Question

type the correct answer in the box. use numerals instead of words. if necessary, use / for the fraction bar
a survey of factories in five northeastern states found that 10% of the 300 workers surveyed were satisfied with the benefits offered by their
employers.
the margin of error for the population proportion, rounded to the nearest hundredth of a percent, is ± %
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Explanation:

Step1: Identify the formula for margin of error for proportion

The formula for margin of error \(E\) for a proportion is \(E = z\sqrt{\frac{p(1 - p)}{n}}\). For a \(95\%\) confidence level (common in surveys when not specified otherwise), \(z = 1.96\). Here, \(p=0.1\) (since \(10\%=0.1\)) and \(n = 300\).

Step2: Substitute the values into the formula

First, calculate \(1-p=1 - 0.1=0.9\). Then \(\frac{p(1 - p)}{n}=\frac{0.1\times0.9}{300}=\frac{0.09}{300}=0.0003\).
Next, \(\sqrt{\frac{p(1 - p)}{n}}=\sqrt{0.0003}\approx0.01732\).
Then \(E = 1.96\times0.01732\).

Step3: Calculate the margin of error

\(E=1.96\times0.01732 = 0.0339472\).
To convert to a percentage, multiply by \(100\). So \(E = 3.39472\%\approx3.39\%\)

Answer:

\(3.39\)