Deck 17: Simple Linear Regression

Question 1

You are given the following pairs of scores on X (Pretest score) and Y (Posttest score).

$\begin{array}{cc}\hline X & Y \\\hline 65 & 74 \\82 & 87 \\70 & 82 \\46 & 53 \\55 & 69 \\75 & 81 \\\hline\end{array}$

a. Find the linear regression model for predicting Y from X.
b. Use the prediction model obtained to predict the value of Y for a new person who scored 80 on the pretest.

Answer 1

a. Intercept a = 15.919, slope b = .892.

The regression model is Y_i = .892X_i + 15.919 + e_i
The prediction equation is Y_i = .892X_i + 15.919

b. When X = 80, Y' = .892X + 15.919 = .892(80) + 15.919 = 87.279Y'
Procedure:
Create a data set with two variables: Pretest (X), Posttest (Y). The data set should have 6 cases.

1) Go to Analyze $\rightarrow$ Regression $\rightarrow$ Linear.
2) Select Posttest to the Dependent list. Select Pretest to the Independent(s) list.
3) Click Statistics. Select Confidence interval under Regression Coefficients. Click Continue.
"4) Click OK.
Selected SPSS Output:
$\begin{array}{l}\text { Model Summary }\\\begin{array}{ccccc}\hline \text { Model } & R & R \text { Square } & \text { Adjusted R Square } & \text { Std. Error of the Estimate } \\\hline 1 & .964 & .930 & .912 & 3.627 \\\hline\end{array}\end{array}$

Question 2

You are given the following pairs of scores on X (Percentage of students whose families are below poverty line) and Y (Percentage of students at or above proficiency level) for nine schools.

$\begin{array}{cc}\hline X & Y \\\hline 92.3 & 18.9 \\0.9 & 87.3 \\25.1 & 48.0 \\67.1 & 45.4 \\24.7 & 80.2 \\90.7 & 13.7 \\44.0 & 26.9 \\65.5 & 58.6 \\40.6 & 46.3 \\\hline\end{array}$

a. Find the linear regression model for predicting Y from X.
b. Use the prediction model obtained to predict the value of Y for a school that has 50% of students whose families are below the poverty line.

Answer 2

a. Intercept a = 15.919, slope b = .892.
The regression model is Y_i = .892X_i + 15.919 + e_i
The prediction equation is Y'_i = .892X_i + 15.919

b. When X = 80, Y' = .892X + 15.919 = .892(80) + 15.919 = 87.279
Procedure:
Create a data set with two variables: Pretest (X), Posttest (Y). The data set should have 6 cases.

1. Go to Analyze $\rightarrow$ Regression $\rightarrow$ Linear.

2. Select Posttest to the Dependent list. Select Pretest to the Independent(s) list.

3. Click Statistics. Select Confidence interval under Regression Coefficients. Click Continue.

"4. Click OK.
Selected SPSS Output:

$\begin{array}{l}\text { Model Summary }\\\begin{array}{ccccc}\hline \text { Model } & R & R \text { Square } & \text { Adjusted R Square } & \text { Std. Error of the Estimate } \\\hline 1 & .964 & .930 & .912 & 3.627 \\\hline\end{array}\end{array}$

Question 3

You are given the following pairs of scores on X (height in inches) and Y (weight in lbs).

$\begin{array}{cc}X & Y \\\hline 66 & 140 \\69 & 155 \\72 & 195 \\74 & 160 \\72 & 155 \\67 & 145 \\66 & 135 \\71 & 170 \\70 & 130 \\68 & 170 \\72 & 190 \\69 & 145 \\73 & 155 \\68 & 150 \\68 & 130 \\69 & 145 \\69 & 150 \\66 & 120 \\62 & 131 \\62 & 120 \\64 & 102 \\68 & 110 \\63 & 116 \\64 & 125 \\62 & 110 \\\hline\end{array}$
Perform the following computations using $\alpha$ = .05.
a. The regression equation of Y predicted by X.
b. Test of the significance of X as a predictor.
c. Plot Y versus X.
d. Compute the residuals.
e. Plot residuals versus X.

Answer 3

a. Intercept a = -199.139, slope b = 5.037.
The prediction equation is Y'_i = 5.037X_i -199.139

b. Height is a good predictor of weight, F(1,23) = 29.262, p < .001. Additionally, the unstandardized slope (5.037) and standardized slope (.748) are statistically significantly different from 0 (t = 5.409, df = 23, p < .001); with every one inch increase in height, weight is expected to increase by 5.037 lbs.

c. Plot of Y versus X.
(d) Residuals e_i = Y_i-Y'_i.

