In: Statistics and Probability
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
Solution :
Given that,
n = 32
Degrees of freedom = df = n - 1 = 32 - 1 = 31
a ) At 99% confidence level the t is ,
a ) The standard error = (s /n)
= 24
Margin of error = E = t/2,df * (s /
n)
= 75.187
The 99% confidence interval estimate of the population mean is,
(322.813, 473.187 )
b )margin of error = E = 16
At 95% confidence level the z is ,
Sample size = n = ((Z/2 *
) / E)2
= (2.576 * 155) / 16)2
= 342.8052
Sample size = 343