Question

A survey of 26 randomly selected "iPhone" owners showed that the purchase price has a mean of $398 with a sample standard deviation of $170. (Use t Distribution Table.) a. Compute the standard error of the sample mean. (Round your answer to the nearest whole number.) Standard error b. Compute the 95% confidence interval for the mean. (Round your answers to 3 decimal places.) The confidence interval is between and . c. To be 95% confident, how large a sample is needed to estimate the population mean within $16? (Roundup your answer to the next whole number.) Sample size

Answer 1

Solution :

Given that,

= 398

s = 170

n = 26

Degrees of freedom = df = n - 1 = 26 - 1 = 25

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,25 = 2.060

a )  The standard error =  (s /n) = 33.34 = 33

Margin of error = E = t_/2,df * (s /n)

= 2.060 * (170 / 26) = 68.680

The 95% confidence interval estimate of the population mean is,

- E < < + E

398 - 68.680 < < 398 + 68.680

329.320 < < 466.680

(329.320, 466.680 )

b )margin of error = E = 16

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

Sample size = n = ((Z_/2 * ) / E)²

= ((1.960 * 170) / 16)²

= 433.6806

Sample size = 434