Question

A student records the repair cost for 39 randomly selected TVs. A sample mean of $52.20 and standard deviation of $22.66 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the TVs. Assume the population is normally distributed. Step 2 of 2 : Construct the 90% confidence interval. Round your answer to two decimal places.

Answer 1

Solution :

Given that,

= $52.20

= $22.66

n = 39

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* ( /n)

= 1.645 * (22.66 / 39)

= 5.969

At 90% confidence interval estimate of the population mean is,

- E < < + E

52.20 - 5.969< < 52.20 +5.969

46.231< < 58.169

(46.231 ,  58.169)