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 (0.496, 0.528) (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 $\hat{p}$

Sample proportion $\hat{p}=\frac{x}{n}$, where $x = 1301$ (number of successes) and $n=2542$ (sample size).
$\hat{p}=\frac{1301}{2542}\approx0.512$

Step2: Calculate $q$

Since $q = 1-\hat{p}$, then $q=1 - 0.512=0.488$

Step3: Find $z$-values

For a 95% confidence interval, the critical value $z_{\alpha/2}$:
The significance level $\alpha=1 - 0.95=0.05$, so $\alpha/2=0.025$. From the standard normal distribution table, $z_{0.025}=1.96$

Step4: Calculate the margin of error $E$

The formula for the margin of error for a proportion is $E = z_{\alpha/2}\sqrt{\frac{\hat{p}q}{n}}$
Substitute $\hat{p}=0.512$, $q = 0.488$, $n = 2542$, and $z_{\alpha/2}=1.96$
$E=1.96\sqrt{\frac{0.512\times0.488}{2542}}$
First, calculate $0.512\times0.488 = 0.250$
Then $\sqrt{\frac{0.250}{2542}}\approx\sqrt{0.0000983}\approx0.010$
$E=1.96\times0.010 = 0.020$

Step5: Calculate the confidence interval

The confidence interval for a proportion is $\hat{p}-ESubstitute $\hat{p}=0.512$ and $E = 0.020$
$0.512-0.020=0.492$ and $0.512 + 0.020=0.532$

Answer:

The 95% confidence interval for the population proportion $p$ is $(0.492,0.532)$