Question

Suppose a recent random sample of employees nationwide that have a 401(k) retirement plan found that 22% of them had borrowed against it in the last year. A random sample of 130 employees from a local company who have a 401(k) retirement plan found that 16 had borrowed from their plan. Based on the sample results, is it possible to conclude, using αα = .05, that the local company had a lower proportion of borrowers from its 401(k) retirement plan than the 22% reported nationwide? Round all numeric results to 3 decimal places.

1. Write the hypotheses to test if the proportion of employees borrowing from their plan is less than the percentage reported nationwide.
H0: The proportion of employees who have borrowed from their plan is （less than /more than / the same as) the 22% reported nationwide. The difference (is/ is not ) due to chance.

HA: The proportion of employees who have borrowed from their plan  ( is less than / more than / the same as )the 22% reported nationwide. The difference ( is/ is not ) due to chance.

2. Calculate the proportion of employees in the sample who borrowed from their retirement plan.
p̂ =

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  ( red/ black )cards in the sample and call it p̂sim. Repeat 3,000 times and plot the resulting sample proportions. The p-value will be the proportion of simulations where p̂simp^sim is  (greater than/ less than/ further from 0.22 than) ( ) .

Answer 1

Solution:-

1)

H0: The proportion of employees who have borrowed from their plan is same as the 22% reported nationwide. The difference is due to chance.

HA: The proportion of employees who have borrowed from their plan is less than the 22% reported nationwide. The difference is not due to chance.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P > 0.22
Alternative hypothesis: P < 0.22

Note that these hypotheses constitute a one-tailed test.  

2)

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation (S.D) and compute the z-score test statistic (z).

S.D = sqrt[ P * ( 1 - P ) / n ]

S.D = 0.03633
z = (p - P) / S.D

z = - 2.67

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is less than -2.67.

Thus, the P-value = 0.004

Interpret results. Since the P-value (0.004) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the local company had a lower proportion of borrowers from its 401(k) retirement plan than the 22% reported nationwide.