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

Explanation:

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

Sample proportion $\hat{p}=\frac{x}{n}$, where $x = 724$ (number of successes) and $n=2059$ (sample size). So, $\hat{p}=\frac{724}{2059}\approx0.352$.

Step2: Calculate $q = 1-\hat{p}$

$q=1 - 0.352=0.648$.

Step3: Find $z$-value for 90% confidence interval

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

Step4: Calculate margin of error $E$ for 90% confidence interval

Margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}q}{n}}$.
Substitute the values: $E=1.645\sqrt{\frac{0.352\times0.648}{2059}}$.
First, calculate $\frac{0.352\times0.648}{2059}=\frac{0.228}{2059}\approx0.0001107$.
Then, $\sqrt{0.0001107}\approx0.0105$.
$E = 1.645\times0.0105\approx0.017$.

Step5: Calculate confidence interval for 90%

Confidence interval is $\hat{p}-E$0.352 - 0.017=0.335$ and $0.352+0.017 = 0.369$.

Answer:

$(0.335,0.369)$