Question

in a survey of 2267 adults in a recent year, 1419 say they have made a new years resolution. 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 (0.609, 0.643). (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 $x = 1419$ (number of successes) and $n=2267$ (sample size). So, $\hat{p}=\frac{1419}{2267}\approx0.626$.

Step2: Calculate $z$-score for 95% confidence interval

For a 95% confidence interval, the significance level $\alpha=1 - 0.95=0.05$, and $\alpha/2=0.025$. The $z$-score $z_{\alpha/2}=z_{0.025}=1.96$.

Step3: Calculate margin of error $E$

The formula for the margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$.
Substitute $\hat{p}=0.626$, $n = 2267$, and $z_{\alpha/2}=1.96$ into the formula:

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Step4: Calculate confidence interval

The confidence interval is $\hat{p}-ESubstitute $\hat{p}=0.626$ and $E = 0.020$:
$0.626-0.020 = 0.606$ and $0.626+0.020=0.646$

Answer:

The 95% confidence interval for the population proportion $p$ is $(0.606,0.646)$