Question

In a random sample of five microwave​ ovens, the mean repair cost was ​$65.00 and the standard deviation was ​$13.50. Assume the population is normally distributed and use a​ t-distribution to construct a 90​% confidence interval for the population mean mu. What is the margin of error of mu​? Interpret the results. The 90​% confidence interval for the population mean mu is ​( nothing​, nothing​). ​(Round to two decimal places as​ needed.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 65

sample standard deviation = s = 13.50

sample size = n = 5

Degrees of freedom = df = n - 1 = 5 - 1 = 4

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,4 = 2.132

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

= 2.132 * (13.50 / 5)

= 12.87

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

- E < < + E

65 - 12.87 < < 65 + 12.87

52.13 < < 77.87

(52.13 , 77.87)