Question

Edwin conducted a survey to find the percentage of people in an area who smoked regularly. He defined the label “smoking regularly” for males smoking 30 or more cigarettes in a day and for females smoking 20 or more. Out of 635 persons who took part in the survey, 71 are labeled as people who smoke regularly.

Edwin wishes to construct a significance test for his data. He finds that the proportion of chain smokers nationally is 14.1%.

What is the standard deviation of the sample proportion? Answer choices are rounded to the thousandths place.

• 3.164

• 0.014

• 0.012

• 1.381

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Answer 1

GIVEN:

Sample size \((n)=635\)

Number of persons who smoke regularly \((x)=71\)

SOLUTION:

The formula for sample proportion \(\left(p_{h a t}\right)\) is given by,

\(\left(p_{\text {hat }}\right)=\) Number of successes / Total number of observations in the sample (or) \(x / n\)

\(=71 / 635\)

\(=0.1118\)

The formula for calculating standard deviation of sample proportion \(\left(\sigma_{p_{h a t}}\right)\) is given by,

\(\sigma_{p_{h a t}}=\left[\left[p_{\text {hat }}\left(1-p_{\text {hat }}\right)\right] / n\right]^{1 / 2}\)

\(=[[0.118(1-0.118)] / 635]^{1 / 2}\)

\(=[0.104076 / 635]^{1 / 2}\)

\(=0.0128\)

Thus the standard deviation of sample proportion \(\left(\sigma_{p_{h a t}}\right)\) is \(0.0128\).

Thus option (c) \(0.012\) is the correct answer.