Question

In a random sample of eleven cell​ phones, the mean full retail price was ​$515.50 and the standard deviation was ​$212.00. Assume the population is normally distributed and use the​ t-distribution to find the margin of error and construct a 90​% confidence interval for the population mean mu. Interpret the results.

Answer 1

Solution :

t_{$\alpha$ /2,df} = 1.812

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

= 1.812 * (212 / $\sqrt$ 11)

Margin of error = E = 115.8

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

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

515.5 - 115.8 < $\mu$ < 515.5 + 115.8

399.7 < $\mu$ < 631.3

A 90​% confidence interval for the population mean mu is : (399.7 , 631.3)