Handout_assignment_IIIrev

Handout_assignment_IIIrev - Maria Luongo Handout assignment...

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Unformatted text preview: Maria Luongo Handout assignment III 1. H&O No. 8.65, 8.66 8.65.a. Descriptive Statistics: LeaveDays Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 LeaveDays 50 0 5.040 0.569 4.020 0.000 2.000 4.000 7.000 Variable Maximum LeaveDays 18.000 This information suggests the shape is skewed to the right because the median is lower than the mean. The mean is pulled out toward the long tail. b. Stem-and-Leaf Display: LeaveDays Stem-and-leaf of LeaveDays N = 50 Leaf Unit = 1.0 9 0 011111111 23 0 22222223333333 (9) 0 444444555 18 0 6667777 11 0 8999 7 1 001 4 1 2 3 1 44 1 1 1 1 8 Yes, the plot confirms my impression about the skewness. 8.66.a. One-Sample T: LeaveDays Test of mu = 5.7 vs not = 5.7 Variable N Mean StDev SE Mean 95% CI T P LeaveDays 50 5.040 4.020 0.569 (3.897, 6.183) -1.16 0.251 P=0.251 > 0.05 so, we must retain the null Hypothesis that the mean is still 5.7 b. The test was two-sided using the computer package because question 8.66.a. asked to test whether the null hypothesis is still 5.7 or not . However, a one-sided test would be more appropriate to determine if the number of leave days has decreased due to the offered bonuses. See below results for one sided test using the computer package: One-Sample T: LeaveDays Test of mu = 5.7 vs < 5.7 95% Upper Variable N Mean StDev SE Mean Bound T P LeaveDays 50 5.040 4.020 0.569 5.993 -1.16 0.126 2. H&O No. 9.69, 9.70 9.69.a. Two-Sample T-Test and CI: DaysRequired, DaysRequired_2 Two-sample T for DaysRequired vs DaysRequired_2 N Mean StDev SE Mean DaysRequired 21 24.0 12.6 2.7 DaysRequired_2 59 27.24 3.54 0.46 b....
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Handout_assignment_IIIrev - Maria Luongo Handout assignment...

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