Question

A survey of 54 randomly selected "iPhone" owners showed that the purchase price has a mean of $400 with a sample standard deviation of $170.

a. Compute the standard error of the sample mean. (Round the final answer to the nearest whole number.)

Standard error           

b. Compute the 80% confidence interval for the mean. (Round the final answers to 2 decimal places.)

The confidence interval is between $  and $  .

c. How large a sample is needed to estimate the population mean within $8 at a 80% degree of confidence? (Round the final answer to the nearest whole number.)       

Sample size           

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 400

sample standard deviation = s = 170

sample size = n = 54

Degrees of freedom = df = n - 1 = 54 - 1 = 53

a) SE = s /$\sqrt$n = 170 /$\sqrt$54 = 23

b) At 80% confidence level

$\alpha$ = 1 - 80%

$\alpha$ =1 - 0.80 =0.20

$\alpha$/2 = 0.10

t$\alpha$/2,df = t0.10,53 = 1.298

Margin of error = E = t$\alpha$/2,df * SE

= 1.298 * 23

Margin of error = E = 29.85

The 80% confidence interval estimate of the population mean is,

$\bar x$  ± E  

= 400 ± 29.85

= ( 370.15, 429.85 )

c) margin of error = E = 8

sample size = n = [t$\alpha$/2,df* s / E]2

n = [1.298 * 170 / 8 ]2

n = 760.79

Sample size = n = 761