Answer with explanation:
As per given , we have
Sample size : n= 5
Degree pf freedom = : df= 5-1=4
[tex]\mu=\$75.00[/tex]
[tex]\sigma=\$13.00[/tex]
Significance level for 90% confidence = [tex]\alpha=1-0.90=0.1[/tex]
Using t-value table , t-critical value for 90% confidence:
[tex]t_{\alpha/2, df}=t_{0.05, 4}=2.132.[/tex]
Margin of error of [tex]\mu[/tex]: [tex]E=t_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\(2.132)\dfrac{13}{\sqrt{5}}=12.3949720129\approx12.39[/tex]
Interpretation : The repair cost will be within $12.39 of the real population mean value [tex]\mu[/tex] 90% of the time.
