Question

A random sample of 20 purchases showed the amounts in the table​ (in $). The mean is ​$50.76 and the standard deviation is ​$19.48 ​a) Construct a 80​% confidence interval for the mean purchases of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met. ​b) How large is the margin of​ error? ​c) How would the confidence interval change if you had assumed that the standard deviation was known to be $20?

60.91
64.3
65.17
66.82
47.34
27.46
9.39
28.39
51.29
52.12
52.26
32.46
53.07
53.35
87.14
37.63
38.46
89.39
58.95
39.23

Answer 1

a)

We need to construct the 80% confidence interval for the population mean \muμ. The following information is provided:

Sample Mean =	50.76
Sample Standard Deviation (s) =	19.48
Sample Size (n) =	20

The critical value for α=0.2 and df=n−1=19 degrees of freedom is . The corresponding confidence interval is computed as shown below:

b)

c)

If the standard deviation was known to be 20, then the width of the confidence interval will decrease.

We need to construct the 80% confidence interval for the population mean μ. The following information is provided:

Sample Mean =	50.76
Population Standard Deviation (σ) =	20
Sample Size (N) =	20

The critical value for α=0.2 is zc​=z1−α/2​=1.282. The corresponding confidence interval is computed as shown below:

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