Question

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?

Answer #1

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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