Question

in a survey of 2059 adults in a recent year, 724 made a new years resolution to eat healthier. construct 90% and 95% confidence intervals for the population proportion. interpret the results and compare the widths of the confidence intervals. the 90% confidence interval for the population proportion p is (0.335, 0.369). (round to three decimal places as needed.) the 95% confidence interval for the population proportion p is (□,□). (round to three decimal places as needed.)

Explanation:

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

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 724$ and $n=2059$. So, $\hat{p}=\frac{724}{2059}\approx0.352$.

Step2: Find $z$-values

For a $95\%$ confidence interval, the significance level $\alpha=1 - 0.95=0.05$, and $\alpha/2=0.025$. From the standard normal distribution table, $z_{\alpha/2}=z_{0.025}=1.96$.

Step3: Calculate 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 $\hat{p}=0.352$, $n = 2059$, and $z_{\alpha/2}=1.96$ into the formula:

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Step4: Calculate confidence interval

The confidence interval for the population proportion $p$ is $\hat{p}-ESubstitute $\hat{p}=0.352$ and $E = 0.0206$:
$0.352-0.0206=0.331$ and $0.352 + 0.0206=0.373$

Answer:

The $95\%$ confidence interval for the population proportion $p$ is $(0.331,0.373)$