Homework 4
Covers Chapters 7
Use the Project Talent data set.
1. Perform a multiple regression by regressing Math on Gender, SES, Sociability,
Reading, and Mechanical Reasoning entering the predictors in this order.
a. What is the interpretation of the t-tests for SES and Sociability? Be sure to
include the null and alternative hypothesis being tested, the degrees of freedom, the pvalue, and your conclusion and decision about Ho at a 10% level of significance.
SES
Ho: B2 = 0
Ha: B2 ≠ 0
t-stat: 1.63
DF: 19
p-value: 0.119
Decision/Conclusion: Since p-value is greater than 0.10 we would not reject Ho.
Plausible that the addition of SES makes no significant linear contribution to
prediction of Math Achievement when Gender, Sociability, Reading, and
Mechanical Reasoning are in the model.
Sociability
Ho: B3 = 0
Ha: B3 ≠ 0
t-stat: - 0.11
DF: 19
p-value: 0.912
Decision/Conclusion: Since p-value is greater than 0.10 we would not reject Ho.
Plausible that the addition of Sociability makes no significant linear contribution to
prediction of Math Achievement when Gender, SES, Reading, and Mechanical
Reasoning are in the model.
Regression Analysis: Math versus Gender, SES, Social, Reading, Mech
The regression equation is
Math = - 19.4 - 0.63 Gender + 0.231 SES - 0.050 Social + 0.237 Reading
+ 1.28 Mech
Predictor
Constant
Gender
Coef
-19.36
-0.635
SE Coef
14.14
3.101
T
-1.37
-0.20
P
0.187
0.840
1
SES
Social
Reading
Mech
0.2312
-0.0501
0.2370
1.2820
0.1415
0.4476
0.1826
0.4641
1.63
-0.11
1.30
2.76
0.119
0.912
0.210
0.012
b. State the null and alternative hypotheses being tested by the ANOVA table in
output. Include the F-statistic, p-value, and your decision and conclusion about Ho.
Ho: All slopes equal 0
Ha: Not all slopes equal 0
F-stat: 8.83
p-value: 0.000
Decision/Conclusion: Reject Ho and conclude at least one slope does not equal 0.
Analysis of Variance
Source
Regression
Residual Error
Total
Source
Gender
SES
Social
Reading
Mech
DF
1
1
1
1
1
DF
5
19
24
SS
1769.93
761.91
2531.84
MS
353.99
40.10
F
8.83
P
0.000
Seq SS
401.28
582.95
48.57
431.08
306.04
c. Create an ANOVA table that shows the contributions of each of these
predictors using the Sequential Sum of Squares from output. The addition of these sum
of squares should equal the SSR (Sum of Squares Regression) found in the ANOVA table
of your output.
Source
Regression
Gender
SES|Gender
Social|Gender,SES
Reading|Gender,SES,Social
Mech|Gender,SES,Social,Reading
Error
Total
SS
1769.93
401.28
582.95
48.57
431.08
306.04
761.91
2531.84
DF
5
1
1
1
1
1
19
24
MS
353.99
401.28
582.95
48.57
431.08
306.04
40.10
d. For each of the Sequential SS conduct the partial F test for that variable. Be
sure to include the null and alternative hypotheses being tested, the degrees of Freedom,
the p-value from both the F-table in text (this will be a range) and by using Calc >
2
Probability Distributions in Minitab, and your conclusion and decision about Ho at
10% level of significance. Note that this will include five (5) partial F tests.
Gender
Ho: B1= 0
Ha: B1 ≠ 0
F-stat: F 
401.28 / 1
401.28

 4.33
(761.91  306.04  431.08  48.57  582.95) / 23 92.63
DF: 1, 23
p-value from table: 0.05 < p < 0.10 p-value from Minitab: 0.0488
Decision/Conclusion: Since p-value is less than 0.10 we reject Ho and conclude that
Gender is a significant linear predictor of Math Achievement.
SES
Ho: B2= 0
Ha: B2 ≠ 0
F-stat: F 
582.95 / 1
582.95

 8.29
(761.91  306.04  431.08  48.57) / 22 70.35
DF: 1, 22
p-value from table: 0.005 < p < 0.01 p-value from Minitab: 0.0087
Decision/Conclusion: Since p-value is less than 0.10 we reject Ho and conclude that
SES is a significant linear predictor of Math Achievement when Gender in model.
Sociability
Ho: B3= 0
Ha: B3 ≠ 0
F-stat: F 
48.57 / 1
48.57

 0.68
(761.91  306.04  431.08) / 21 71.38
DF: 1, 21
p-value from table: 0.50 < p < 0.10 p-value from Minitab: 0.4189
Decision/Conclusion: Since p-value is greater than 0.10 we fail to reject Ho and
conclude that Sociability is not a significant linear predictor of Math Achievement
when Gender and SES are in model.
Reading
Ho: B4= 0
Ha: B4 ≠ 0
3
F-stat: F 
431.08 / 1
431.08

 8.07
(761.91  306.04) / 20 53.40
DF: 1, 20
p-value from table: 0.01 < p < 0.025 p-value from Minitab: 0.0102
Decision/Conclusion: Since p-value is less than 0.10 we reject Ho and conclude that
Reading is a significant linear predictor of Math Achievement when Gender, SES
and Sociability are in model.
Mechanical Reasoning
Ho: B5= 0
Ha: B5 ≠ 0
F-stat: F 
306.04 / 1
306.04

