Question

39 29 20 33 28 25 23 34 37 39 26 34 33 42 32 42...

39 29 20 33 28

25 23 34 37 39

26 34 33 42 32

42 45 24 32 41

48 36 16 13 40

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?

Homework Answers

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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