60 61 40 70 68 60 45 5. Employees are supposed to take a 60-minute lunch. The manager is concerned that they are taking longer, so he has his secretary select a random sample of 30 employees shown below. Minutes at lunch 63 What is the mean and standard deviation of the data? Mean 62.47 s= 7.88 Construct an 85% confidence interval for the mean tension readings. Show formulas even if using Excel. T* =T.Inv.2T (.15,29)=1.48705 62.47 -1.478*(7.88/sqrt(30)), 64.47 + 1.478*(7.88/sqrt(30)) (60.34003, 64.59331) 65 66 69 63 65 Conduct an appropriate test of significance (at the 5% level) to decide if the data support the manager's claim that employees are taking too long of a lunch break. 66 70 Ho: u = 60 The average lunch time of employees is 60 Minutes 62 63 64 68 Hau > 60 The average lunch time of employees is greater than 60 Minutes N 30 10 SRS so meets conditions. 67 55 67 66 Since 60 is not in the confidence interval we have evidence to reject the Null Hypothesis. There is evidence that the average lunch time of employees is greater than 60 Minutes 40 60 64 Could also calulate the test statistic t= (62.47-60)/(7.88/sqrt(30) = 1.7168 68 69 p-value = T.DIST.RT (1.7168, 29) = 0.0483 63 67 0.0483 <0.05 Therefore we reject the null hypothesis, There is evidence that the average lunch time of employees is greater than 60 Minutes
60 61 40 70 68 60 45 5. Employees are supposed to take a 60-minute lunch. The manager is concerned that they are taking longer, so he has his secretary select a random sample of 30 employees shown below. Minutes at lunch 63 What is the mean and standard deviation of the data? Mean 62.47 s= 7.88 Construct an 85% confidence interval for the mean tension readings. Show formulas even if using Excel. T* =T.Inv.2T (.15,29)=1.48705 62.47 -1.478*(7.88/sqrt(30)), 64.47 + 1.478*(7.88/sqrt(30)) (60.34003, 64.59331) 65 66 69 63 65 Conduct an appropriate test of significance (at the 5% level) to decide if the data support the manager's claim that employees are taking too long of a lunch break. 66 70 Ho: u = 60 The average lunch time of employees is 60 Minutes 62 63 64 68 Hau > 60 The average lunch time of employees is greater than 60 Minutes N 30 10 SRS so meets conditions. 67 55 67 66 Since 60 is not in the confidence interval we have evidence to reject the Null Hypothesis. There is evidence that the average lunch time of employees is greater than 60 Minutes 40 60 64 Could also calulate the test statistic t= (62.47-60)/(7.88/sqrt(30) = 1.7168 68 69 p-value = T.DIST.RT (1.7168, 29) = 0.0483 63 67 0.0483 <0.05 Therefore we reject the null hypothesis, There is evidence that the average lunch time of employees is greater than 60 Minutes
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:60
61
40
70
68
60
45
5. Employees are supposed to take a 60-minute lunch. The manager is concerned that they are taking
longer, so he has his secretary select a random sample of 30 employees shown below.
Minutes
at lunch
63
What is the mean and standard deviation of the data?
Mean 62.47 s= 7.88
Construct an 85% confidence interval for the mean tension readings. Show formulas
even if using Excel.
T* =T.Inv.2T (.15,29)=1.48705
62.47 -1.478*(7.88/sqrt(30)), 64.47 + 1.478*(7.88/sqrt(30))
(60.34003, 64.59331)
65
66
69
63
65
Conduct an appropriate test of significance (at the 5% level) to decide if the data
support the manager's claim that employees are taking too long of a lunch break.
66
70
Ho: u = 60 The average lunch time of employees is 60 Minutes
62
63
64
68
Hau > 60 The average lunch time of employees is greater than 60 Minutes
N 30 10 SRS so meets conditions.
67
55
67
66
Since 60 is not in the confidence interval we have evidence to reject the Null
Hypothesis. There is evidence that the average lunch time of employees is greater
than 60 Minutes
40
60
64
Could also calulate the test statistic t= (62.47-60)/(7.88/sqrt(30) = 1.7168
68
69
p-value = T.DIST.RT (1.7168, 29) = 0.0483
63
67
0.0483 <0.05 Therefore we reject the null hypothesis, There is evidence that the
average lunch time of employees is greater than 60 Minutes
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