### Introduction

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 percent of the vote within 3 percentage points (if the sample is large enough). Often, election polls are calculated with 95 percent confidence, so the pollsters would be 95 percent 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 error bound for a population proportion (EBP)**, 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 data that you are collecting is categorical, consisting of two categories: Success or Failure, Yes or No. Examples of situations where you are trying to estimate the true population proportion are the following: What proportion of the population smoke? What proportion of the population will vote for candidate A? What proportion of the population has a college-level education?

The distribution of the sample proportions (based on samples of size *n*), is denoted by *P′* (read “P prime”).

The central limit theorem for proportions asserts that the sample proportion distribution *P′* follows a normal distribution with mean value *p*, and standard deviation $\sqrt{\frac{p\u2022q}{n}}$, where *p* is the population proportion and *q = 1 – p*.

The confidence interval has the form (*p′* – *EBP*, *p′* + *EBP*). *EBP* is error bound for the proportion.

*p′* = the *estimated proportion* of successes (*p′* is a *point estimate* for *p*, the true proportion.)

*x* = the *number* of successes

*n* = the size of the sample

*The error bound for a proportion is*

$EBP=\left({z}_{\frac{\alpha}{2}}\right)\left(\sqrt{\frac{{p}^{\prime}{q}^{\prime}}{n}}\right)\text{,}$ where *q′* = 1 – *p′*.

This formula is similar to the error bound 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 error bound formula, we use $\sqrt{\frac{{p}^{\prime}{q}^{\prime}}{n}}$ as the standard deviation, instead of $\sqrt{\frac{pq}{n}}$.

In the error bound formula, the *sample proportions p′ and q′, are estimates of the unknown population proportions p and q*. The estimated proportions

*p′*and

*q′*are used because

*p*and

*q*are not known. The sample proportions

*p′*and

*q′*are calculated from the data:

*p′*is the estimated proportion of successes, and

*q′*is the estimated proportion of failures.

The confidence interval can be used only if the number of successes *np′* and the number of failures *nq′* are both greater than five.

That is, in order to use the formula for confidence intervals for proportions, you need to verify that both $n{p}^{\text{'}}\ge 5$ and $n{q}^{\text{'}}\ge 5$.

### Example 8.10

Suppose that a market research firm is hired to estimate the percentage of adults living in a large city who have cell phones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes, they own cell phones. Using a 95 percent confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones.

- The first solution is step-by-step (Solution A).
- The second solution uses a function of the TI-83, 83+, or 84 calculators (Solution B).

Solution A Let *X* = the number of people in the sample who have cell phones. *X* is binomial. $X~B\left(500,\frac{421}{500}\right)$.

To calculate the confidence interval, you must find *p′*, *q′*, and *EBP*.

*n* = 500

*x* = the number of successes = 421

*p′* = 0.842 is the sample proportion; this is the point estimate of the population proportion.

Because *CL* = 0.95, then *α* = 1 – *CL* = 1 – 0.95 = 0.05 $\left(\frac{\alpha}{2}\right)$ = 0.025.

Then ${z}_{\frac{\alpha}{2}}={z}_{0.025}=1.96\text{.}$

Use the TI-83, 83+, or 84+ calculator command invNorm(0.975,0,1) to find *z*_{0.025}. Remember that the area to the right of *z*_{0.025} is 0.025, and the area to the left of *z*_{0.025} is 0.975. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table.

The confidence interval for the true binomial population proportion is (*p′* – *EBP*, *p′* + *EBP*) = (0.810, 0.874).

InterpretationWe estimate with 95 percent confidence that between 81 percent and 87.4 percent of all adult residents of this city have cell phones.

Explanation of 95 percent Confidence LevelNinety-five percent of the confidence intervals constructed in this way would contain the true value for the population proportion of all adult residents of this city who have cell phones.

Solution B

### Using the TI-83, 83+, 84, 84+ Calculator

Press `STAT`

and arrow over to `TESTS`

.

`A:1-PropZint`

. Press `ENTER`

.Arrow down to $x$ and enter 421.

Arrow down to $n$ and enter 500.

Arrow down to

`C-Level`

and enter .95.Arrow down to

`Calculate`

and press `ENTER`

.Suppose 250 randomly selected people are surveyed to determine whether they own tablets. Of the 250 surveyed, 98 reported owning tablets. Using a 95 percent confidence level, compute a confidence interval estimate for the true proportion of people who own tablets.

### Example 8.11

For a class project, a political science student at a large university wants to estimate the percentage of students who are registered voters. He surveys 500 students and finds that 300 are registered voters. Compute a 90 percent confidence interval for the true percentage of students who are registered voters, and interpret the confidence interval.

- The first solution is step-by-step (Solution A).
- The second solution uses a function of the TI-83, 83+, or 84 calculators (Solution B).

Solution A

Because *CL* = 0.90, then *α* = 1 – *CL* = 1 – 0.90 = 0.10$\left(\frac{\alpha}{2}\right)$ = 0.05.

Use the TI-83, 83+, or 84+ calculator command invNorm(0.95,0,1) to find *z*_{0.05}. Remember that the area to the right of *z*_{0.05} is 0.05, and the area to the left of *z*_{0.05} is 0.95. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table.

The confidence interval for the true binomial population proportion is (*p′* – *EBP*, *p′* + *EBP*) = (0.564, 0.636).

**Interpretation**

- We estimate with 90 percent confidence that the true percentage of all students who are registered voters is between 56.4 percent and 63.6 percent.
- Alternate wording: We estimate with 90 percent confidence that between 56.4 percent and 63.6 percent of all students are registered voters.

Explanation of 90 percent Confidence LevelNinety percent of all confidence intervals constructed in this way contain the true value for the population percentage of students who are registered voters.

Solution B

### Using the TI-83, 83+, 84, 84+ Calculator

Press `STAT`

and arrow over to `TESTS`

.

`A:1-PropZint`

. Press `ENTER`

.Arrow down to $x$ and enter 300.

Arrow down to $n$ and enter 500.

Arrow down to

`C-Level`

and enter 0.90.Arrow down to

`Calculate`

and press `ENTER`

.A student polls her school to determine whether students in the school district are for or against the new legislation regarding school uniforms. She surveys 600 students and finds that 480 are against the new legislation.

a. Compute a 90 percent confidence interval for the true percentage of students who are against the new legislation, and interpret the confidence interval.

b. In a sample of 300 students, 68 percent said they own an iPod and a smartphone. Compute a 97 percent confidence interval for the true percentage *p* of students who own an iPod and a smartphone.