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.

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]