Question

(7 points) Suppose a recent random sample of employees nationwide that have a 401(k) retirement plan found that 20% of them had borrowed against it in the last year. A local company is interested in assessing whether this proportion adequately describes the borrowing behavior of their employees. They take a random sample of 150 employees and find that 17 had borrowed from their plan. Conduct a hypothesis test to investigate whether the borrowing rate at this company is less than the nationally reported figure of 20%. Round all numeric results to 4 decimal places. 1. Write the hypotheses to test if the proportion of employees borrowing from their plan is less than the percentage reported nationwide. Ho: The proportion of employees who have borrowed from their plan ? the 20% reported nationwide. The difference chance. ? due to ? the 20% reported nationwide. The difference ? due to HA: The proportion of employees who have borrowed from their plan chance. 2. Calculate the proportion of employees in the sample who borrowed from their retirement plan. p = 3.Describe a setup for a simulation that would be appropriate in this situation and how the p-value can be calculated using the simulation results. To setup a simulation for this situation, we let each employee be represented with a card. We take 100 cards, black cards represent employees who have borrowed from their plan and red cards represent employees who have not. Shuffle the cards and draw with replacement cards representing the random sample of customers. Calculate the proportion of ? cards in the sample and call it Psim. Repeat 3,000 times and plot the resulting sample proportions. The p-value will be the proportion of simulations where psim is ? 4. Use the One Proportion Resampling Test app to conduct the simulation. Draw at least 3000 samples, then report your p-value from the app. Use this external link to the One Proportion Resampling Test if the app does not load properly on your computer. (Right click to open in a new tab or window.)

Answer 1

solution . com 1) Here we find the null and alternative hypotheses test. Ho : The peo portion of employes who have borrowed from their plan is same as the 20% reported nationwide. "The difference is due to chance. HA : The proportion of employes who have borrowed from their plan is less than the 20% reported nationwide. The difference is not due to chance. Ho ;P >0.20 Ha: P20.20 From the given data n': 150 's 17 2) Now we find Þ value ? PE $ = 0.1 33 Here d = 0.05 Z-critical value = -1.64 3) We find test sta stistics:- Po (1-Po/n

where Po = 0.20 0.1133-0.20 Jo.20 (1-0.20)/150 ; -0.0867 10.16/150 -0.0867 10.00106 minutes -0.0867 0.03256 Z -[-2.6628] 4) P-value = 0.00391 P-value Pavalue La Z=-2.6628 2 Zc =-1.64. So, we reject null hypothesis.