Question

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.

Answer 1

Answer:

$CI=(66.54,78.46)$                                  

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 , $\bar{X}=72.5\%$ , 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.

$\bar X\pm t_{n-1,\frac{\alpha}{2}}\frac{s}{\sqrt n}$

From the t-table with 9 degree of freedom for $\frac{\alpha}{2}=0.005$

$t_{n-1,\frac{\alpha}{2}}=t_{9,0.005}$

$t_{n-1,\frac{\alpha}{2}}=3.250$

The 99% confidence interval is given by,

$CI=72.5 \pm 3.25\frac{5.8}{\sqrt{10}}$

$CI=72.5 \pm 5.96$

$CI=(72.5+5.96),(72.5-5.96)$

$CI=(66.54,78.46)$