estimate s_{$\varepsilon$} will be 0.0.

In simple linear regression, the coefficient of correlation r and the least squares estimate b₁ of the population slope $\beta$ ₁:

Given the least squares regression line $\hat { y } = 2.48 - 1.63 x$ , and a coefficient of determination of 0.81, the coefficient of correlation is:

In the least squares regression line $\tilde { y } = 3 - 2 x$ , the predicted value of y equals:

A regression analysis between sales (in $1,000) and advertising (in $1,000) resulted in the following least squares line: $\hat { y } = 80 + 5 x$ . This implies that:

The plot of residuals vs. predicted values should show no patterns if the conditions of a regression analysis are met.

The probability distribution of the error variable $\varepsilon$ is normal, with mean E( $\varepsilon$ ) = 0, and standard deviation $\sigma$ _{$\varepsilon$}=1.

In simple linear regression, most often we perform a two-tail test of the population slope $\beta$ ₁ to determine whether there is sufficient evidence to infer that a linear relationship exists. The null hypothesis is stated as:

An outlier is an observation that is unusually small or unusually large.

Exam 16: Simple Linear Regression and Correlation

One method of diagnosing heteroscedasticity is to plot the residuals against the predicted values of y, then look for a change in the spread of the plotted values.

The vertical spread of the data points about the regression line is measured by the y-intercept.

If all the points in a scatter diagram lie on the least squares regression line, then the coefficient of correlation must be 1.0.

The confidence interval estimate of the expected value of y will be narrower than the prediction interval for the same given value of x and confidence level. This is because there is less error in estimating a mean value as opposed to predicting an individual value.

The confidence interval estimate of the expected value of y for a given value x, compared to the prediction interval of y for the same given value of x and confidence level, will be:

When the actual values y of a dependent variable and the corresponding predicted values y^\hat { y }y^​ are the same, the standard error of estimate s ε\varepsilonε will be 0.0.

In simple linear regression, the coefficient of correlation r and the least squares estimate b1 of the population slope β\betaβ 1:

Given the least squares regression line y^=2.48−1.63x\hat { y } = 2.48 - 1.63 xy^​=2.48−1.63x , and a coefficient of determination of 0.81, the coefficient of correlation is:

In the least squares regression line y~=3−2x\tilde { y } = 3 - 2 xy~​=3−2x , the predicted value of y equals:

A regression analysis between sales (in $1,000) and advertising (in $1,000) resulted in the following least squares line: y^=80+5x\hat { y } = 80 + 5 xy^​=80+5x . This implies that:

The plot of residuals vs. predicted values should show no patterns if the conditions of a regression analysis are met.

The probability distribution of the error variable ε\varepsilonε is normal, with mean E( ε\varepsilonε ) = 0, and standard deviation σ\sigmaσ ε\varepsilonε =1.

In simple linear regression, most often we perform a two-tail test of the population slope β\betaβ 1 to determine whether there is sufficient evidence to infer that a linear relationship exists. The null hypothesis is stated as:

An outlier is an observation that is unusually small or unusually large.

When the actual values y of a dependent variable and the corresponding predicted values $\hat { y }$ are the same, the standard error of estimate s_{$\varepsilon$} will be 0.0.

In simple linear regression, the coefficient of correlation r and the least squares estimate b₁ of the population slope $\beta$ ₁:

Given the least squares regression line $\hat { y } = 2.48 - 1.63 x$ , and a coefficient of determination of 0.81, the coefficient of correlation is:

In the least squares regression line $\tilde { y } = 3 - 2 x$ , the predicted value of y equals:

A regression analysis between sales (in $1,000) and advertising (in $1,000) resulted in the following least squares line: $\hat { y } = 80 + 5 x$ . This implies that:

The probability distribution of the error variable $\varepsilon$ is normal, with mean E( $\varepsilon$ ) = 0, and standard deviation $\sigma$ _{$\varepsilon$}=1.

In simple linear regression, most often we perform a two-tail test of the population slope $\beta$ ₁ to determine whether there is sufficient evidence to infer that a linear relationship exists. The null hypothesis is stated as: