Question

A Manufacturing Engineer wanted to study whether doing preventive maintenance affected the reliability of a machine.  The machine was observed from 2220 shifts with preventive maintenance and 2280 shifts without preventive maintenance and the number of shifts where the machine failed was recorded.  The following results data was obtained:

	# shifts failed	Total shifts
Preventive	95	2220
No preventive	139	2280

Test at a 5% significance level if there is a difference in doing preventive maintenance on this machine or not using a critical value approach.  Clearly state the hypothesis and conclusions. Let preventive maintenance be population 1

Answer 1

For sample 1, we have that the sample size is 2220, the number of favorable cases is X1 -95 0.0428 so then the sample proportion is p,- For sample 2, we have that the sample size is N2- 2280, the number of favorable cases is x2-139, so then the sample proportion is p,- =畾-0.061 The value of the pooled proportion is computed as2230-1250 0.052 Also, the given significance level is α 0.05 12220 2280 95+139 The following null and alternative hypotheses need to be tested This corresponds to a two-tailed test, for which a z-test for two population proportions needs to be conducted (2) Rejection Region Based on the information provided, the significance level is α 0.05, and the critical value for a two- tailed test is z 1.96 The rejection region for this two-tailed test is R {: > 1.96} The z-statistic is computed as follows P1 P2 0.0428 0.061 VP(1 p(1/nı +1/n2) /0.052 (1 - 0.052) (1/2220 +1/2280) Since it is observed that2.745 ze 1.96, it is then concluded that the null hypothesis is rejected Using the P-value approach: The p-value is p 0.0061, and since p 0.0061 < 0.05, it is concluded that the null hypothesis is rejected (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population proportion pi is different than p2, at the 0.05 significance level