 ## Introduction

### 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 $p•qnp•qn$, 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

$EBP=(zα2)(p′q′n),EBP=(zα2)(p′q′n),$ 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 $σnσn$. For a proportion, the appropriate standard deviation is $pqnpqn$.

However, in the error bound formula, we use $p′q′np′q′n$ as the standard deviation, instead of $pqnpqn$.

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 $np'≥5np'≥5$ and $nq'≥5nq'≥5$.

### 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.

Solution 8.10
• 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. $X~B(500,421500)X~B(500,421500)$.

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

n = 500

x = the number of successes = 421

$p′=xn=421500=0.842p′=xn=421500=0.842$

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 $(α2)(α2)$ = 0.025.

Then $zα2=z0.025=1.96.zα2=z0.025=1.96.$

Use the TI-83, 83+, or 84+ calculator command invNorm(0.975,0,1) to find z0.025. Remember that the area to the right of z0.025 is 0.025, and the area to the left of z0.025 is 0.975. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table.

$EBP=(zα2)p′q′n=(1.96)(0.842)(0.158)500=0.032EBP=(zα2)p′q′n=(1.96)(0.842)(0.158)500=0.032$
$p′+EBP=0.842+0.032=0.874p′+EBP=0.842+0.032=0.874$

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 8.10

Solution B

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

Press STAT and arrow over to TESTS.

Arrow down to A:1-PropZint. Press ENTER.

Arrow down to $xx$ and enter 421.

Arrow down to $nn$ and enter 500.

Arrow down to C-Level and enter .95.

Arrow down to Calculate and press ENTER.

The confidence interval is (0.81003, 0.87397).
Try It 8.10

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.

Solution 8.11
• 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

$p′=xn=300500=0.600p′=xn=300500=0.600$

Because CL = 0.90, then α = 1 – CL = 1 – 0.90 = 0.10$(α2)(α2)$ = 0.05.

Use the TI-83, 83+, or 84+ calculator command invNorm(0.95,0,1) to find z0.05. Remember that the area to the right of z0.05 is 0.05, and the area to the left of z0.05 is 0.95. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table.

$EBP=(zα2)p′q′n=(1.645)(0.60)(0.40)500=0.036EBP=(zα2)p′q′n=(1.645)(0.60)(0.40)500=0.036$
$p′– EBP=0.60–0.036=0.564p′–EBP=0.60−0.036=0.564$
$p′+EBP=0.60+0.036=0.636p′+EBP=0.60+0.036=0.636$

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 8.11

Solution B

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

Press STAT and arrow over to TESTS.

Arrow down to A:1-PropZint. Press ENTER.

Arrow down to $xx$ and enter 300.

Arrow down to $nn$ and enter 500.

Arrow down to C-Level and enter 0.90.

Arrow down to Calculate and press ENTER.

The confidence interval is (0.564, 0.636).
Try It 8.11

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.

## Plus-Four Confidence Interval for p

### Plus-Four Confidence Interval for p

There is a certain amount of error introduced into the process of calculating a confidence interval for a proportion. Because we do not know the true proportion for the population, we are forced to use point estimates to calculate the appropriate standard deviation of the sampling distribution. Studies have shown that the resulting estimation of the standard deviation can be flawed.

Fortunately, there is a simple adjustment that allows us to produce more accurate confidence intervals: We simply pretend that we have four additional observations. Two of these observations are successes, and two are failures. The new sample size, then, is n + 4, and the new count of successes is x + 2.

Computer studies have demonstrated the effectiveness of the plus-four confidence interval for p method. It should be used when the confidence level desired is at least 90 percent and the sample size is at least ten.

### Example 8.12

A random sample of 25 statistics students was asked: “Have you used a product in the past week?” Six students reported using the product within the past week. Use the plus-four method to find a 95 percent confidence interval for the true proportion of statistics students who use the product weekly.

Solution 8.12
Solution A

Solution ASix students out of 25 reported using a product within the past week, so x = 6 and n = 25. Because we are using the plus-four method, we will use x = 6 + 2 = 8, and n = 25 + 4 = 29.

