Question

The firm selects a random sample of 1,000 personal computers sold this year, and the sample shows that 400 of them come from relatively unknown manufacturers. A 90% confidence interval for the real proportion is?

Answer 1

Solution :

Given that,

n = 1000

x = 400

Point estimate = sample proportion = $\hat p$ = x / n = 400/1000=0.4

1 - $\hat p$ = 1 -0.4=0.6

At 90% confidence level

$\alpha$ = 1 - 90%  

$\alpha$ = 1 - 0.90 =0.10

$\alpha$/2 = 0.05

Z$\alpha$/2 = Z_{0.05 = 1.645} ( Using z table )

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 1.645 ($\sqrt$((0.4*0.6) /1000 )

E = 0.0255

A 90% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.4 -0.0255 < p <0.4 + 0.0255

0.3745< p < 0.4255

The 90% confidence interval for the population proportion p is : 0.3745, 0.4255