Homework 4
Covers Chapters 7 and 8
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. Use the Sequential SS to conduct the following partial F tests. Be sure to
include the null and alternative hypotheses 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 at
10% level of significance.
Reading and Mechanical Reasoning to model with Gender, SES, and Sociability
Ho: B4= B5= 0 (both slopes are 0)
Ha: at least one of these two slopes does not equal 0.
F-stat: F =
(431.08+ 306.04 / 2 368.56
=
= 9.19
(761.91) /19
40.10
DF: 2, 19
p-value from table: 0.001< p < 0.005 p-value from Minitab: 0.0016
Decision/Conclusion: Since p-value is less than 0.10 we reject Ho and conclude that
at least one of these predictors, Reading and Mechanical Reasoning, is a significant
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linear predictor of Math Achievement when Gender, SES, and Sociability are in
model.
Reading to model with Gender, SES, and Sociability
Ho: B4= 0
Ha: B4 ≠ 0
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.
d. Compute the partial coefficient of determination (i.e. partial R2) and partial
correlation for adding Mechanical Reasoning to model already containing Gender, SES,
Sociability and Reading. Be sure to correctly identify the direction of this correlation.
Partial r: 0.535
Partial R2: 28.6%
Interpretation: This indicates that the effect of Mechanical Reasoning in reducing
the variability in Math Achievement when Gender, SES, Sociability, and Reading
are already in the model is 28.6%
e. Use Minitab > Basic Statistics > Correlation to get the correlation matrix of the
six variables. Which variable has the strongest relationship with Math Achievement?
The weakest?
Strongest: Mechanical Reasoning (r = 0.797
Weakest: Sociability (r = - 0.219)
f. Compute the test statistic to determine whether the correlation between Math
Achievement 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 of part e.
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Ho: p = 0
Ha: p ≠ 0
t-stat: t = r√(n-2)/ √(1-r2) = -0.219√(25-2)/ √(1-(-0.2192)) = - 1.076
DF: 23
p-value: 0.292
Decision/Conclusion: Since p-value is greater than 0.05 we would fail to reject Ho
and conclude that the population correlation between Math Achievement and
Sociability is not different from 0.
g. 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.
h. 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.
2. The variable School Size is interpreted as follows:
1 = number of students is less than 100
2 = number of students is from 100 to 399
3 = number of students is 400 or more
A lack-of-fit test regressing Math on Gender, SES, Sociability, Reading, Mechanical
Reasoning, and School Size indicates that a possible curvature exists for the variables
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Reading and Mechanical Reasoning. Without centering, create new, second-order
variables for Reading and for Mechanical Reasoning. Repeat the multiple regression
analysis including these second order terms. Select Options and select Variance Inflation
Factor and Lack-of-Fit > Data Subsetting and answer the following questions.
a. What are the VIF values for all of the predictors?
Predictor
Constant
Gender
Reading
Mech
Social
SES
School_Size
Reading2
Mech2
Coef
12.255
-0.955
-1.1096
-2.4246
0.5326
0.12741
2.9210
0.02624
0.17232
SE Coef
8.689
1.557
0.5938
0.9587
0.2436
0.08062
0.9586
0.01083
0.05258
T
1.41
-0.61
-1.87
-2.53
2.19
1.58
3.05
2.42
3.28
P
0.178
0.548
0.080
0.022
0.044
0.134
0.008
0.028
0.005
VIF
1.547
68.549
44.186
1.366
1.836
1.316
76.321
54.056
b. Does this indicate the presence of Multicollinearity? Why or why not?
Yes, since the VIF are greater than 10 for the first and second order terms
c. What is the p-value of the lack-of-fit tests and what does this indicate about model fit?
P >= 0.1 so no evidence of lack of fit.
3. Create new second order terms for Reading and Mechanical Reasoning after centering
both the first order terms. Repeat the multiple regression analysis including these
centered first and second order terms. Select Options and select Variance Inflation Factor
and Lack-of-Fit > Data Subsetting and answer the following questions.
a. What are the VIF values for all of the predictors?
Predictor
Constant
Gender
Social
SES
School_Size
cReading
cMech
cReading2
cMech2
Coef
-2.591
-0.955
0.5326
0.12741
2.9210
0.5045
1.4354
0.02624
0.17232
SE Coef
7.368
1.557
0.2436
0.08062
0.9586
0.1214
0.3610
0.01083
0.05258
T
-0.35
-0.61
2.19
1.58
3.05
4.15
3.98
2.42
3.28
P
0.730
0.548
0.044
0.134
0.008
0.001
0.001
0.028
0.005
VIF
1.547
1.366
1.836
1.316
2.867
6.264
2.895
3.047
b. Does this indicate the presence of Multicollinearity and support your answer?
No, there are no VIF values greater than 10
c. What is the p-value of the lack-of-fit test and what does this indicate about model fit?
P >= 0.1 so no evidence of lack of fit.
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