Question

translate the statement into a confidence interval. approximate the level of confidence.
in a survey of 1048 adults in a country, 71% said being able to speak the language is at the core of national identity. the surveys margin of error is ±3.1%.
the confidence interval for the proportion is (0.679, 0.741).
(round to three decimal places as needed.)
the interval is a □% confidence interval.
(round to one decimal place as needed.)

Explanation:

Step1: Recall the formula for margin of error \(E = z_{\alpha/2}\sqrt{\frac{p(1 - p)}{n}}\)

Given \(n=1048\), \(p = 0.71\), \(E=0.031\)
First, calculate \(\sqrt{\frac{p(1 - p)}{n}}=\sqrt{\frac{0.71\times(1 - 0.71)}{1048}}=\sqrt{\frac{0.71\times0.29}{1048}}\approx\sqrt{\frac{0.2059}{1048}}\approx\sqrt{0.0001965}\approx0.014\)

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

Since \(E = z_{\alpha/2}\sqrt{\frac{p(1 - p)}{n}}\), then \(z_{\alpha/2}=\frac{E}{\sqrt{\frac{p(1 - p)}{n}}}\)
Substitute \(E = 0.031\) and \(\sqrt{\frac{p(1 - p)}{n}}\approx0.014\) into the formula: \(z_{\alpha/2}=\frac{0.031}{0.014}\approx2.21\)

Step3: Find the confidence level

If \(z_{\alpha/2}\approx2.21\), then from the standard normal table, the area to the left of \(z = 2.21\) is \(0.9864\) and the area to the left of \(z=- 2.21\) is \(0.0136\)
The confidence level \(C=1-\alpha\), where \(\alpha = 2\times(1 - 0.9864)=2\times0.0136 = 0.0272\)
So \(C=1 - 0.0272=0.9728\) or \(97.3\%\)

Answer:

\(97.3\)