Question

a simple random sample of 225 college students was taken in order to estimate the proportion of college students that agrees with the "no teacher left alone act". of those surveyed, 142 agreed with the law. construct the 95% confidence interval estimate of the proportion of all college students that agrees with this law

Answer 1

Solution :

Given that,

n = 225

x = 142

= x / n = 142 / 225 = 0.631

1 - = 1 - 0.631 = 0.369

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

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

= 1.96 * (((0.631 * 0.369) / 225)

= 0.063

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.631 - 0.063 < p < 0.631 + 0.063

0.568 < p < 0.694

The 95% confidence interval for the population proportion p is : (0.568 , 0.694)