Question

In a random sample of five mobile​ devices, the mean repair cost was ​$90.00 and the standard deviation was ​$11.00. 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 ​(76.34,103.66).

​(Round to two decimal places as​ needed.)

The margin of error is ​$____?? I NEED THIS ANSWER, OR EXPLAINED HOW TO FIGURE THIS PART OUT PLEASE

​(Round to two decimal places as​ needed.)

Answer 1

Solution :

Given that,

=$ 90.00

s =$11.00

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

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,4 = 2.776 ( using student t table)

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

=2.776 * (11.00 / 5)

= 13.66

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

- E < < + E

90.00 - 13.66< < 90.00+ 13.66

76.34 < < 103.66

(76.34 , 103.66 )

Margin of error = E = 13.66