Question
A production supervisor at major chemical company wishes to determine whether a new catalyst, catalyst XA-100, increases the mean hourly yield chemical process beyond the current mean hourly yield, which is known to be roughly equal to, but no m ore 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 yield 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 trail runs for which s = 19.97 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 confidant that pi, the sample mean hourly yield, is within a margin of error of 8 pounds of the population mean hourly yield mu when catalyst XA-100 is used. n ______ trial runs (b) 99 percent confident that x is within a margin of error of 5 pounds of mu. n ______ trial runs
Explanation / Answer
a) For 95% Interval
n = (1.960*19.97/8)^2= 24 (nearest integer)
So, answer is 24
B) For 99% Interval
n = (2.5758*19.97/5)^2= 106 (nearest integer)
So, answer is 106
