Question

In a random sample of four mobile​ devices, the mean repair cost was ​$85.00 and the standard deviation was ​$12.00. Assume the population is normally distributed and use a​ t-distribution to find the margin of error and construct a 95​% confidence interval for the population mean. Interpret the results. The 95​% confidence interval for the population mean mu is ​( ​,​). ​(Round to two decimal places as​ needed.) The margin of error is ​$ nothing. ​(Round to two decimal places as​ needed.) Interpret the results. Choose the correct answer below.

A. If a large sample of mobile devices are taken approximately 95​% of them will have repair costs between the bounds of the confidence interval.

B. It can be said that 95​% of mobile devices have a repair cost between the bounds of the confidence interval.

C. With 95​% ​confidence, it can be said that the population mean repair cost is between the bounds of the confidence interval.

D. With 95​% ​confidence, it can be said that the repair cost is between the bounds of the confidence interval. Click to select your answer(s).

Answer 1

Solution :

degrees of freedom = n - 1 = 4 - 1 = 3

t/2,df = t_0.025,3 = 3.182

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

= 3.182 * ( 12 / 4 )

Margin of error = E = 19.09

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

  ± E  

= 85  ± 19.09

= $ 65.91, $ 104.09 )

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

= 3.182 * ( 12 / 4 )

Margin of error = E = $ 19.09

C. With 95​% ​confidence, it can be said that the population mean repair cost is between the bounds of the confidence interval.