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In a random sample of six microwave​ ovens, the mean repair cost was ​$80.00 and the standard deviation was ​$11.00. Assume the population is normally distributed and use a​ t-distribution to construct a 90% confidence interval for the population mean mμ. What is the margin of error of mμ​? Interpret the results.

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Answer:

E=14.1641

Explanation:

We have given that


\bar{X} = 80 \\s= 13.5 \\n= 6

Assuming a confidence interval of 95% for \mu we have that,


\bar{X}-E< \mu < \bar{X}+E

Where is Margin of error given by,


E=t_c ((s)/(√(n)))

For 95% in the T-Table we have that
t_c = 2.57, so


E=2.57*((13.5)/(6)) = 14.1641

So our Margin of error (E) is 14.1641

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