According to the National Institute for Occupational Safety and Health, job stress poses a major threat to the health of workers. A news report claims that 75% of restaurant employees feel that work stress has a negative impact on their personal lives. Managers of a large restaurant chain wonder whether this claim is valid for their employees. A random sample of 100 employees finds that 68 answer “Yes” when asked, “Does work stress have a negative impact on your personal life?”

1. Do these data provide convincing evidence at the α = 0.10 significance level that the proportion of all employees in this chain who would say “Yes” differs from 0.75? 2. A 90% confidence interval for the restaurant worker data was also created and found to be (0.603272, 0.756728). Explain how the confidence interval is consistent with, but gives more information than, the test.

Question

jarrod.schamberger · Accepted Answer

Hypothesis test on the proportion of employees, gives an evidence based conclusion which is consistent with the confidence interval. Response:There is insufficient or no convincing statistical evidence to conclude that the proportion differs from 0.75The confidence interval indicates that the proportion of workers that feel that work stress has a negative impact on their lives is between 0.603272 and 0.756728, a proportion that includes 0.75, which is consistent with the test, and also provides the minimum and maximum expected proportion at 95% confidence level.How can the test of the hypothesis be performed?1. The percentage of workers that the news report claims work stress have a negative impact on their personal lives, p = 75%Number of employees in the sample, n = 100Number that answered yes = 68The null hypothesis is, H: p = 0.75The alternative hypothesis is, H: p 0.75The proportion of the employees that said yes, [tex]\hat p[/tex] = 68 100 = 0.68The z-test formula is; [tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p \cdot q}{n}}}[/tex]q = 1 - pWhich gives;[tex]z= \mathbf{\dfrac{0.68-0.75}{\sqrt{\dfrac{0.75 	imes (1 - 0.75)}{100}}}} \approx -1.62[/tex]The p-value 2 p(Z &lt; -1.62) = 2 0.0526 = 0.1052The significance level, = 0.10Given that the p-value is larger than the significance level, we fail to reject the null hypothesis, and conclude that as follows;There is insufficient convincing statistical evidence that the proportion differs from 0.75.2. The 90% confidence interval is (0.603272, 0.756728)The confidence interval indicates that at 90% confidence level, the true proportion of employees that feel that work stress has a negative impact on their personal lives is between 0.603272 and 0.756728.The report claim of 75% or 0.75 is within the confidence interval, which consistent with the hypothesis testThe information in the confidence interval also indicates that at 90% confidence level, the proportion of the employee that say yes is between (0.603272, 0.756728), thereby giving more informationLearn more about hypothesis testing here:https://brainly.com/question/23082833

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