Question

8.2 one - sample interval for the proportion. construct and interpret a confidence interval for the population proportion. assume that a sample is used to estimate a population proportion, p. find the 99% confidence interval for a sample of size 272 with 95 successes. enter your answer as an inequality using decimals (not percents) accurate to three decimal places. < p < question help: video message instructor post to forum submit question

Explanation:

Step1: Calculate sample proportion $\hat{p}$

$\hat{p}=\frac{x}{n}$, where $x = 95$ (number of successes) and $n=272$ (sample - size). So, $\hat{p}=\frac{95}{272}\approx0.349$.

Step2: Find $z -$score for 99% confidence level

The confidence level is 99%, so the significance level $\alpha=1 - 0.99 = 0.01$. Then $\alpha/2=0.005$. Looking up in the standard normal distribution table, $z_{\alpha/2}=z_{0.005} = 2.576$.

Step3: Calculate the margin of error $E$

The formula for the margin of error for a proportion is $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.349$, $n = 272$, and $z_{\alpha/2}=2.576$ into the formula. First, $1-\hat{p}=1 - 0.349 = 0.651$. Then $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.349\times0.651}{272}\approx\frac{0.227199}{272}\approx0.000835$. And $E = 2.576\sqrt{0.000835}\approx2.576\times0.0289\approx0.074$.

Step4: Construct the confidence interval

The confidence - interval for the population proportion $p$ is given by $\hat{p}-E

Answer:

$0.275 < p < 0.423$