A number of cashiers at a chain of grocery stores complained about the working conditions (shift length, design of cash registers, etc.), claiming that they resulted in back pain in at least 80 percent of cashiers. The employer, trying to show that that claim is not true, carries out a study involving 135 cashiers. It showed that 102 of them did indeed experience back pain that was due to the working conditions.
Is the sample data consistent with the cahiers’ claim? Please answer “yes” or “no” and provide a brief explanation.
Using a significance level of 0.05, can we reject the cashiers’ claim? Please make sure to show your work!
Look at the answers to Parts (a) and (b). At first glance, they may seem contradictory. Explain how that seeming contradiction can be resolved in the context of hypothesis testing.
Statistics - MathGuru, Friday, May 25, 2012 at 6:28pm
Here are a few hints:
Ho: p ≥ .80 -->null hypothesis
Ha: p < .80 -->alternate hypothesis
You can try a proportional one-sample z-test for this one since this problem is using proportions.
Using a formula for a proportional one-sample z-test with your data included, we have:
z = .76 - .80 -->test value (102/135 is approximately .76) minus population value (.80) divided by
√[(.80)(.20)/135] --> .20 represents 1-.80 and 135 is sample size.
Finish the calculation. Remember if the null is rejected, then p < .80 . If the null is not rejected, then p could very well be greater than or equal to .80 in the population of interest.
I hope this will help get you started.