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 in the Answer box.

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

n = number of employees selected randomly = 200

x = number of employees logged onto Facebook that day = 76

i) Sample proportion:

Sample proportion = = 0.38

ii)

Confidence level = c = 0.95

z critical value for (1+c)/2 = (1+0.95)/2 = 0.975 is

zc = 1.96 (From statistical table of z values)

Critical value z = 1.96

iii)

Standard error:

SE = 0.0343 (Round to 4 decimal)

Standard error = 0.0343

iv)

Margin of error (e) :

e = 1.96 * 0.0343

e = 0.0672 (Round to 4 decimal)

Margin of error = 0.0672

v)

Lower limit = - e

= 0.38 - 0.0672

= 0.3128 (Round to 4 decimal)

Lower limit = 0.3128

vi)

Upper limit = + e

= 0.38 + 0.0672

= 0.4472 (Round to 4 decimal)

Upper limit = 0.4472