Question

A random sample of 100 employees at a company shows that 62 are female. Form a 90% confidence interval for the proportion of all employees who are female. The interval is 0.62  what margin of error? Round your answer to 3 decimal places.

Answer 1

Solution :

Given that,

n = 100

x = 62

Point estimate = sample proportion = = x / n = 62/100=0.62

1 -   = 1- 0.62 =0.38

At 90% confidence level the z is

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E= Z/2 * (( * (1 - )) / n)

= 1.645 *((0.62*0.38) /100 )

E = 0.0798

Margin of error = E = 0.0798