Question

A sociologist sampled 200 people who work in computer related jobs, and found that 42 of them have changed jobs in the past 6 months. Construct a 85% confidence interval for the proportion of those who work in computer-related jobs who have changed jobs in the past six months.

Answer 1

Solution :

Given that,

n = 200

x = 42

Point estimate = sample proportion = $\hat p$ = x / n = 42 / 200 = 0.210

1 - $\hat p$ = 1 -0.210 = 0.790

At 95% confidence level

$\alpha$ = 1-0.85% =1-0.85 =0.15

$\alpha$/2 =0.15/ 2= 0.075

Z$\alpha$/2 = Z0.075 = 1.44

Z$\alpha$/2 = 1.44

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

= 1.44 * ($\sqrt$(0.210*(0.790) / 200)

= 0.041

A 85% confidence interval for population proportion p is ,

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

0.210 -0.041 < p < 0.210 +0.041

0.169 < p < 0.251  

(0.169 , 0.251 )

A 85% confidence interval for population proportion p is=(0.169 , 0.251 )