The concentration of an active ingredient in the output of chemical reaction is

Question

The concentration of an active ingredient in the output of chemical reaction is strongly influenced by the catalyst that is used in the reaction. It is felt that when catalyst A is used, the population mean concentration exceeds 65%. The standard deviation is known be, sigma = 5%. (a) To check the average value of reaction, a test was carried with a sample of 30 independent experiments. The average concentration was found to be, X_A = 64.5%. Can we still believe that the sample average is 65%? (use alpha = 0.05) (b) Suppose a similar experiment is done with the use of another catalyst B. The standard deviation s is still assumed to be 5% and X_a turns out to be 70%. Is we truly greater than mu_A?

Explanation / Answer

4. a. The hypotheses are as follows:

H0: mu=65 (mean concentration of active ingredient is 65% when catalyst A is used)

H1:mu>65 (mean concentration of active ingredient exceeds 65% when catalyst A is used)

For a known population standard deviation, use 1-sample Z test.

Test statistic:

Z=(xbar-mu)/(sigma/sqrt n), where, xbar is sample mean, mu is population mean, sigma is population standard deviation, and n is sample size.

=(64.5-65)/(5/sqrt 30)

=-0.55

p value is: 0.29116.

Conclusion: Per rule, reject H0, if p value is less than 0.05. Here, p value is not less than 0.05, therefore, fail to reject H0. There is insufficient sample evidence to warrant the rejection of null hypothesis.

b. State the hypotheses:

H0:muA-muB=0 (there is no difference in mean concentration of active ingredient while using catalyst A and B)

H1:muA-muB<0 (the mean concentration of active ingredient is higher for catalyst B than catalyst A)

Compute the test statistic.

Z=(xAbar-xBbar)-(muA-muB)/sqrt[sigmaA^2/nA-sigma^2B/nB], where, xbar is sample mean, mu is population mean, sigma is population standard deviation, and n is sample size, A and B denote catalyst A and B respectively.

=(64.5-70)-0/sqrt[5^2/30-5^2/30]

=0

p value is: 0.5.

Conclusion: Per rule, reject H0, if p value is less than 0.05. Here, p value is not less than 0.05, therefore, fail to reject H0, insufficient sample evidence to warrant the rejection of null hypothesis.

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