$\begin{array}{|c|c|c|}\hline X_{i} & Y_{i} & e_{i} \\\hline 66 & 140 & 6.705 \\\hline 69 & 155 & 6.594 \\\hline 72 & 195 & 31.484 \\\hline 74 & 160 & -13.590 \\\hline 72 & 155 & -8.516 \\\hline 67 & 145 & 6.668 \\\hline 66 & 135 & 1.705 \\\hline 71 & 170 & 11.520 \\\hline 70 & 130 & -23.443 \\\hline 68 & 170 & 26.631 \\\hline 72 & 190 & 26.484 \\\hline 69 & 145 & -3.406 \\\hline 73 & 155 & -13.553 \\\hline 68 & 150 & 6.631 \\\hline 68 & 130 & -13.369 \\\hline 69 & 145 & -3.406 \\\hline 69 & 150 & 1.594 \\\hline 66 & 120 & -13.295 \\\hline 62 & 131 & 17.852 \\\hline 62 & 120 & 6.852 \\\hline 64 & 102 & -21.221 \\\hline 68 & 110 & -33.369 \\\hline 63 & 116 & -2.184 \\\hline 64 & 125 & 1.779 \\\hline 62 & 110 & -3.148 \\\hline\end{array}$
(e) Plot of residuals versus X.


Procedure:
Create a data set with two variables: Height (X), Weight (Y). The data set should have 25 cases.

1) Go to Analyze $\rightarrow$ Regression $\rightarrow$ Linear.
2) Select Weight to the Dependent list. Select Height to the Independent(s) list.
3) Click Statistics. Select Confidence interval under Regression Coefficients. Click Continue.
4) To save residuals for the residual plot, click Save. Select Unstandardized under Residuals. Click Continue. Click OK.
5) To plot Y versus X, go to Graphs $\rightarrow$ Legacy Dialogs $\rightarrow$ Scatter/Dot. Select Simple Scatter. Click Define. Select Weight to Y axis, and Height to X axis. Click OK.
"6) To obtain the residual plot, go to Graphs $\rightarrow$ Legacy Dialogs $\rightarrow$ Scatter/Dot. Select Simple Scatter. Click Define. Select RES_1 to Y axis, and Height to X axis. Click OK.
Selected SPSS Output:

$\begin{array}{l}\text { Model Summary }\\\begin{array}{ccccc}\hline \text { Model } & R & R \text { Square } & \text { Adjusted RSquare } & \text { Std. Error of the Estimate } \\\hline 1 & .748 & .560 & .541 & 16.196 \\\hline\end{array}\end{array}$ $ANOVA^{b}$
$\begin{array}{cccccc}\hline \text { Model } & \text { Sum of Squares } & d f & \text { Mean Square } & F & \text { Sig. } \\\hline \text { Regression } & 7676.012 & 1 & 7676.012 & 29.262 & .000^{a} \\1\text { Residual } & 6033.348 & 23 & 262.319 & & \\\text { Total } & 13709.360 & 24 & & & \\\hline\end{array}$

a. Predictors: (Constant), height
b. Dependent Variable: weight

Question 4

Complete this sentence by selecting one of the following statements: "In simple linear regression, if the slope is found to be ?0.002, . . ."

A) the value of Y is equal to ?.002 when X is 0.
B) the value of Y is equal to ?.002 when X is 1.
C) the value of Y will decrease by 0.002 units when X increases by 1 unit.
D) there is a negative, but very weak relationship between X and Y.

Question 5

In which of the following situations is it most appropriate to use the simple linear regression model?

A)
B)
C)
D)

Question 6

Which one of the following reflects variables appropriate for a simple linear regression model?

A) One categorical dependent variable and one continuous independent variable
B) One continuous dependent variable and one continuous or categorical independent variable
C) One continuous dependent variable and two or more continuous independent variables
D) Two or more continuous dependent variables and one continuous or categorical independent variable

Question 7

Which of the following is that part of the dependent variable that is not predicted by the independent variable?

A) Covariate
B) Intercept
C) Residual
D) Slope

Question 8

The sample intercept is which of the following? Select all that apply.

A) The point where the regression line crosses the Y axis
B) The predicted change in Y for a one-unit change in X
C) The unstandardized regression coefficient
D) The value of the dependent variable when the independent variable is zero