Question

in a survey of 2542 adults in a recent year, 1301 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 $x = 1301$ and $n=2542$.
$\hat{p}=\frac{1301}{2542}\approx0.512$

Step2: Find $z$-score for 90% confidence interval

For a 90% confidence interval, the significance level $\alpha=1 - 0.90=0.10$, and $\alpha/2=0.05$. The $z$-score $z_{\alpha/2}=z_{0.05}\approx1.645$ (from standard normal distribution table).

Step3: Calculate margin of error $E$

The formula for margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$.
Substitute $\hat{p}=0.512$, $n = 2542$, and $z_{\alpha/2}=1.645$:

$$
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Step4: Calculate confidence interval

The confidence interval is $\hat{p}-E$0.512-0.016 < p<0.512 + 0.016$
$0.496

Answer:

$(0.496,0.528)$