# Question: Referring to Exercise 8 21 page 307 regard the sample of

Referring to Exercise 8.21 (page 307), regard the sample of five trial runs (which has standard deviation 19.65) as a preliminary sample. Determine the number of trial runs of the chemical process needed to make us:

In exercise 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. The resulting yields for the trial runs (in pounds per hour) are 801, 814, 784, 836, and 820. 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.

a. 95 percent confident that the sample mean hourly yield, is within a margin of error of eight pounds of the population mean hourly yield m when catalyst XA-100 is used.

b. 99 percent confident that is within a margin of error of five pounds of μ.

In exercise 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. The resulting yields for the trial runs (in pounds per hour) are 801, 814, 784, 836, and 820. 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.

a. 95 percent confident that the sample mean hourly yield, is within a margin of error of eight pounds of the population mean hourly yield m when catalyst XA-100 is used.

b. 99 percent confident that is within a margin of error of five pounds of μ.

**View Solution:**## Answer to relevant Questions

Referring to Exercise 8.20, regard the sample of 10 sales figures (which has standard deviation 32.866) as a preliminary sample. How large a sample of sales figures is needed to make us 95 percent confident that x-bar, the ...In each of the following cases, compute 95 percent, 98 percent, and 99 percent confidence intervals for the population proportion p. a. p̂ = .4 and n = 100 b. p̂ = .1 and n = 300 c. p̂ = .9 and n = 100 d. p̂ = .6 and ...Consumer Reports (January 2005) indicates that profit margins on extended warranties are much greater than on the purchase of most products. In this exercise we consider a major electronics retailer that wishes to increase ...In an article in Accounting and Business Research, Beattie and Jones investigate the use and abuse of graphic presentations in the annual reports of United Kingdom firms. The authors found that 65 percent of the sampled ...Define each of the following: Type I error, α, Type II error, β.Post your question