Question

High Tech, Inc randomly tests its employees about company policies. Last year in the 400 random tests conducted, 14 employees failed the test.A. Develop a 99% confidence interval for the proportion of applicants that fail the test. (round answers to 3 decimal places) Confidence interval for the proportion mean is between _____ and ____B. Would it be reasonable to conclude 5% of the employees cannot pass the employee test? Yes or No

Answer 1

Answer:

Confidence interval for the proportion mean is between 0.0113 and 0.0587. B. Yes, it is reasonable to conclude that 5% of the employees cannot pass the employee test.

Step-by-step explanation:

We have a large sample size of n = 400 random tests conducted. Let p be the true proportion of employees who failed the test. A point estimate of p is $\hat{p} = 14/400 = 0.035$, we can estimate the standard deviation of $\hat{p}$ as $\sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{0.035(1-0.035)/400}=0.0092$. A $100(1-\alpha)%$ confidence interval is given by $\hat{p}\pm z_{\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n$, then, a 99% confidence interval is $0.035\pm z_{0.005}0.0092$, i.e., $0.035\pm (2.5758)(0.0092)$, i.e., (0.0113, 0.0587). $z_{0.005} = 2.5758$ is the value that satisfies that there is an area of 0.005 above this and under the standard normal curve. B. Yes, it is reasonable to conclude that 5% of the employees cannot pass the employee test, because this inverval contain 0.05.