Question

The CEO of a company wants to estimate the percent of employees that use company computers to go on Facebook during work hours. He selects a random sample of 200 of the employees and finds that 76 of them logged onto Facebook that day. Construct a 95% confidence interval for the population proportion.

Answer the following questions

a. i) Sample proportion = ?    a. ii) critical value Z = ? a. iii) Standard error = ?    a. iv) Margin of error = ?    a. v) Lower limit = ?    a. vi) upper limit = ?

Answer 1

Solution :

Given that,

n = 200

x = 76

i) Point estimate = sample proportion = = x / n = 76 / 200 = 0.38

1 - = 1 - 0.38 = 0.62

ii) At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025  = 1.96

iii) =  [p ( 1 - p ) / n] =   [(0.38 * 0.62) / 200 ] = 0.0343

iv) Margin of error = E = Z / 2 *

= 1.96 * 0.0343

= 0.067

A 95% confidence interval for population proportion p is ,

± E

= 0.38 ± 0.067

= ( 0.313, 0.447 )

v) Lower limit = 0.313

vi) Upper limit = 0.447