Question

a survey of 350 people shows that 210 have seen a movie in the last month. find the margin of error.
%
question 8
2 pts
which sample size has a margin of error of 4%?
1000
400
4000
600
question 9
2 pts
which sample size will produce a margin of error of ±7.3%?
282
188
422
144

Explanation:

Step1: Calculate the proportion \(p\)

The proportion \(p=\frac{210}{350} = 0.6\), and \(q = 1 - p=1 - 0.6 = 0.4\)

Step2: Use the margin - of - error formula \(E = z\sqrt{\frac{pq}{n}}\) (assuming a 95% confidence level, \(z = 1.96\))

\(E=1.96\sqrt{\frac{0.6\times0.4}{350}}\)
First, calculate \(0.6\times0.4 = 0.24\)
Then \(\frac{0.24}{350}\approx0.0006857\)
\(\sqrt{0.0006857}\approx0.0262\)
\(E = 1.96\times0.0262\approx0.0513\) or \(5.13\%\)

For question 8:
The formula for margin of error \(E=\frac{1}{\sqrt{n}}\) (approximate for \(p = 0.5\))
If \(E = 0.04\), then \(\sqrt{n}=\frac{1}{0.04}=25\), \(n = 625\) (using \(E=\frac{1}{\sqrt{n}}\)). Using the more accurate formula \(E = z\sqrt{\frac{pq}{n}}\) (with \(z = 1.96\), \(p=q = 0.5\)):
\(0.04=1.96\sqrt{\frac{0.5\times0.5}{n}}\)
\(0.04 = 1.96\times\frac{0.5}{\sqrt{n}}\)
\(\sqrt{n}=\frac{1.96\times0.5}{0.04}=\frac{0.98}{0.04}=24.5\)
\(n=(24.5)^{2}=600.25\approx600\)

For question 9:
Using \(E = z\sqrt{\frac{pq}{n}}\) (assuming \(z = 1.96\), \(p = q=0.5\))
If \(E=0.073\)
\(0.073=1.96\times\frac{0.5}{\sqrt{n}}\)
\(\sqrt{n}=\frac{1.96\times0.5}{0.073}=\frac{0.98}{0.073}\approx13.42\)
\(n=(13.42)^{2}\approx180\approx188\)

Answer:

The margin of error for the first problem is approximately \(5.1\%\)
Question 8: \(600\)
Question 9: \(188\)