Question

A steel factory produces iron rods that are supposed to be 36 inches long. The machine that makes
these rods does not produce each rod exactly 36 inches long. The lengths of these rods vary slightly.
It is known that when the machine is working properly, the mean length of the rods is 36 inches.
According to design, the standard deviation of the lengths of all rods produced on this machine is
always equal to .05 inches. The quality control department at the factory takes a sample of 40 such
rods each week, calculates the mean length of these rods, and tests the null hypothesis µ = 36 inches
against the alternative hypothesis µ 6= 36 inches using 1% significance level. If the null hypothesis
is rejected, the machine is stopped and adjusted. A recent sample of 40 such rods produced a mean
length of 36.015 inches. Answer the following questions.
(a) Write down the null and alternative hypotheses. Is the alternative one-sided or two-sided?
(b) State the Type I and Type II errors for this problem and tell why Type I error is more serious.
(c) State the Z-test statistic for this problem and tell its distribution.
(d) Is the small, large or both small and large value of the Z-test statistic that will lead us to reject
the null hypothesis?
(e) State the wanted significance level, determine the critical value for the rejection region and
write down the decision rule.
(f) Calculate the Z-test statistic from the available data.
(g) Determine whether or not the null hypothesis is rejected at the significance level wanted and
tell why.
(h) Calculate the P-value and do the hypothesis testing based on the P-value.
(i) Express the conclusion in the context of the problem, using common English

Answer 1

The provided sample mean is X = 36.015 and the known population standard deviation is σ = 0.05, and the sample size is n = 40 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: 36 Ha: 36 This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used. Based on the information provided, the significance level is a 0.01, and the critical value for a two- tailed test is z 2.58. The rejection region for this two-tailed test is R = {z : |z| > 2.58) (3) Test Statistics The z-statistic is computed as follows X36.015-36 1.897 (4) Decision about the null hypothesis Since it is observed that lz-1.897 < z.-2.58, it is then concluded that the null hypothesis is not rejected Using the P-value approach. The p-value is p 0.0578, and since p 0.0578 > 0.01, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is different than 36, at the 0.01 significance level.