Question

A production supervisor at a major chemical company wishes to determine whether a new catalyst, catalyst XA-100, increases the mean hourly yield of a chemical process beyond the current mean hourly yield, which is known to be roughly equal to, but no more than, 750 pounds per hour.
To test the new catalyst, five trial runs using catalyst XA-100 are made.
Assuming that all factors affecting yields of the process have been held as constant as possible during the test runs, it is reasonable to regard the five yields obtained using the new catalyst as a random sample from the population of all possible yields that would be obtained by using the new catalyst.
Furthermore, we will assume that this population is approximately normally distributed. Regard the sample of 5 trial runs for which s = 19.62 as a preliminary sample.

Determine the number of trial runs of the chemical process needed to make us: (Round up your answers to the next whole number.)

(a) 95 percent confident that 1formula102.mml, the sample mean hourly yield, is within a margin of error of 8 pounds of the population mean hourly yield µ when catalyst XA-100 is used.

(b) 99 percent confident that 1formula102.mml is within a margin of error of 5 pounds of µ.

Answer 1

Answer:

a. n= 47

b. n= 128

Step-by-step explanation:

Hello!

The objective of this experiment is to test if the new catalyst, XA-100, increases the mean hourly yield of a chemical process, that is known to be μ=750 (pounds per hour) with the current process.

You need to calculate the sample size to estimate the population with determined error margins.

To do so, since you have no population information, only that it is approximately normal distributed, you'll use the Student t statistic to get the sample size.

The formula of the margin of error (d) is:

d= $t_{n-1: 1-\alpha/2}$ * ($\frac{S}{\sqrt{n} }$)

I've  cleared the sample size of the formula

$n= (S*\frac{t_{n-1; 1-\alpha /2} }{d} )^{2}$

You need a sample size for the t-Student value and a standard deviation, that's why the information of a pilot study with n=5 and S= 19.62 is given.

a)

95% CI

d= 8 pounds

$t_{n-1; 1-\alpha/2 } = t_{5-1;1-0.025}  = t_{4;0.975} =2.776$

$n= (19.62*\frac{2.776 }{8} )^{2}$

n= 46.35 ≅ 47

b)

99% CI

d= 5 pounds

$t_{n-1; 1-\alpha/2 } = t_{5-1;1-0.005}  = t_{4;0.995} =4.604$

$n= (19.62*\frac{4.604}{8} )^{2}$

n= 127.49 ≅ 128

I hope it helps!