Question

We are interested to estimate the proportion of the population who favor a candidate. Suppose that 210 of the people in a sample of 500 favored the candidate.

(a) What is the proportion estimate, p-hat, and the standard error?

(b) Find the 90% confidence interval for the proportion of the population who favor the candidate. Interpret result.

Answer 1

Solution :

Given that,

n = 500

x = 210

Point estimate = sample proportion = = x / n = 210/500=0.42

1 -   = 1- 0.42 =0.58

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 (((0.42*0.58) /500 )

E = 0.043

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.42-0.043 < p < 0.42+0.043

0.377< p < 0.463