Question

Suppose 300 randomly selected people are surveyed to determine whether or not they rent their home. Of the 300 surveyed, 60 reported renting their home. Find the confidence interval for the population proportion with a 99% confidence level.

Answer 1

.141, .259

p′=60300=0.2.

The standard error of the sampling distribution of the sample proportion is

σp′=p^(1−p′)n−−−−−−−−√=0.2(1−0.2)300−−−−−−−−−−√≈0.023.

With zα2=2.576 for a confidence level of 99%, the margin of error is thus

marginof error=(zα2)σp′≈(2.576)(0.023)≈0.059.

With point estimate p′=0.2=0.200 and a margin of error of 0.059, the confidence interval is

(0.200−0.059,0.200+0.059)(0.141,0.259).

We estimate with 99% confidence that the true population proportion of people who rent their home is between 0.141 and 0.259.

Answer 2

Answer: (0.143, 0.257)

Explanation:

Given : Level of significance :
`1-\alpha:0.99`

Then , significance level :
`\alpha: 1-0.99=0.01`

Since , sample size :
`n=300` .

Critical value :
`z_(\alpha/2)=2.576`

Also, the proportion of people said that they were fans of the visiting team :-

`\hat{p}=( 60)/(300)\approx0.2`

The confidence interval for population proportion is given by :-

`\hat{p}\pm z_(\alpha/2)\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}`

`0.2\pm(2.576)\sqrt{(0.2(1-0.2))/(329)}\\\\\approx0.2\pm0.057\\\\=(0.2-0.057, 0.2+0.056=(0.143,\ 0.257)`

Hence, the confidence interval for the population proportion with a 99% confidence level= (0.143, 0.257)