A survey of 25 randomly selected customers found the ages shown (in years). The

Question

A survey of 25 randomly selected customers found the ages shown (in years). The mean is 32.72 years and the standard deviation is 10.06 30 37 49 24 38 years. a) Construct a 90% confidence interval for the mean age of all customers, assuming that the assumptions and conditions for the confidence 41 29 25 31 39 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 11.0 years? 24 18 47 10 47 31 35 29 31 44 49 34 24 21 31 a) What is the confidence interval? (Round to two decimal places as needed.)

Explanation / Answer

(a) 90% confidence interval = x +- tdf, 0.10 (s/ n)

Here n = 25

s = 10.06

x = 32.72 years

dF = 25 -1 = 24

t24,0.10 = 1.71088

90% confidence interval = x +- tdf, 0.10 (s/ n)

= 32.72 +- 1.71088 * (10.06/ 25)

= 32.72 + 1.71088 * 2.012

= (29.28, 36.16)

(b) Margin of error = (36.16 - 29.28)/2 = 3.44

(c) If we assume that standard deviation is 11.0 years, the standard error of mean would increase that would increase the width of confidence interval.

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