Question

In a random sample of six microwave​ ovens, the mean repair cost was ​$70.00 and the standard deviation was ​$12.00. Assume the variable is normally distributed and use a​ t-distribution to construct a 99​% confidence interval for the population mean u. What is the margin of error of u​?

The 99​% confidence interval for the population mean u is (_, _ ) ​(Round to two decimal places as​ needed.)

The margin of error is ___. ​(Round to two decimal places as​ needed.)

Answer 1

Solution :

Given that,

= $70.00

s = $12.00

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)   

E = 4.032 * (12.00 / 6) = 19.75

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

- E < < + E

70.00 - 19.75 < < 70.00 + 19.75

50.25 < < 89.75

(50.25 ,  89.75 )