 7.63
(761.91) / 19 40.10
DF: 1, 19
p-value from table: 0.01 < p < 0.025 p-value from Minitab: 0.0124
Decision/Conclusion: Since p-value is less than 0.10 we reject Ho and conclude that
Mechanical Reasoning is a significant linear predictor of Math Achievement when
Gender, SES, Sociability, and Reading are in model.
NOTE: This last partial F-test is related to the conditional t-test in the regression
portion of output. That is, the t-statistic for Mechanical Reasoning is 2.76 with a pvalue of 0.012. Squaring 2.76 gets you 7.62 which is the partial F-statistic (with
rounding error). The p-values are the same (again including some error for
rounding). This relationship is only true for the last variable entered in the model.
e. Compute by hand the partial coefficient of determination (i.e. partial R2) and
partial correlation for adding Reading to model already containing Gender, SES, and
Sociability. Be sure to correctly identify the direction of this correlation.
SSE ( x1, x 2, x3)  SSE ( x1, x 2, x3, x 4)
SSE ( x1, x 2, x3)
Partial R2:
SeqSS (Re ading )
431.08


 0.288  28.8%
(761.91  306.04  431.08) 1499.03
R y24|1, 2,3 
Interpretation: This indicates that the effect of Reading in reducing the variability
in Math Achievement when Gender, SES, and Sociability are already in the model is
28.8%
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Partial r: square root of partial R2 which is √0.288 = 0.537 and is positive since slope
for Reading is positive when these four variables are in model to predict Math
Achievement.
f. Repeat part d using Minitab to find the partial correlation and paste your
Minitab output of this correlation below.
Pearson correlation of RESI3 and RESI4 = 0.536
P-Value = 0.006
g. Conduct the partial F test to determine whether Gender, SES, and Sociability
together account for a significant amount of variability in Math Achievement. Be sure to
include the null and alternative hypothesis being tested, the degrees of Freedom, the pvalue from both the F-table in text (this will be a range) and by using Calc >
Probability Distributions in Minitab, and your conclusion and decision about Ho.
Ho: B1 = B2 = B3 = 0
Ha: At least one of the slopes does not equal 0
F-stat: F 
(401 .28  582 .95  48 .57 ) / 3
344 .27

 4.82
(761 .91  306 .04  431 .08) / 21 71 .38
DF: 3,21
p-value from table: 0.01 < p < 0.025
p-value from Minitab: 0.0105
Decision/Conclusion: Since p-value is less than 0.10 we would reject Ho and
conclude that at least one of the slopes does not equal 0 meaning at least one of
Gender, SES, and Sociability is a significant linear predictor of Math Achievement.
h. Use Minitab > Basic Statistics > Correlation to get the correlation matrix of the
six variables. Copy and past your output. Which variable has the strongest relationship
with Math Achievement? The weakest?
Strongest: Mechanical Reasoning (r = 0.797
Weakest: Sociability (r = - 0.219)
Gender
-0.091
0.666
Reading
-0.528
0.007
0.506
0.010
0.121
0.566
-0.317
0.123
-0.202
0.332
SES
-0.225
0.279
0.146
0.487
0.502
0.011
-0.095
0.652
Math
-0.398
0.049
0.528
0.007
0.797
0.000
-0.219
0.294
Reading
Mech
Social
Mech
Social
SES
0.557
0.004
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i. Compute the test statistic to determine whether the correlation between Reading
and Sociability in the population is equal to zero with a two-tailed test and alpha of 5%.
Use Minitab Calc > Probability Distributions to get p-value. Check your conclusion (i.e.
p-value) to that found in the correlation matrix.
Ho: p = 0
Ha: p ≠ 0
t-stat: t = r√(n-2)/ √(1-r2) = -0.317√(25-2)/ √(1-(-0.3172)) = - 1.603
DF: 23
p-value: 0.122
Decision/Conclusion: Since p-value is greater than 0.05 we would fail to reject Ho
and conclude that the population correlation between Reading and Sociability is not
different from 0.
j. Compute the 95% confidence interval for the correlation between Reading and
Sociability. Remember to transform your interval back from the Fisher’s value to the
population correlation using Table B8.
z '
z(1 / 2)
z '
z (1 / 2)
n3
where z’ =
(n  3)
 0.325 
1  1  r12 
ln 

2  1  r12 
where z ' 
1  1  r  1  1  (0.317) 
  0.325
ln 
  ln 
2  1  r  2  1  (0.317) 
1.96
 0.325  0.418  0.743  z '  0.093
(25  3)
From Table B8 this converts to  0.63  p  0.09 Since this interval contains 0 we
would fail to reject Ho at the 5% level of significance.
k. How would you interpret the correlation between Gender and Mechanical
Reasoning (remember that Males were coded as 0 and Females as 1). How could we
have made this correlation positive?
The correlation of – 0.528 implies that Males scored higher on Mechanical
Reasoning than Females (i.e as the value of Gender increased from 0 to 1 the
Mechanical Reasoning scores decreased). If one wanted to change the direction you
could simply recode Females as 0 and Males as 1.
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