Question

A survey of mining companies is to be conducted to estimate p, the proportion of companies that anticipate hiring engineers during the coming year. (a) How large a sample is required to estimate p to within 0.04 with 95% confidence? (b) A sample of 500 yields 105 companies that plan to hire such engineers. Give the 95% confidence interval for p.

Answer 1

Solution:

a) Given that:

E= 0.04

= 0.5

1 - = 1 - 0.5 = 0.5

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95= 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

sample size = ( Z/2 / E)2 * * (1 - )     

   = ( 1.96/ 0.04 )2 *0.5 *0.5

                   = 600.25

sample size = 601

b)

Given that,

n = 500

x = 105

= x / n = 0.21

1 - = 1 - 0.21 = 0.79

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

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

= 1.96 * (((0.21 * 0.79) / 500)

= 0.036

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.21 - 0.036 < p < 0.21 + 0.036

0.174 < p < 0.246

(0.174,0.246)

The 95% confidence interval is 0.174 to 0.246