supervisor records the repair cost for 25 randomly selected dryers. A sample mean of $93.36 and standard deviation of $19.95 are subsequently computed. Determine the 98% confidence interval for the mean repair cost for the dryers. Assume the population is approximately normal. Find the critical value that should be used in constructing the confidence interval.

Answer with explanation:The confidence interval for population mean (when population standard deviation is unknown) is given by :-[tex]\overline{x}-t^*\dfrac{s}{\sqrt{n}}< \mu<\overline{x}+z^*\dfrac{s}{\sqrt{n}}[/tex], where n= sample size[tex]\overline{x}[/tex] = Sample mean s= sample sizet* = Critical value.Given : n= 25Degree of freedom : [tex]df=n-1=24[/tex][tex]\overline{x}= \$93.36[/tex] [tex]s=\ $19.95[/tex]Significance level for 98% confidence interval : [tex]\alpha=1-0.98=0.02[/tex]Using t-distribution table ,Two-tailed critical value for 98% confidence interval : [tex]t^*=t_{\alpha/2,\ df}=t_{0.01,\ 24}=2.4922[/tex]⇒ The critical value that should be used in constructing the confidence interval = 2.4922Then, the 95% confidence interval would be :-[tex]93.36-(2.4922)\dfrac{19.95}{\sqrt{25}}< \mu<93.36+(2.4922)\dfrac{19.95}{\sqrt{25}}[/tex][tex]=93.36-9.943878< \mu<93.36+9.943878[/tex][tex]=93.36-9.943878< \mu<93.36+9.943878[/tex][tex]=83.416122< \mu<103.303878\approx83.4161<\mu<103.3039[/tex]Hence, the 98% confidence interval for the mean repair cost for the dryers. = [tex]83.4161<\mu<103.3039[/tex]