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 (0.340, 0.378). (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 the sample proportion

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

Step2: Calculate the standard error

The standard error \(SE=\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\)
Substitute \(\hat{p}=0.358\) and \(n = 2007\) into the formula:
\(SE=\sqrt{\frac{0.358\times(1 - 0.358)}{2007}}=\sqrt{\frac{0.358\times0.642}{2007}}\approx\sqrt{\frac{0.2298}{2007}}\approx0.0107\)

Step3: Find the critical value for 95% confidence interval

For a 95% confidence interval, the critical value \(z_{\alpha/2}\) is \(1.96\)

Step4: Calculate the margin of error

The margin of error \(E=z_{\alpha/2}\times SE\)
Substitute \(z_{\alpha/2}=1.96\) and \(SE = 0.0107\) into the formula: \(E=1.96\times0.0107\approx0.021\)

Step5: Calculate the confidence interval

The confidence interval for the population proportion \(p\) is \(\hat{p}-ESubstitute \(\hat{p}=0.358\) and \(E = 0.021\) into the formula: \(0.358-0.021 < p<0.358 + 0.021\)

Answer:

\((0.337,0.379)\)