30 A Population Proportion

During an election year, we see articles in the newspaper that state confidence intervals in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show that the candidate has 40% of the vote within three percentage points (if the sample is large enough). Often, election polls are calculated with 95% confidence, so, the pollsters would be 95% confident that the true proportion of voters who favored the candidate would be between 0.37 and 0.43: (0.40 – 0.03,0.40 + 0.03).

Investors in the stock market are interested in the true proportion of stocks that go up and down each week. Businesses that sell personal computers are interested in the proportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers.

The procedure to find the confidence interval, the sample size, the margin of error or margin of error, and the confidence level for a proportion is similar to that for the population mean, but the formulas are different.

How do you know you are dealing with a proportion problem? First, the underlying distribution is a binomial distribution. (There is no mention of a mean or average.) If X is a binomial random variable, then X ~ B(n, p) 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,

$\hat P = \frac{X}{n}$.

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

$X \sim N(np, \sqrt{npq})$

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

$\frac{X}{n} = \hat P \sim N \left( \frac{np}{n}, \frac{\sqrt{npq}}{n} \right)$

Using algebra to simplify : npq√n=

$\frac{np}{n} = p$ and  $\frac{\sqrt{npq}}{n} = \sqrt{\frac{pq}{n}}$

$\hat P$ follows a normal distribution for proportions: $\hat P \sim N \left(p, \sqrt{\frac{pq}{n}}\right)$

The confidence interval has the form $(\hat p – E, \hat p +E)$ where $E$ is the margin of error for the proportion.

$\hat p = \frac{x}{n}$

$\hat p$ = the estimated proportion of successes ($\hat p$ is a point estimate for p, the true proportion.)

x = the number of successes

n = the size of the sample

The margin of error for a proportion is

$E = z_{\alpha/2}\cdot \sqrt{\frac{\hat p \hat q}{n}}$ where $\hat q = 1 – \hat p$

This formula is similar to the margin of error formula for a mean, except that the “appropriate standard deviation” is different. For a mean, when the population standard deviation is known, the appropriate standard deviation that we use is $\frac{\sigma}{\sqrt{n}}$. For a proportion, the appropriate standard deviation is $\sqrt{\frac{pq}{n}}$

However, in the margin of error formula, we use $\sqrt{\frac{\hat p \hat q}{n}}$ (with the hats)  as the standard deviation, instead of $\sqrt{\frac{pq}{n}}$.

In the margin of error formula, the sample proportions $\hat p$ and $\hat q$ are estimates of the unknown population proportions p and q. The estimated proportions $\hat p$ and $\hat q$ are used because p and q are not known. The sample proportions $\hat p$ and $\hat q$ are calculated from the data: $\hat p$ is the estimated proportion of successes, and $\hat q$ is the estimated proportion of failures.

The confidence interval can be used only if the number of successes $n\hat p$ and the number of failures $n \hat q$ are both greater than five.