Question
In: Statistics and Probability
Using the Sample Hypothesis Test Data and Chi-Square Data with a .05 level of significance, provide...
Using the Sample Hypothesis Test Data and Chi-Square Data with a .05 level of significance, provide a summary report including the following information:
Two-Sample Hypothesis Test: Discuss the hypothesis test assumptions and test used. Provide the test statistic and p-value in your response. Evaluate the results of the hypothesis test with the scenario.
Chi-square Hypothesis Test: Discuss the hypothesis test assumptions and test used. Provide the test statistic and p-value in your response. Evaluate the results of the hypothesis test with the scenario.
Chi-Square Data:
| Location | ||||
| Error Type | Midtown | Uptown | Total | |
| Caller Verification | 15 | 12 | 27 | |
| Provided Correct Information | 17 | 11 | 28 | |
| Correct Update | 14 | 16 | 30 | |
| Other Error Types | 10 | 16 | 26 | |
| Total | 56 | 55 | 111 | |
Two sample Hypothesis test data:
| Shift | Call Time |
| AM | 6.5 |
| AM | 6.5 |
| AM | 6.5 |
| AM | 6.5 |
| AM | 7 |
| AM | 7 |
| AM | 7 |
| AM | 7 |
| AM | 7 |
| AM | 8 |
| AM | 8 |
| AM | 8.5 |
| AM | 8.5 |
| AM | 9 |
| AM | 9 |
| AM | 9 |
| AM | 9 |
| AM | 9.5 |
| AM | 9.5 |
| AM | 10 |
| AM | 10 |
| AM | 10 |
| AM | 10 |
| AM | 10 |
| AM | 10.5 |
| AM | 10.5 |
| AM | 10.5 |
| AM | 10.5 |
| AM | 10.5 |
| AM | 10.5 |
| AM | 10.5 |
| AM | 10.5 |
| AM | 10.5 |
| AM | 11.5 |
| AM | 11.5 |
| AM | 11.5 |
| AM | 12 |
| AM | 12 |
| AM | 12 |
| AM | 12 |
| AM | 12.5 |
| AM | 12.5 |
| AM | 13 |
| AM | 13 |
| AM | 13.5 |
| AM | 15.5 |
| AM | 16 |
| AM | 16.5 |
| AM | 17 |
| AM | 18 |
| PM | 6 |
| PM | 9 |
| PM | 9.5 |
| PM | 10 |
| PM | 10.5 |
| PM | 10.5 |
| PM | 11 |
| PM | 11 |
| PM | 11 |
| PM | 11 |
| PM | 11.5 |
| PM | 11.5 |
| PM | 11.5 |
| PM | 12 |
| PM | 12 |
| PM | 12 |
| PM | 12 |
| PM | 12 |
| PM | 12 |
| PM | 12 |
| PM | 12.5 |
| PM | 12.5 |
| PM | 12.5 |
| PM | 12.5 |
| PM | 13 |
| PM | 13 |
| PM | 13.5 |
| PM | 13.5 |
| PM | 14 |
| PM | 14 |
| PM | 14 |
| PM | 14 |
| PM | 14 |
| PM | 14.5 |
| PM | 14.5 |
| PM | 14.5 |
| PM | 15 |
| PM | 15 |
| PM | 15.5 |
| PM | 16 |
| PM | 16.5 |
| PM | 17 |
| PM | 17.5 |
| PM | 18 |
| PM | 18 |
| PM | 18 |
| PM | 18.5 |
| PM | 19 |
| PM | 19.5 |
| PM | 19.5 |
| AM | 5.25 |
| PM | 5.25 |
| AM | 5.25 |
| PM | 5.25 |
| AM | 5.75 |
| PM | 5.75 |
| AM | 5.75 |
| PM | 5.75 |
| AM | 5.75 |
| PM | 6.75 |
| AM | 6.75 |
| PM | 7.25 |
| AM | 7.25 |
| PM | 7.75 |
| AM | 7.75 |
| PM | 7.75 |
| AM | 7.75 |
| PM | 8.25 |
| AM | 8.25 |
| PM | 8.25 |
| AM | 8.75 |
| PM | 8.75 |
| AM | 8.75 |
| PM | 8.75 |
| AM | 8.75 |
| PM | 9.25 |
| AM | 9.25 |
| PM | 9.25 |
| AM | 9.25 |
| PM | 9.25 |
| AM | 9.25 |
| PM | 9.25 |
| AM | 9.25 |
| PM | 10.25 |
| AM | 10.25 |
| PM | 10.25 |
| AM | 10.75 |
| PM | 10.75 |
| AM | 10.75 |
| PM | 10.75 |
| AM | 11.25 |
| PM | 11.25 |
| AM | 11.75 |
| PM | 11.75 |
| AM | 12.25 |
| PM | 14.25 |
| AM | 14.75 |
| PM | 15.25 |
| AM | 15.75 |
| PM | 16.75 |
Solutions
Expert Solution
Two sample t test:
From above probability plots we observed that two samples come from two independent normal populations. Hence assumption of normality holds here.
Test and CI for Two Variances: Call Time (AM Shift), Call Time (PM Shift)
Method
Null hypothesis
Variance(Call Time (AM Shift)) / Variance(Call Time (PM Shift)) =
1
Alternative hypothesis Variance(Call Time (AM Shift)) /
Variance(Call Time (PM Shift)) not = 1
Significance level Alpha = 0.05
Statistics
Variable
N StDev Variance
Call Time (AM Shift) 75 2.858 8.166
Call Time (PM Shift) 75 3.537 12.508
Ratio of standard deviations = 0.808
Ratio of variances = 0.653
Test
Method
DF1 DF2 Statistic P-Value
F Test
(normal)
74 74
0.65 0.069
Since p-value>0.05 so we can assume population variance of two populations are same.
Now we use two sample 2 test with equal variances.
orchestra answered 1 year agoRelated Solutions
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Location
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Type
Midtown
Uptown
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Caller
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15
12
27
Provided
Correct Information
17
11
28
Correct
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