Question

In a random sample of fivefive microwave​ ovens, the mean repair cost was ​$75.0075.00 and the standard deviation was ​$13.0013.00. Assume the population is normally distributed and use a​ t-distribution to construct a 9090​% confidence interval for the population mean muμ. What is the margin of error of muμ​? Interpret the results.

Answer 1

Answer with explanation:

As per given , we have

Sample size : n= 5

Degree pf freedom = : df= 5-1=4

$\mu=\ 75.00$

$\sigma=\ 13.00$

Significance level for 90% confidence = $\alpha=1-0.90=0.1$

Using t-value table , t-critical value for 90% confidence:

$t_{\alpha/2, df}=t_{0.05, 4}=2.132.$

Margin of error of $\mu$: $E=t_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\(2.132)\dfrac{13}{\sqrt{5}}=12.3949720129\approx12.39$

Interpretation : The repair cost will be within $12.39 of the real population mean value $\mu$ 90% of the time.