Question

The employee benefits manager of a medium size business would like to estimate the proportion of full-time employees who prefer adopting plan A of three available health care plans in the coming annual enrollment period. A reliable frame of the company’s employees and their tentative health care preferences are available. Using Excel, the manager chose a random sample of size 50 from the frame. There were 17 employees in the sample who preferred plan A. Construct a 99% confidence interval for the proportion of company employees who prefer plan A. Assume that the population consists of the preferences of all employees in the frame. The lower boundary is _______________ and the upper bound is___________________ . (Please keep three decimal points.)

Answer 1

Here, n=sample size=50,

x=Number of employees in the sample who preferred plan A=17,

p=Sample proportion of employees who preferred plan A=x/n=17/50= 0.34,

Level of significance==0.01

In general.100*(1-)% confidence interval for population proportion is ,

[p-Z_/2*,p+Z_/2*]

Here,Z_/2=Z_0.01/2=Z_0.005=2.575829............From standrad normal distibution ytable

Here, Lower  boundary= p-Z_/2*

=0.34-2.575829*

= 0.1674387

&

upper boundary= p+Z_/2*

=0.34+2.575829*

=0.5125613

Thus, a 99% confidence interval for the proportion of company employees who prefer plan A is

[0.1674387, 0.5125613].