The owner of a retail outlet sold and provided service for copying machines. One type of machine, DW 140 model, needed service often. The owner was interested in the mean time required to service this model and decided not to carry the model if the mean repair time appeared to be more than 40 minutes. The repairman provided the 39 times (in minutes) which is on BlackBoard in SERVICE TIME. IGNORE OUTLIERS
a. b. c.
Construct a 90% confidence interval. Based on the
information.
With the given information would the retailer keep this model and
why?
If the retailer wanted the margin of error to be plus or minus 2
minutes how many observations from the repairman would she need?
Use the sample standard deviation as a proxy for the population
standard deviation and keep the same significance level.
23
41
47
27
42
47
28
42
48
33
42
49
35
43
50
37
44
51
39
44
52
39
44
53
39
45
53
39
46
56
40
46
57
40
46
61
41
47
62
Sample size = n = 39
Sample mean = = 44.0513
Standard deviation = s = 8.4572
We have to construct 90% confidence interval.
Formula is
Here E is a margin of error.
Degrees of freedom = n - 1 = 39 - 1 = 38
Level of significance = 0.10
tc = 1.686 ( Using t table)
So confidence interval is ( 44.0513 - 2.2832 , 44.0513 + 2.2832) = > ( 41.7681 , 46.3345)
If the retailer wanted the margin of error to be plus or minus 2 minutes how many observations from the repairman would she need? Use the sample standard deviation as a proxy for the population standard deviation and keep the same significance level.
In this question, we have to find the sample size(n).
Confidence level = 0.90
tc = 1.686
Margin of error = E = 2
Population standard deviation = = 8.4572
We have to find sample size (n)
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