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6.3: The Central Limit Theorem for Sample Proportions

  • Page ID
    125703
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    If \(X\) is a binomial random variable, then

    \[X \sim B(n, p)\nonumber \]

    where \(n\) is the number of trials and \(p\) is the probability of a success.

    To form a proportion, take \(X\), the random variable for the number of successes and divide it by \(n\), the number of trials (or the sample size). The random variable \(\hat{P}\) (read "P hat") is that proportion,

    \[\hat{P} = \dfrac{X}{n}\nonumber \]

    When \(n\) is large and \(p\) is not close to zero or one, we can use the normal distribution to approximate the binomial.

    \[X \sim N(np, \sqrt{npq})\nonumber \]

    If we divide the random variable, the mean, and the standard deviation by \(n\), we get a normal distribution of proportions with \(\hat{P}\), called the estimated proportion, as the random variable. (Recall that a proportion as the number of successes divided by \(n\).)

    \[\dfrac{X}{n} = \hat{P} \sim N\left(\dfrac{np}{n}, \dfrac{\sqrt{npq}}{n}\right)\nonumber \]

    Using algebra to simplify:

    \[\dfrac{\sqrt{npq}}{n} = \sqrt{\dfrac{pq}{n}}\nonumber \]

    \(\hat{P}\) follows a normal distribution for proportions:

    \[\dfrac{X}{n} = \hat{P}\ \sim N\left(p, \sqrt{\dfrac{pq}{n}}\right)\nonumber \]

    Review

    The Central Limit Theorem for Proportions:

    If 

    1. Sample is random with independent observations.

    2. Sample is large:  The sample size, n, is considered large enough when the sample expects at least 10 successes (yes) and 10 failures (no); i.e. 

     \(n\cdot\ p\)≥10 and \(n\cdot(1-p)\)≥10

    3. Big population:  If sampling is done without replacement, the population must be at least 10 times larger than the sample size.

    then \(\hat{P}\) follows a normal distribution with

    1. The mean of sampling distribution of sample proportions \(\mu_{\hat{P}}\) is the population proportion, p, \(\mu_{\hat{P}}=p\)
    2. The standard deviation of the sampling distribution of sample proportions, \(\sigma_{\hat{P}}\)=\(\sqrt{\dfrac{pq}{n}}=\sqrt{\dfrac{p(1-p)}{n}}\)

    Formula Review


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