Question

In a random sample of five microwave​ ovens, the mean repair cost was ​$60.00 and the standard deviation was ​$14.00. Assume the variable is normally distributed and use a​ t-distribution to construct a 95​% confidence interval for the population mean u.

What is the margin of error of u​?

The 95​% confidence interval for the population mean u is l(_, _ ) ​(Round to two decimal places as​ needed.

Answer 1

Solution :

Given that,

= 60

s = 14

n = 5

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

) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,4 =2.776

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

= 2.776 * (14 / 5) = 17.38

The 95% confidence interval estimate of the population mean is,

- E < < + E

60-17.38 < < 60+17.38

42.62 < < 77.38

(42.62,77.38)