Question

in a survey of 2007 adults in a recent year, 718 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 $\hat{p}$

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 718$ (number of successes) and $n=2007$ (sample size). So, $\hat{p}=\frac{718}{2007}\approx0.358$.

Step2: Find $z$-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 $z_{\alpha/2}=z_{0.05}\approx1.645$ (from standard normal distribution table).

Step3: Calculate the 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 $\hat{p}=0.358$, $n = 2007$, and $z_{\alpha/2}=1.645$:

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Step4: Construct the confidence interval

The confidence interval for a proportion is $\hat{p}-ESubstitute $\hat{p}=0.358$ and $E = 0.0176$:
$0.358-0.0176 < p<0.358 + 0.0176$
$0.340

Answer:

$(0.340,0.376)$