Question

years. A survey of 25 randomly selected customers found the ages shown (in years). The mean is 32.84 years and the standard deviation is 8.84 25 39 43 25 42p a) Construct a 99% confidence interval for the mean age of all customers, assuming that the assumptions and conditions for the confidence 23 28 42 45 31 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? 31 23 34 36 17 34 18 23 33 34 48 39 47 33 28 A) What is the confidence interval B) What is the margin of error? C) What is the confidence interval using the given population standard deviation?

Answer 1

A )

Given, $\bar{x}$ = 32.84, S = 8.84

99% Confidence interval for $\mu$ is

$\bar{x}$ - t_$\alpha$/2 * S / sqrt(n) < $\mu$ < $\bar{x}$ + t_$\alpha$/2 * S / sqrt(n)

32.84 - 2.797 * 8.84 / sqrt(25) < $\mu$ < 32.84 + 2.797 * 8.84 / sqrt(25)

27.8949 < $\mu$ < 37.7851

99% CI for $\mu$ is (27.8949 ,37.7851)

b)

Margin of error = t_$\alpha$/2 * S / sqrt(n)

= 2.797 * 8.84 / sqrt(25)

= 4.9451

c)

Standard deviaiton is known that is $\sigma$ = 9

99% Confidence interval for $\mu$ is

$\bar{x}$ - Z_$\alpha$/2 * $\sigma$ / sqrt(n) < $\mu$ < $\bar{x}$ + Z_$\alpha$/2 * $\sigma$ / sqrt(n)

32.84 - 2.5758 * 9 / sqrt(25) < $\mu$ < 32.84 + 2.5758 * 9 / sqrt(25)

28.2035 < $\mu$ < 37.4765

99% CI for $\mu$ is (28.2035 , 37.4765)