8.6 Confidence intervals: confidence interval for a population  (Page 21/7)

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 votewithin 3 percentage points. Often, election polls are calculated with 95% confidence. So, the pollsters would be 95% confident that the true proportion of voters whofavored 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 theproportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up ordown 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, and the confidence level for a proportion is similar to that for the population mean. The formulas are different.

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

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

(Sometimes the random variable is denoted as $p\text{'}$ , read "P prime".)

When $n$ is large and p is not close to 0 or 1, we can use the normal distribution to approximate the binomial.

$X$ ~ $N(n\cdot p,\sqrt{n\cdot p\cdot q})$

If we divide the random variable by $n$ , the mean by $n$ , 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 = thenumber of successes divided by $n$ .)

$\frac{X}{n}=\hat{p}$ ~ $N(\frac{n\cdot p}{n},\frac{\sqrt{n\cdot p\cdot q}}{n})$

Using algebra to simplify : $\frac{\sqrt{n\cdot p\cdot q}}{n}=\sqrt{\frac{p\cdot q}{n}}$

$\hat{p}$ follows a normal distribution for proportions : $\hat{p}$ ~ $N(p,\sqrt{\frac{p\cdot q}{n}})$

The confidence interval has the form $(\hat{p}-\text{ME},\hat{p}+\text{ME})$ .

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

$\hat{p}$ = 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

$\text{ME}=z_{\frac{\alpha }{2}}\cdot \sqrt{\frac{\hat{p}\cdot \hat{q}}{n}}$ $where\hat{q}=1-\hat{p}$

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 $\sqrt{\frac{p\cdot q}{n}}$ .

However, in the error bound formula, we use $\sqrt{\frac{\hat{p}\cdot \hat{q}}{n}}$ as the standard deviation, instead of $\sqrt{\frac{p\cdot q}{n}}$

However, in the margin of error formula, the standard deviation is $\sqrt{\frac{\hat{p}\cdot \hat{q}}{n}}$ .