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.
Part (a)
Is the sample data consistent with the cahiers’ claim? Please answer “yes” or “no” and provide a brief explanation.
Part (b)
Using a significance level of 0.05, can we reject the cashiers’ claim? Please make sure to show your work!
Part (c)
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.
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.
Part (a)
Yes, the sample data is consistent with the cashiers' claim. Out of the 135 cashiers included in the study, 102 of them experienced back pain due to the working conditions. This means that approximately 75.56% of the cashiers in the sample reported back pain, which is greater than the 80% claimed by the cashiers.
Part (b)
To determine if we can reject the cashiers' claim, we need to perform a hypothesis test. Our null hypothesis, denoted as H0, would be that the proportion of cashiers experiencing back pain is equal to or less than 80%, while the alternative hypothesis, denoted as Ha, would be that the proportion is greater than 80%.
The significance level (α) is given as 0.05, which means we are willing to accept a 5% chance of making a Type I error (rejecting the null hypothesis when it is true).
To conduct the hypothesis test, we can use a one-sample proportion test.
Let p̂ be the sample proportion of cashiers experiencing back pain, which is calculated as p̂ = 102/135 ≈ 0.756.
The test statistic can be calculated as:
z = (p̂ - p) / √(p * (1-p) / n),
where p is the proportion claimed by the cashiers (0.80), and n is the sample size (135).
Plugging in the values, we get:
z = (0.756 - 0.80) / √(0.80 * (1-0.80) / 135)
z ≈ -1.62
Using a standard normal distribution table or a statistical software, we can find the critical z-value for a one-sided test at a significance level (α) of 0.05. For α = 0.05, the critical z-value is approximately 1.645.
Since -1.62 is less than -1.645, we fail to reject the null hypothesis. Therefore, we do not have sufficient evidence to conclude that the proportion of cashiers experiencing back pain is greater than 80%.
Part (c)
The seeming contradiction between parts (a) and (b) can be resolved by understanding the concept of hypothesis testing and the role of statistical significance.
In part (a), we observed that 75.56% of the sample cashiers experienced back pain, which is greater than the 80% claimed by the cashiers. This supports their claim based on the sample data.
However, in part (b), we failed to reject the null hypothesis, which states that the proportion of cashiers experiencing back pain is equal to or less than 80%. This means that we do not have strong enough evidence to conclude that the cashiers' claim is true for the entire population of cashiers.
In statistical hypothesis testing, we make inferences about a population based on sample data. The conclusion is drawn based on the statistical evidence and the predetermined significance level. In this case, our sample data provided some evidence supporting the claim, but it was not strong enough to reach a statistically significant conclusion.
Therefore, the seeming contradiction can be resolved by understanding that the sample data suggests a certain pattern, but statistical hypothesis testing considers the overall population and requires stronger evidence to reject the null hypothesis.