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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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cjmejiab

Answer:

E=14.1641

Step-by-step explanation:

We have given that

[tex]\bar{X} = 80 \\s= 13.5 \\n= 6[/tex]

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

[tex]\bar{X}-E< \mu < \bar{X}+E[/tex]

Where is Margin of error given by,

[tex]E=t_c (\frac{s}{\sqrt{n}})[/tex]

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

[tex]E=2.57*(\frac{13.5}{6}) = 14.1641[/tex]

So our Margin of error (E) is 14.1641

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