Statistics for the Behavioral Sciences
Statistics for the Behavioral Sciences
3rd Edition
ISBN: 9781506386256
Author: Gregory J. Privitera
Publisher: SAGE Publications, Inc
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Chapter 16, Problem 28CAP

1.

To determine

Complete the F table.

Make a decision to retain or reject the null hypothesis that the multiple regression equation can be used to significantly predict health.

1.

Expert Solution
Check Mark

Answer to Problem 28CAP

The completed F table is,

Source of variationSSdfMSFobt
Regression126.22263.1114.48
Residual (error)21.7854.36
Total148.007

The decision is to reject the null hypothesis.

Theregression equation significantly predicts variance in criterion variable (Y) health [BMI].

Explanation of Solution

Calculation:

The given information is that, the researcher has tested whether ‘daily intake of fat (in grams)’ and ‘amount of exercise (in minutes)’ can predict health (measured using a body mass index [BMI] scale.

The formula for test statistic is,

Fobt=MSRegressionMSResidual

Decision rules:

  • If the test statistic value is greater than the critical value, then reject the null hypothesis
  • If the test statistic value is smaller than the critical value, then retain the null hypothesis

Null hypothesis:

H0: The regression equation does not significantly predict variance in criterion variable (Y) health [BMI].

Alternative hypothesis:

H1: The regression equation significantly predicts variance in criterion variable (Y) health [BMI].

Software procedure:

Step by step procedure to obtain test statistic value using SPSS software is given as,

  • Choose Variable view.
  • Under the name, enter the name as Fat, Exercise, Health.
  • Choose Data view, enter the data.
  • Choose Analyze>Regression>Linear.
  • In Dependents, enter the column of Health.
  • In Independents, enter the column of Fat, and Exercise.
  • Click OK.

Output using SPSS software is given below:

Statistics for the Behavioral Sciences, Chapter 16, Problem 28CAP , additional homework tip  1

The table of F is,

Source of variationSSdfMSFobt
Regression126.22263.1114.48
Residual (error)21.7854.36
Total148.007

Table 1

Critical value:

The considered significance level is α=0.05.

The degrees of freedom for regression are 2, the degrees of freedom for residual are 5.

From the Appendix C: Table C.3-Critical values for F distribution:

  • Locate the value 2 in degrees of freedomnumerator row.
  • Locate the value 5 in degrees of freedomdenominator row.
  • Locate the 0.05 level of significance (value in lightface type) in combined row.
  • The intersecting value that corresponds to the (2, 5) with level of significance 0.05 is 5.79.

Thus, the critical value for df=(2,5) with 0.05, level of significance is 5.79.

Conclusion:

The value of test statistic is 14.48.

The critical value is 5.79.

The test statistic value is greater than the critical value.

The test statistic value falls under critical region.

Hence the null hypothesis is rejected.

Thus, the regression equation significantly predicts variance in criterion variable (Y) health [BMI].

2.

To determine

Determine which predictor variable or variables, when added to the multiple regression equation, significantly contributed to predictions in Y (health).

2.

Expert Solution
Check Mark

Answer to Problem 28CAP

The predictor variabledaily intake of fat significantly contributed to predictions in Y (health) when added to the multiple regression equation.

Explanation of Solution

Calculation:

The given information is that, a sample of 8 scores is recorded. The predictor variables are ‘daily intake of fat (in grams)(X1)’ and ‘amount of exercise (in minutes) (X2)’ and criterion variable is health (measured using a body mass index [BMI] scale (Y).

Relative contribution to fat(X1):

Null hypothesis:

H0: Adding fat (X1) does not improve prediction of variance in Y (health) beyond that already predicted by exercise (X2).

Alternative hypothesis:

H1: Adding fat (X1) does improve prediction of variance in Y (health) beyond that already predicted by exercise (X2).

The predictor variable fat is tested. The other predictor variable that is not tested is ‘Exercise’. Calculate the correlation between ‘Exercise’ and ‘Health[BMI]’.

Software procedure:

Step by step procedure to obtain correlation using SPSS software is given as,

  • Choose Variable view.
  • Under the name, enter the name as Health, Exercise.
  • Choose Data view, enter the data.
  • Choose Analyze>Correlate>Bivariate.
  • In variables, enter the Health, and Exercise.
  • Click OK.

Output using SPSS software is given below:

Statistics for the Behavioral Sciences, Chapter 16, Problem 28CAP , additional homework tip  2

The correlation between ‘Exercise’ and ‘Health [BMI]’ is –0.789.

