MGMT2263
Worksheet #6
Solutions
We begin with the KPK output:
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SUMMARY
OUTPUT |
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Regression Statistics |
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Multiple
R |
0.8376 |
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R Square |
0.7016 |
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Adjusted
R Square |
0.6220 |
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Standard
Error |
9.8975 |
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Observations |
20 |
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ANOVA |
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df |
SS |
MS |
F |
Significance F |
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Regression |
4 |
3454.6016 |
863.6504 |
8.8164 |
0.0007 |
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Residual |
15 |
1469.3984 |
97.9599 |
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Total |
19 |
4924.0000 |
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Coefficients |
Standard Error |
t Stat |
P-value |
Lower 95% |
Upper 95% |
VIF |
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Intercept |
24.3923 |
11.9934 |
2.0338 |
0.0601 |
-1.1710 |
49.9556 |
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study |
3.8911 |
0.7286 |
5.3403 |
0.0001 |
2.3380 |
5.4441 |
1.0850 |
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sleep |
2.9052 |
1.2416 |
2.3399 |
0.0335 |
0.2589 |
5.5515 |
1.0700 |
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gender |
-3.8384 |
4.5543 |
-0.8428 |
0.4126 |
-13.5456 |
5.8688 |
1.0481 |
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income |
-2.6087 |
4.8213 |
-0.5411 |
0.5964 |
-12.8851 |
7.6676 |
1.0797 |
1) From the output, grade = 24.3923 + 3.8911(study) + 2.9052(sleep) – 3.8384(gender) – 2.6087(income)
2) The percentage of variation in grade explained by the model is 70.16%.
3) Ho: B1 = B2 = B3 = B4 = 0
Ha: not all coefficients equal 0
Reject Ho if test statistic > 3.06
Test statistic = 8.8164.
Reject Ho and conclude the model is significant.
4)
From the t test
section, we see the p-value for study is 0.01%, that of sleep is 3.35%, that of
gender is 41.26% and that of income is 59.64%. Since study and sleep are the
only variables with p-values less than 5%, these are the only variables that
are significant.
5)
The highest VIF is
1.085. Since all the VIF are less than 10, there are no collinearity problems.
Here is the KPK output for the second model:
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SUMMARY
OUTPUT |
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Regression Statistics |
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Multiple
R |
0.8247 |
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R Square |
0.6802 |
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Adjusted
R Square |
0.6426 |
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Standard
Error |
9.6247 |
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Observations |
20 |
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ANOVA |
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df |
SS |
MS |
F |
Significance F |
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Regression |
2 |
3349.2104 |
1674.6052 |
18.0775 |
6.18997E-05 |
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Residual |
17 |
1574.7896 |
92.6347 |
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Total |
19 |
4924 |
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Coefficients |
Standard Error |
t Stat |
P-value |
Lower 95% |
Upper 95% |
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Intercept |
20.0198 |
10.9148 |
1.8342 |
0.0842 |
-3.0083 |
43.0480 |
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study |
3.9576 |
0.6876 |
5.7560 |
0.0000 |
2.5070 |
5.4083 |
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sleep |
3.0188 |
1.1798 |
2.5589 |
0.0203 |
0.5298 |
5.5079 |
6) From the output, grade = 20.0198 + 3.9576(study) + 3.0188(sleep)
7) The KPK output gives 80.728 which rounds to 81.
8) The 95% confidence interval given by KPK ranges from 76 to 86 after rounding.
9) The 95% prediction interval given by KPK ranges from 60 to 102. Since the upper limit cannot be more than 100, the appropriate upper limit is 100.
10) Here are the results:
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Adj. r2 |
ANOVA p-value |
t tests (all variables significant?) |
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#1 |
62.2% |
0.0007 |
no |
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#2 |
64.26% |
6.19 x 10-5 |
yes |
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Best |
#2 |
#2 |
#2 |
Model #2 is best since it has the highest adjusted r2, the lowest ANOVA p-value and all its variables are significant.
11) Ho: age does not depend on gender
Ha: age does depend on gender
Degrees of freedom = 1 x 2 = 2; Reject Ho if test statistic > 5.9915
Here is the crosstab of expected values:
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22.09 |
9.82 |
13.09 |
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31.91 |
14.18 |
18.91 |
Test statistic = 0.165 + 0.068 + 0.091 + 0.114 + 0.047 + 0.063 = 0.548.
Do not reject Ho.
Conclude that the client’s age does not depend on gender.
12) Cramer’s V = sqrt(0.548/110) = 0.0706 = 7.06%
13) When we work out the expected values for the 20+ category, this is what we get:
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working |
semi-retired |
retired |
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1.33 |
3.62 |
5.05 |
As we see, 2 of the 3 expected values are less than 5. Thus, we need to collapse this category with the 10 to 19 category.
14) Here is the new crosstab of observed values:
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< 10 |
10 or more |
Total |
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Working |
10 |
4 |
14 |
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Semi-retired |
13 |
25 |
38 |
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Retired |
15 |
38 |
53 |
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Total |
38 |
67 |
105 |
Ho: hours worked does not depend on work status
Ha: hours worked does depend on work status
Degrees of freedom = 2 x 1 = 2; Reject Ho if test statistic > 5.9915.
Here is the crosstab of expected values:
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5.067 |
8.933 |
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13.752 |
24.248 |
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19.181 |
33.819 |
Test statistic = 4.803 + 2.724 + 0.041 + 0.023 + 0.911 + 0.517 = 9.019
Reject Ho.
Conclude hours worked depends on work status
15) Cramer’s V = sqrt(9.019/105) = 0.2931 = 29.31%
16) Ho:
the data is normally distributed
Ha: it is not
Reject Ho if test statistic >
0.285
The mean of the data is 278.125 and
the standard deviation is 436.8551. This spreadsheet summarizes the results:
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Value |
Z score |
F(z) |
S(z) |
S'(z) |
Max diff |
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50 |
-0.52 |
0.3015 |
0.1250 |
0 |
0.3015 |
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75 |
-0.46 |
0.3228 |
0.2500 |
0.125 |
0.1978 |
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100 |
-0.41 |
0.3409 |
0.3750 |
0.25 |
0.0909 |
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120 |
-0.36 |
0.3594 |
0.5000 |
0.375 |
0.1406 |
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140 |
-0.32 |
0.3745 |
0.6250 |
0.5 |
0.2505 |
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150 |
-0.29 |
0.3859 |
0.7500 |
0.625 |
0.3641 |
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240 |
-0.09 |
0.4641 |
0.8750 |
0.75 |
0.4109 |
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1350 |
2.45 |
0.9929 |
1.0000 |
0.875 |
0.1179 |