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)

Explanation:

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

Sample size : n= 300

Proportion of people proffered Candidate A :
`\hat{p}=(69)/(300)=0.23`

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

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

`=\sqrt{(0.23*0.77)/(300)}=0.02429677619\approx0.0242`

Margin of error :
`E=z_(\alpha/2)* S.E.`

`=z_(0.025)*0.0242=1.96*0.0242=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)`