Question

A survey of 32 randomly selected "iPhone" owners showed that the purchase price has a mean of $398 with a sample standard deviation of $155. (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 99% confidence interval for the mean. (Round your answers to 3 decimal places.) The confidence interval is between and . c. To be 99% 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 = 155

n = 32

Degrees of freedom = df = n - 1 = 32 - 1 = 31

a ) At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

  / 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,31 = 2.744

a )  The standard error =  (s /n)

= (155 / 32 )

= 24

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

= 2.744 * (155 / 32)

= 75.187

The 99% confidence interval estimate of the population mean is,

- E < < + E

398 - 75.187 < < 398 + 75.187

322.813 < < 473.187

(322.813, 473.187 )

b )margin of error = E = 16

At 95% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Sample size = n = ((Z_/2 * ) / E)²

= (2.576 * 155) / 16)²

= 342.8052

Sample size = 343