Question

A survey of 41 randomly selected "iPhone" owners showed that the purchase price has a mean of $414 with a sample standard deviation of $180.

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 $13 at a 80% degree of confidence? (Round the final answer to the nearest whole number.)       

Sample size           

Answer 1

Solution :

Given that,

$\bar x$ =414

s = 180

n = 41

Degrees of freedom = df = n - 1 =41- 1 = 40

At 80% confidence level the t is ,

$\alpha$ = 1 - 80% = 1 - 0.80 = 0.20

$\alpha$ / 2 = 0.20 / 2 = 0.10

t$\alpha$ /2,df = t0.10,40 =1.303

a ) The standard error = (s /$\sqrt$n)

= (180/ $\sqrt$ 41)

= 28.11

Margin of error = E = t$\alpha$/2,df * (s /$\sqrt$n)

= 1.303 * (180/ $\sqrt$ 41)

= 36.63

Margin of error = 36.63

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

$\bar x$ - E < $\mu$ < $\bar x$ + E

414 - 36.63 < $\mu$ < 414 + 36.63

377.37 < $\mu$ < 450.63

(377.37, 450.63)

c ) Given that,

standard deviation = S =180

margin of error = E = 13

At 80% confidence level the z is ,

$\alpha$ = 1 - 80% = 1 - 0.80 = 0.20

$\alpha$ / 2 = 0.20 / 2 = 0.10

Z$\alpha$/2 = Z0.10 = 1.282

Sample size = n = ((Z$\alpha$/2 * S) / E)2

= ((1.282 * 6250) / 500)2

= 315

Sample size = 315