A production supervisor at a major chemical company wishes to determine whether a new catalyst, catalyst XA-100, increases themean hourly yield of a chemical process beyond the current mean hourly yield, which is known to be roughly equal to, but no morethan, 750 pounds per hour. To test the new catalyst, five trial runs using catalyst XA-100 are made. Assuming that all factors affectingyields of the process have been held as constant as possible during the test runs, it is reasonable to regard the five yields obtainedusing 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= 19.71 as a preliminary sample. Determine the number of trial runs of the chemicalprocess needed to make us: (Round up your answers to the next whole number.)(a) 95 percent confident that ī, the sample mean hourly yield, is within a margin of error of 8 pounds of the population mean hourlyyield u when catalyst XA-100 is used.ntrial runs(b) 99 percent confident that a is within a margin of error of 5 pounds of pu.ntrial runs

(a) The sample size is 5, and sample standard deviation is 19.71, margin of error is 8, and confidence level is 95%. The degrees of freedom is, df=n-1=5-1=4 The degrees of freedom is 4. Computation of critical value: The critical value of t-distribution for 0.05 level of significance at 4 degrees of freedom can be obtained using the excel formula “=T.INV.2T(0.05,4)”. The critical value is 2.7764. The number of trial runs of the chemical process needed is, n=tα2sE2=2.7764×19.7182=6.84035552=46.79≈47 Thus, the number of trial runs of the chemical process needed is 47.(b) Computation of critical value: The critical value of t-distribution for 0.01 level of significance at 4 degrees of freedom can be obtained using the excel formula “=T.INV.2T(0.01,4)”. The critical value is 4.6041. The number of trial runs of the chemical process needed is, n=tα2sE2=4.6041×19.7182=11.3433513752=128.6716≈129 Thus, the number of trial runs of the chemical process needed is 129.

Math

Statistics

(a) 95 percent confident that , the sample mean hourly yield, is within a margin of error of 8 pounds of the population mean hourly yield u when catalyst XA-100 is used. trial runs (b) 99 percent confident that is within a margin of error of 5 pounds of u. n trial runs

(a) 95 percent confident that , the sample mean hourly yield, is within a margin of error of 8 pounds of the population mean hourly yield u when catalyst XA-100 is used. trial runs (b) 99 percent confident that is within a margin of error of 5 pounds of u. n trial runs

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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