In: Statistics and Probability
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
Solution :
Given that,
n = 26
Degrees of freedom = df = n - 1 = 26 - 1 = 25
a ) At 95% confidence level the t is ,
a ) The standard error = (s /n) = 33.34 = 33
Margin of error = E = t/2,df * (s /
n)
The 95% confidence interval estimate of the population mean is,
(329.320, 466.680 )
b )margin of error = E = 16
At 95% confidence level the z is ,
Sample size = n = ((Z/2
*
) / E)2
= ((1.960 * 170) / 16)2
= 433.6806
Sample size = 434