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 (). (round to three decimal places as needed.)

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 1419$ and $n=2267$.
$\hat{p}=\frac{1419}{2267}\approx0.626$.

Step2: Calculate $z$-value for 90% confidence interval

For a 90% confidence interval, the significance level $\alpha=1 - 0.90=0.10$, and $\alpha/2=0.05$. The $z$-value $z_{\alpha/2}=z_{0.05}\approx1.645$.

Step3: Calculate margin of error

The margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$.
Substitute $\hat{p}=0.626$, $n = 2267$, and $z_{\alpha/2}=1.645$:

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Step4: Calculate confidence interval

The confidence interval is $\hat{p}-ESubstitute $\hat{p}=0.626$ and $E = 0.017$:
$0.626-0.017 < p<0.626 + 0.017$, so $0.609

Answer:

$(0.609,0.643)$