Question

In a random sample of four microwave​ ovens, the mean repair cost was $85.00 and the standard deviation was $12.00. Assume the population is normally distributed and use a​ t-distribution to construct a 99​% confidence interval for the population mean mu. What is the margin of error of mu? Interpret the results. The 99​% confidence interval for the population mean mu is?

Answer 1

Solution :

Given that,

= $85.00

s = $12.00.

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} = t_0.005,3 = 5.841

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

= 5.841 * (12.00 / 4)

= 35.05

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

- E < < + E

85.00 - 35.05 < < 85.00 + 35.05

49.95 < < 120.05