Question

suppose we take a poll (random sample) of 3604 students classified as juniors and find that 3059 of thembelieve that they will find a job immediately after graduation. what is the 99 % confidence interval for the proportion of gsu juniors who believe that they will,immediately, be employed after graduation.

Answer 1

The 99% confidence interval for the population proportion p is 0.833< p < 0.863.

Given that,

x = 3059

n = 3604

Point estimate = sample proportion = x / n =3059/3604=0.848

$1 - \hat p$  = $1-0.848 =0.152$

At 99% confidence level the z is ,

$\alpha$ = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$/ 2 = 0.01 / 2 = 0.005

$Z\alpha$/2 = $Z*0.005$ = 2.576  (  Using z table   )

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

                              = $2.576* (\sqrt((0.848*0.152) /3604 )$

                               =0.015

A 99% confidence interval for population proportion p is ,

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

0.848-0.015< p <0.848+ 0.015

0.833< p < 0.863

Hence, the 99% confidence interval for the population proportion p is 0.833< p < 0.863.

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