Question

# In a random sample of six microwave​ ovens, the mean repair cost was ​\$65.00 and the...

In a random sample of six microwave​ ovens, the mean repair cost was ​\$65.00 and the standard deviation was ​\$12.50. Assume the population is normally distributed and use a​ t-distribution to construct a 90​% confidence interval for the population mean mu. What is the margin of error of mu​? Interpret the results. The 90​% confidence interval for the population mean mu is ​( nothing​, nothing​). ​(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. It can be said that 90​% of microwaves have a repair cost between the bounds of the confidence interval. B. If a large sample of microwaves are taken approximately 90​% of them will have repair costs between the bounds of the confidence interval. C. With 90​% ​confidence, it can be said that the population mean repair cost is between the bounds of the confidence interval. D. With 90​% ​confidence, it can be said that the repair cost is between the bounds of the confidence interval.

Solution :

Given that,

Point estimate = sample mean = = 65.00

sample standard deviation = s = 12.50

sample size = n = 6

Degrees of freedom = df = n - 1 = 6 - 1 = 5

At 90% confidence level = 1 - 90% =1 - 0.90 =0.10 /2 = 0.05

t /2,df = t0.05,5 = 2.015

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

= 2.015 * (12.50 / 6)

Margin of error = E = 10.28

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

= 65.00 ± 10.28

= ( 54.72, 75.28 )

C. With 90​%​confidence, it can be said that the population mean repair cost is between the bounds of the confidence interval.

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