Mathematics

Case Answers

1
Using ANOVA, we reject the null hypothesis that all ratings are the same; so at least one differs from the rest.
Anova: Single Factor
SUMMARY
Groups Count Sum Average Variance
Quality 200 879 4.395 0.5818844221
Ease of Use 200 833 4.165 0.6108291457
Price 200 734 3.67 1.1367839196
Service 200 828 4.14 0.7943718593
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 55.505 3 18.5016666667 23.6907048446 0 2.6160889565
Within Groups 621.65 796 0.7809673367
Total 677.155 799
2
The proportion of on time deliveries in 2010 was 0.9850.
We may test the null hypothesis that the proportion of on time deliveries in 2012 is < 0.985 to determine if it has improved (the alternate hypothesis is p > 0.985)
The sample proportion for 2014 is 0.9907
z = (0.9907 – 0.985)/SQRT(.985*(1-0.985)) = 0.046893334
Critical value = 1.645
p-value = 0.4812991205
Therefore, we cannot conclude a significant improvement
3 A chart of the average number of defects by year shows a declining trend.
We may test for differences between 2010 and 2014 (assuming the samples are the monthly data since we don’t know the actual number of shipments)
t-Test: Two-Sample Assuming Unequal Variances
2008 2012
Mean 826.3333333333 496.25
Variance 135.3333333333 2940.0227272727
Observations 12 12
Hypothesized Mean Difference 0
df 12
t Stat 20.6189486406
P(T<=t) one-tail 0
t Critical one-tail 1.7822875556
P(T<=t) two-tail 0.0000000001
t Critical two-tail 2.1788128297
The test clearly shows a significant difference in the mean defect rates.
4 Testing hypotheses that the mean cost has improved for one of the new processes, we cannot conclude a signficant improvement.
t-Test: Two-Sample Assuming Unequal Variances
Current Process A
Mean 289.6 285.5
Variance 2061.1448275862 4217.6379310345
Observations 30 30
Hypothesized Mean Difference 0
df 52
t Stat 0.2834045089
P(T<=t) one-tail 0.3889960249
t Critical one-tail 1.6746891537
P(T<=t) two-tail 0.7779920499
t Critical two-tail 2.0066468051
t-Test: Two-Sample Assuming Unequal Variances
Current Process B
Mean 289.6 298.4333333333
Variance 2061.1448275862 435.3574712644
Observations 30 30
Hypothesized Mean Difference 0
df 41
t Stat -0.9683208014
P(T<=t) one-tail 0.1692809435
t Critical one-tail 1.6828780021
P(T<=t) two-tail 0.338561887
t Critical two-tail 2.0195409704
5 Conduct two sample tests on mean years at PLE for each factor.
t-Test: Two-Sample Assuming Unequal Variances
Female Male
Mean 5.5307692308 5.5407407407
Variance 12.2506410256 6.4494301994
Observations 13 27
Hypothesized Mean Difference 0
df 18
t Stat -0.0091747587
P(T<=t) one-tail 0.4963903133
t Critical one-tail 1.7340636066
P(T<=t) two-tail 0.9927806266
t Critical two-tail 2.1009220402
CONCLUSION: NO SIGNIFICANT DIFFERENCE BY GENDER
t-Test: Two-Sample Assuming Unequal Variances
College Grad N College Grad Y
Mean 4.8923076923 5.8481481481
Variance 5.8191025641 9.1095156695
Observations 13 27
Hypothesized Mean Difference 0
df 29
t Stat -1.0788151945
P(T<=t) one-tail 0.1447808862
t Critical one-tail 1.6991270265
P(T<=t) two-tail 0.2895617724
t Critical two-tail 2.0452296421
CONCLUSION: NO SIGNIFICANT DIFFERENCE BY COLLEGE GRAD STATUS
t-Test: Two-Sample Assuming Unequal Variances
Local N Local Y
Mean 3.6294117647 6.947826087
Variance 5.6909558824 5.2726086957
Observations 17 23
Hypothesized Mean Difference 0
df 34
t Stat -4.4186414248
P(T<=t) one-tail 0.0000480512
t Critical one-tail 1.6909242552
P(T<=t) two-tail 0.0000961023
t Critical two-tail 2.0322445093
CONCLUSION: SIGNIFICANT DIFFERENCE BY LOCAL AREA
EMPLOYEES FROM THE LOCAL AREA HAVE GREATER RETENTION

826.33333333333337 837.41666666666663 785.91666666666663 669.08333333333337 496.25

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