A new catalyst is being investigated for use in the production of a plastic chemical. Ten batches of the chemical are produced. The mean yield of the 10 batches is 72.5% and the standard deviation is 5.8%. Assume the yields are independent and approximately normally distributed. Find a 99% confidence interval for the mean yield when the new catalyst is used.
Accepted Solution
A:
Answer:[tex]CI=(66.54,78.46)[/tex] Step-by-step explanation:Given : A new catalyst is being investigated for use in the production of a plastic chemical. Ten batches of the chemical are produced. The mean yield of the 10 batches is 72.5% and the standard deviation is 5.8%. Assume the yields are independent and approximately normally distributed.To find : A 99% confidence interval for the mean yield when the new catalyst is used ?Solution : Let X be the yield of the batches.We have given, n=10 , [tex]\bar{X}=72.5\%[/tex] , s=5.8%Since the size of the sample is small.We will use the student's t statistic to construct a 995 confidence interval.[tex]\bar X\pm t_{n-1,\frac{\alpha}{2}}\frac{s}{\sqrt n}[/tex]From the t-table with 9 degree of freedom for [tex]\frac{\alpha}{2}=0.005[/tex][tex]t_{n-1,\frac{\alpha}{2}}=t_{9,0.005}[/tex][tex]t_{n-1,\frac{\alpha}{2}}=3.250[/tex]The 99% confidence interval is given by,[tex]CI=72.5 \pm 3.25\frac{5.8}{\sqrt{10}}[/tex][tex]CI=72.5 \pm 5.96[/tex][tex]CI=(72.5+5.96),(72.5-5.96)[/tex][tex]CI=(66.54,78.46)[/tex]