A sociologist sampled 200 people who work in computer related jobs, and found that 42 of...

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

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Expert Solution

Solution :

Given that,

n = 200

x = 42

Point estimate = sample proportion = = x / n = 42/200=0.21

1 - = 1-0.21 =0.79

At 85% confidence level the z is ,

= 1 - 85% = 1 - 0.85 = 0.15

/ 2 = 0.075

Z/2 = Z0.075 = 1.44 ( Using z table )

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

= 1.44( $\sqrt$ ((0.21*0.79) /200 )

E = 0.0415

A 85% confidence interval for population proportion p is ,

- E < p < + E

0.21-0.0415< p <0.21+ 0.0415

0.1685< p < 0.2515

The 85% confidence interval for the population proportion p is :0.1685, 0.2515

orchestra answered 1 year ago

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