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7.3: The Central Limit Theorem for Sample Proportions

  • Page ID
    26202
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    The Central Limit Theorem can also be applied to Sample Proportions.

    The first step in any of these problems will be to find the mean and standard deviation of the sampling distribution. For a proportion the formula for the sampling mean is

    \[\mu_{\hat{p}}=p \nonumber\]

    And the standard deviation or standard error of the sampling distribution is

    \[\sigma_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \nonumber\]

    where \(p\) is the population proportion given and \(n\) is the sample size.

    If you are finding the probability of something, then the rest of the problem will follow similarly to problems involving means.

    Example \(\PageIndex{1}\)

    A survey found that 27% of people prefer having houseplants instead of a pet. A simple random sample of 181 people are selected from the population. What is the probability that the sample proportion of people who prefer having houseplants instead of a pet is less than 0.29?

    Solution

    Start by finding the sampling mean and sampling standard deviation.

    The sampling mean is

    \[\mu_{\hat{p}}=p \nonumber\]

    The population proportion is given as 27%. Always change a percent to a decimal before doing any calculations. A percent can be changed to a decimal by moving the decimal point to the right two places, \(27%=0.27\).

    So, \(p=0.27\).

    \[\mu_{\hat{p}}=p=0.27 \nonumber\]

    The sampling standard deviation is

    \[\sigma_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.27(1-0.27)}{181}}=\sqrt{\frac{0.27(0.73)}{181}}=\sqrt{\frac{0.1971}{181}}=\sqrt{0.001089}=0.0330 \nonumber\]

    Next, to find the probability something has occured, we need to find the \(z\)-score so we can find the probability using the Standard Normal Distribution in Excel.

    \[z=\frac{x-\mu_{\hat{p}}}{\sigma_{\hat{p}}}=\frac{0.29-0.27}{0.0330}=\frac{0.02}{0.0330}=0.6061 \nonumber\]

    \(P(x<0.29)=P(z<0.6061)\)

    Use the Excel formula for a left-tailed probability, \(=\text{NORM.S.DIST}()\)

    \[=\text{NORM.S.DIST}(0.6061,\text{true})=0.7278 \nonumber\]

    There is a probability of \(0.7278\) or a \(72.78%\) chance the sample proportion of people who prefer having houseplants instead of a pet is less than 0.29.


    7.3: The Central Limit Theorem for Sample Proportions is shared under a CC BY license and was authored, remixed, and/or curated by LibreTexts.

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