The value of r2 is,

r2=(0.789)2=0.6225

The formula for SSY is,

SSY=Y2(Y)2n

The calculation of sums of squares is,

Health [BMI]

(Y)

Y2
321,024
341,156
23529
331,089
28784
27729
25625
22484
Y=224Y2=6,420

Table 2

Substitute, Y=224,Y2=6,420,n=8 in SSY formula.

SSY=6,420(224)28=6,42050,1768=6,4206,272=148

The contribution of SSY predicted by ‘Exercise.’ alone is,

r2SSY=0.6225×148=92.13

From the F table, the value of SSRegression=126.22, that is the SSY predicted by ‘Fat (X1) and Exercise (X2)’ is 126.22.

The contribution of SSY predicted by ‘Fat (X1)’ alone is,

SSRegression SSYpredicted by (X2)=126.2292.13=34.09

Reproduce the F table by replacing, SSRegression=126.22 with SSY predicted by ‘Fat (X1)’ alone 34.09.

Since there would be only one predictor variable, the degrees of freedom for regression would be 1.

The mean sums of squares for regression is,

MSRegression=SSRegressiondf=34.091=34.09

The F statistic value is,

Fobt=MSRegressionMSResidual=34.094.36=7.82

The change F table with (X1) alone is,

Source of variationSSdfMSFobt
Regression34.09134.097.82
Residual (error)21.7854.36
Total148.007

Critical value:

The considered significance level is α=0.05.

The degrees of freedom for regression are 1, the degrees of freedom for residual are 5.

From the Appendix C: Table C.3-Critical values for F distribution:

  • Locate the value 1 in degrees of freedom numerator row.
  • Locate the value 5 in degrees of freedom denominator row.
  • Locate the 0.05 level of significance (value in lightface type) in combined row.
  • The intersecting value that corresponds to the (1, 5) with level of significance 0.05 is 6.61.

Thus, the critical value for df=(2,5) with 0.05, level of significance is 6.61.

Conclusion:

The value of test statistic is 7.82.

The critical value is 6.61.

The test statistic value is greater than the critical value.

The test statistic value falls under critical region.

Hence the null hypothesis is rejected.

Thus, adding fat (X1) does improve prediction of variance in Y (health) beyond that already predicted by exercise.

Relative contribution to exercise (X2):

Null hypothesis:

H0: Adding exercise (X2) does not improve prediction of variance in Y (health) beyond that already predicted by fat (X1).

Alternative hypothesis:

H1: Adding exercise (X2) does improve prediction of variance in Y (health) beyond that already predicted by fat (X1).

The predictor variable exercise is tested. The other predictor variable that is not tested is ‘fat’. Calculate the correlation between ‘fat’ and ‘Health [BMI]’.

Software procedure:

Step by step procedure to obtain correlation using SPSS software is given as,

  • Choose Variable view.
  • Under the name, enter the name as Health, Fat.
  • Choose Data view, enter the data.
  • Choose Analyze>Correlate>Bivariate.
  • In variables, enter the Health, and Fat.
  • Click OK.

Output using SPSS software is given below:

Statistics for the Behavioral Sciences, Chapter 16, Problem 28CAP , additional homework tip  3

The correlation between ‘fat’ and ‘Health [BMI]’ is 0.923.

The value of r2 is,

r2=(0.923)2=0.8519

The value for SSY is calculated as 148. The contribution of SSY predicted by ‘fat’ alone is,

r2SSY=0.8519×148=126.08

From the F table, the value of SSRegression=126.22, that is the SSY predicted by ‘Fat (X1) and Exercise (X2)’ is 126.22.

The contribution of SSY predicted by ‘Exercise (X2)’ alone is,

SSRegression SSYpredicted by (X1)=126.22126.08=0.14

Reproduce the F table by replacing, SSRegression=126.22 with SSY predicted by ‘Exercise (X2)’ alone 0.14.

Since there would be only one predictor variable, the degrees of freedom for regression would be 1.

The mean sums of squares for regression is,

MSRegression=SSRegressiondf=0.141=0.14

The F statistic value is,

Fobt=MSRegressionMSResidual=0.144.36=0.03

The change F table with (X2) alone is,

Source of variationSSdfMSFobt
Regression0.1410.140.03
Residual (error)21.7854.36
Total148.007

The critical value for F table with (1,5) degrees of freedom, at 0.05 significance level is located as 6.61.

Conclusion:

The value of test statistic is 0.03.

The critical value is 6.61.

The test statistic value is less than the critical value.

The test statistic value does not fall under critical region.

Hence the null hypothesis is retained.

Thus, adding exercise (X2) does not improve prediction of variance in Y (health) beyond that already predicted by fat (X1).

Hence, the predictor variable daily intake of fat significantly contributed to predictions in Y (health) when added to the multiple regression equation.

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