Question

            Find a 90% confidence interval for the proportion p if the sample has = 0.23 with n=300 and standard error SE=0.024.

Answer 1

Given that,

n = 300

SE=0.024

Point estimate = sample proportion = =0.23

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E Z/2 *SE

= 1.645 *0.024

E = 0.0395

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.23-0.0395 < p <0.23+ 0.0395

0.1905< p < 0.2695

The 90% confidence interval for the population proportion p is : 0.1905, 0.2695