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
= t_{0.05,5} = 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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