Question

The employee benefits manager of a large public university would like to estimate the proportion of full-time employees who prefer adopting the first (plan A) of three available health care plans in the next annual enrollment period. A random sample of the university’s employees and their tentative health care preferences are given in the file Healthcare.xlsx below. Calculate a 88% confidence interval for the proportion of all the university’s employees who favor plan A. What are the values of lower limit and upper limit? Round your answer to 3 decimal places. Healthcare.xlsxPreview the document Group of answer choices (0.217, 0.330) (0.211, 0.323) (0.242, 0.358) (0.180, 0.287)

Answer 1

Answer:

The 88% confidence interval for the population proportion of full-time employees who favor plan A is (0.208, 0.344).

Step-by-step explanation:

The question is incomplete: it lacks the sample data.

We will work with a sample size n=105 and a count of X=29 that prefer adopting the plan A.

We have to calculate a 88% confidence interval for the proportion.

The sample proportion is p=0.276.

$p=X/n=29/105=0.276$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.276*0.724}{105}}\\\\\\ \sigma_p=\sqrt{0.001903}=0.0436$

The critical z-value for a 88% confidence interval is z=1.555.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=1.555 \cdot 0.0436=0.0678$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.276-0.0678=0.208\\\\UL=p+z \cdot \sigma_p = 0.276+0.0678=0.344$

The 88% confidence interval for the population proportion is (0.208, 0.344).