Question

In a random sample of six microwave​ ovens, the mean repair cost was ​$75.00 and the standard deviation was ​$11.00. Assume the population is normally distributed and use a​ t-distribution to construct a 99​% confidence interval for the population mean μ. What is the margin of error of μ​? Interpret the results.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 75

sample standard deviation = s = 11

sample size = n = 6

Degrees of freedom = df = n - 1 = 6 - 1 = 5

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,5 = 4.032

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

= 4.032 * (11 / 6)

= 18.1

Margin of error = E = 18.1

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

- E < < + E

75 - 18.1 < < 75 + 18.1

56.9 < < 93.1

(56.9 , 93.1)