Question

Out of 300 people sampled, 69 preferred Candidate A. Based on this, find a 95% confidence level for the true proportion of the voting population ( p ) prefers Candidate A. Give your answers as decimals, to three places.

Answer 1

Answer: (0.183,0.277)

Step-by-step explanation:

Given : Out of 300 people sampled, 69 preferred Candidate A.

Sample size : n= 300

Proportion of  people proffered Candidate A : $\hat{p}=\dfrac{69}{300}=0.23$

Significance level : $\alpha =1-0.95=0.05$

Standard error : $S.E.=\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$=\sqrt{\dfrac{0.23\times0.77}{300}}=0.02429677619\approx0.0242$

Margin of error : $E=z_{\alpha/2}\times S.E.$

$=z_{0.025}\times0.0242=1.96\times0.0242=0.047432$

The confidence interval for the population proportion is given by :-

$\hat{p}\pm E$

$=0.23\pm0.047432=(0.23-0.047432,0.23+0.047432)=(0.182568,0.277432)\approx(0.183,0.277)$