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A survey of 25 randomly selected customers found the ages shown​ (in years). The mean is 32.44 years and the standard deviation is 8.97 years. ​a) Construct a 99​% confidence interval for the mean age 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 9.0 ​years?

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Answer #1

Solution:
Given no. of sample = 25
at alpha = 0.01, so Zalpha/2 = 2.575
So 99% confidence interval is
32.44 +/- 2.575*8.97/sqrt(25)
32.44 +/- 4.62
So 99% confidence interval is
27.82 to 37.06
Margin of error can be calculated as
talpha/2 * SD/sqrt(n)
2.575*8.97/sqrt(25) = 4.62

Solution(c)
Now standard deviation = 9
So 99% confidence interval is
32.44 +/- 2.575*9/sqrt(25)
32.44 +/- 4.635
27.8 to 37.07
We can see that Confidence interval length has increased

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