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

`CI=(66.54,78.46)`

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,(\alpha)/(2)}(s)/(\sqrt n)`

From the t-table with 9 degree of freedom for
`(\alpha)/(2)=0.005`

`t_{n-1,(\alpha)/(2)}=t_(9,0.005)`

`t_{n-1,(\alpha)/(2)}=3.250`

The 99% confidence interval is given by,

`CI=72.5 \pm 3.25(5.8)/(√(10))`

`CI=72.5 \pm 5.96`

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

`CI=(66.54,78.46)`