Question

translate the statement into a confidence interval. approximate the level of confidence.
in a survey of 3373 adults in a country, 47% believe the economy is getting better. three weeks prior to this survey, 53% believed that
the economy was getting better. the surveys margin of error is ±1%.
use the results from the newer survey.
the confidence interval for the proportion is (0.460, 480)
(round to three decimal places as needed.)
use the results from the newer survey.
the interval is a % confidence interval.
(round to one decimal place as needed.)

Explanation:

Step1: Recall the formula for margin of error for proportion

The margin of error \(E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\). Here, \(\hat{p}=0.47\), \(n = 3373\), and \(E=0.01\).

Step2: Solve for \(z_{\alpha/2}\)

First, calculate \(\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.47\times(1 - 0.47)}{3373}}=\sqrt{\frac{0.47\times0.53}{3373}}\approx\sqrt{\frac{0.2491}{3373}}\approx\sqrt{0.000074}\approx0.0086\).
Then, since \(E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\), we have \(z_{\alpha/2}=\frac{E}{\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}}\). Substituting \(E = 0.01\) and \(\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\approx0.0086\), we get \(z_{\alpha/2}=\frac{0.01}{0.0086}\approx1.16\).

Step3: Find the confidence level

If \(z_{\alpha/2}\approx1.16\), then using the standard normal distribution table, the area to the right of \(z = 1.16\) is \(P(Z>1.16)=1 - P(Z\leq1.16)\). From the standard - normal table, \(P(Z\leq1.16)=0.8770\), so \(P(Z > 1.16)=1 - 0.8770 = 0.123\). And \(\alpha=2\times0.123 = 0.246\). The confidence level \(C=1-\alpha=1 - 0.246=0.754 = 75.4\%\).

Answer:

\(75.4\)