Question

In a random sample of five microwave​ ovens, the mean repair cost was $60.00 and the standard deviation was $11.50

Assume the population is normally distributed and use a​ t-distribution to construct a 90​% confidence interval for the population mean μ. What is the margin of error of μ​? Interpret the results.

(a)The confidence interval for the population mean is _, _   ​(Round to one decimal place as​ needed.)

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

(c) Interpret the results. Choose the correct answer below:

1. It can be said that 95​% of microwaves have a repair cost between the bounds of the confidence interval.

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

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

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

Answer 1

Solution :

Given that,

a) Point estimate = sample mean = = 60.00

sample standard deviation = s = 11.50

sample size = n = 5

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

At 90% confidence level

= 1 - 90%

=1 - 0.90 =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 * ( 11.50 / 5)

Margin of error = E = 10.96

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

  ± E

= 60.00 ± 10.96

= (49.04, 70.96)

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

= 2.132 * ( 11.50 / 5)

Margin of error = E = 10.96

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