Question

A manager records the repair cost for 4 randomly selected TVs. A sample mean of $88.46 and standard deviation of $17.20 are subsequently computed. Determine the 99% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal. Step 2 of 2: Construct the 99% confidence interval. Round your answer to two decimal places.

Answer 1

solution

Given that,

= $88.46

s =$17.20

n = 4

Degrees of freedom = df = n - 1 = 4- 1 = 3

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2  df = t0.005,3 =5.841    ( using student t table)

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

= 5.841 * ( 17.20/ 4) =50.23

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

- E < < + E

88.46 -50.23   < <88.46 + 50.23

38.23 < < 138.69

(38.23 ,138.69)