Question

translate the statement into a confidence interval. approximate the level of confidence.
in a survey of 1100 adults in a country, 67% think teaching is one of the most important jobs in the country today. the surveys margin of error is ±3%.
the confidence interval for the proportion is (0.64, 0.70).
(round to three decimal places as needed.)
the interval is a % confidence interval.
(round to one decimal place as needed.)

Explanation:

Step1: Calculate the sample proportion \(p\) and sample size \(n\)

Given \(p = 0.67\) (since \(67\%=0.67\)) and \(n = 1100\).
The formula for the margin of error \(E\) for a proportion is \(E = z_{\alpha/2}\sqrt{\frac{p(1 - p)}{n}}\).
We know \(E=0.03\) (since \(3\% = 0.03\)).

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

First, calculate \(\sqrt{\frac{p(1 - p)}{n}}=\sqrt{\frac{0.67\times(1 - 0.67)}{1100}}=\sqrt{\frac{0.67\times0.33}{1100}}\approx\sqrt{\frac{0.2211}{1100}}\approx\sqrt{0.000201}\approx0.0142\).
Then, since \(E = z_{\alpha/2}\sqrt{\frac{p(1 - p)}{n}}\), we can find \(z_{\alpha/2}=\frac{E}{\sqrt{\frac{p(1 - p)}{n}}}\).
Substituting the values, \(z_{\alpha/2}=\frac{0.03}{0.0142}\approx2.11\).

Step3: Find the confidence level

Looking up the \(z -\) value in the standard normal distribution table.
If \(z_{\alpha/2}\approx2.11\), then \(\alpha/2=1 - P(Z\leq z_{\alpha/2})\).
From the standard normal table, \(P(Z\leq2.11) = 0.9826\). So \(\alpha/2=1 - 0.9826=0.0174\) and \(\alpha=2\times0.0174 = 0.0348\).
The confidence level \(C=1-\alpha=1 - 0.0348 = 0.9652\approx96.5\%\).

Answer:

\(96.5\)