Question

A consumer affairs investigator records the repair cost for 4 randomly selected TVs. A sample mean of $75.89 and standard deviation of $13.53 are subsequently computed. Determine the 80% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal.

Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Step 2 of 2: Construct the 80%80% confidence interval. Round your answer to two decimal places.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 75.89

sample standard deviation = s = 13.53

sample size = n = 4

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

Step 1

At 80% confidence level the t is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

t _/2,df = t_{0.10 , 3} = 1.638

Step 2

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

= 1.638 * (13.53 / 4)

= 11.08

The 80% confidence interval estimate of the population mean is,

- E < < + E

75.89 - 11.08 < < 75.89 + 11.08

64.81 < < 86.97

(64.81 , 86.97)