8.3 A population proportion  (Page 5/24)

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Chapter review

Some statistical measures, like many survey questions, measure qualitative rather than quantitative data. In this case, the population parameter being estimated is a proportion. It is possible to create a confidence interval for the true population proportion following procedures similar to those used in creating confidence intervals for population means. The formulas are slightly different, but they follow the same reasoning.

Let p′ represent the sample proportion, x/n , where x represents the number of successes and n represents the sample size. Let q′ = 1 – p′ . Then the confidence interval for a population proportion is given by the following formula:

(lower bound, upper bound) $=(p^{\prime }–EBP,p^{\prime } +EBP)= \left(p^{\prime }–z\sqrt{\frac{p^{\prime }{q}^{\prime }}{n}},p^{\prime }+z\sqrt{\frac{p^{\prime }{q}^{\prime }}{n}}\right)$

The “plus four” method for calculating confidence intervals is an attempt to balance the error introduced by using estimates of the population proportion when calculating the standard deviation of the sampling distribution. Simply imagine four additional trials in the study; two are successes and two are failures. Calculate $p^{\prime }=\frac{x+2}{n+4}$ , and proceed to find the confidence interval. When sample sizes are small, this method has been demonstrated to provide more accurate confidence intervals than the standard formula used for larger samples.

Formula review

p′ = x / n where x represents the number of successes and n represents the sample size. The variable p ′ is the sample proportion and serves as the point estimate for the true population proportion.

q ′ = 1 – p ′

$p^{\prime }~N\left(p,\sqrt{\frac{pq}{n}}\right)$ The variable p′ has a binomial distribution that can be approximated with the normal distribution shown here.

EBP = the error bound for a proportion = $z_{\frac{\alpha }{2}}\sqrt{\frac{p^{\prime }{q}^{\prime }}{n}}$

Confidence interval for a proportion:

$(\text{lower bound, upper bound)}=(p^{\prime }–EBP,p^{\prime }+EBP)=\left(p^{\prime }–z\sqrt{\frac{p^{\prime }{q}^{\prime }}{n}}, p^{\prime }+z\sqrt{\frac{p^{\prime }{q}^{\prime }}{n}}\right)$

$n= \frac{z_{\frac{\alpha }{2}}{}^2{p}^{\prime }{q}^{\prime }}{EB{P}^2}$ provides the number of participants needed to estimate the population proportion with confidence 1 - α and margin of error EBP .

Use the normal distribution for a single population proportion $p\prime =\frac{x}{n}$