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translate the statement into a confidence interval. approximate the lev…

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 75.4% confidence interval.
(round to one decimal place as needed.)
use the results from three weeks prior to the newer survey.
the confidence interval for the proportion is (0.520, 0.540).
(round to three decimal places as needed.)
use the results from three weeks prior to 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

The margin of error \(E = z\sqrt{\frac{p(1 - p)}{n}}\). For a proportion \(p\), in the new - er survey \(p = 0.47\), \(n=3373\), \(E = 0.01\). We know that \(z=\frac{E}{\sqrt{\frac{p(1 - p)}{n}}}\).
Substitute \(p = 0.47\), \(n = 3373\), \(E=0.01\) into the formula:
\(\sqrt{\frac{p(1 - p)}{n}}=\sqrt{\frac{0.47\times(1 - 0.47)}{3373}}=\sqrt{\frac{0.47\times0.53}{3373}}\approx\sqrt{\frac{0.2491}{3373}}\approx\sqrt{0.000074}\approx0.0086\)
\(z=\frac{0.01}{0.0086}\approx1.16\)

For the prior - survey, \(p = 0.53\), \(n = 3373\), \(E = 0.01\)
\(\sqrt{\frac{p(1 - p)}{n}}=\sqrt{\frac{0.53\times(1 - 0.53)}{3373}}=\sqrt{\frac{0.53\times0.47}{3373}}\approx\sqrt{\frac{0.2491}{3373}}\approx0.0086\)
\(z=\frac{0.01}{0.0086}\approx1.16\)

Step2: Find the confidence level

We know that if \(z\) is the critical value, the confidence level \(C\) is related to \(z\) by the formula \(C = 2\Phi(z)-1\), where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution.
Using a standard - normal table or a calculator with a normal - distribution function, if \(z = 1.16\), then \(\Phi(1.16)=0.8770\)
\(C=2\times0.8770 - 1=0.754 = 75.4\%\)

Answer:

\(75.4\)