$p′=xn=829≈0.276p′=xn=829≈0.276$

Because CL = 0.95, we know α = 1 – 0.95 = 0.05, and $α2α2$ = 0.025.

$z0.025=1.96z0.025=1.96$
$EPB=(zα2)p′q′n=(1.96)0.276(0.724)29≈0.163EPB=(zα2)p′q′n=(1.96)0.276(0.724)29≈0.163$

We are 95 percent confident that the true proportion of all statistics students who use the product is between 0.113 and 0.439.

Solution 8.12

Solution B

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

Press STAT and arrow over to TESTS.

Arrow down to A:1-PropZint. Press ENTER.

Arrow down to x and enter eight.

Arrow down to n and enter 29.

Arrow down to C-Level and enter 0.95.

Arrow down to Calculate and press ENTER.

The confidence interval is (0.113, 0.439).

#### Reminder

Remember that the plus-four method assumes an additional four trials: two successes and two failures. You do not need to change the process for calculating the confidence interval; simply update the values of x and n to reflect these additional trials.

Try It 8.12

Out of a random sample of 65 freshmen at State University, 31 students have declared their majors. Use the plus-four method to find a 96 percent confidence interval for the true proportion of freshmen at State University who have declared their majors.

### Example 8.13

A group of researchers recently conducted a study analyzing the privacy management habits of teen internet users. In a group of 50 teens, 13 reported having more than 500 friends on a social media site. Use the plus-four method to find a 90 percent confidence interval for the true proportion of teens who would report having more than 500 online friends.

Solution 8.13

Solution AUsing plus-four, we have x = 13 + 2 = 15, and n = 50 + 4 = 54.

Because CL = 0.90, we know α = 1 – 0.90 = 0.10, and $α2α2$ = 0.05.

$z0.05=1.645z0.05=1.645$
$EPB=(zα2)(p′q′n)=(1.645)((0.278)(0.722)54)≈0.100EPB=(zα2)(p′q′n)=(1.645)((0.278)(0.722)54)≈0.100$

We are 90 percent confident that between 17.8 percent and 37.8 percent of all teens would report having more than 500 friends on a social media site.

Solution 8.13

Solution B

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

Press STAT and arrow over to TESTS.

Arrow down to A:1-PropZint. Press ENTER.

Arrow down to x and enter 15.

Arrow down to n and enter 54.

Arrow down to C-Level and enter 0.90.

Arrow down to Calculate and press ENTER.

The confidence interval is (0.178, 0.378).
Try It 8.13

The research group referenced in Example 8.13 talked to teens in smaller focus groups but also interviewed additional teens over the phone. When the study was complete, 588 teens had answered the question about their social media site friends, with 159 saying that they have more than 500 friends. Use the plus-four method to find a 90 percent confidence interval for the true proportion of teens who would report having more than 500 online friends based on this larger sample. Compare the results to those in Example 8.13.

## Calculating the Sample Size n

### Calculating the Sample Size n

If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size.

The margin of error formula for a population proportion is

• $EBP=zα2×p'•q'nEBP=zα2×p'•q'n$, where p′ is the sample proportion, q′ = 1 – p′, and n is the sample size.
• Solving for n gives you an equation for the sample size.
• $n=(zα2)2(p′q′)EBP2n=(zα2)2(p′q′)EBP2$. This formula tells us that we can compute the sample size n required for a confidence level of $Cl=1−αCl=1−α$ by taking the square of the critical value $za2za2$, multiplying by the point estimate p′, and by q′ = 1 – p′ and finally dividing the result by the square of the margin of error. Always remember to round up the value of n.

### Example 8.14

A mobile phone company wants to determine the current percentage of customers ages 50+ who use text messaging on their cell phones. How many customers ages 50+ should the company survey in order to be 90 percent confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion of customers ages 50+ who use text messaging on their cell phones? Assume that p′ = 0.5.

Solution 8.14

From the problem, we know that EBP = 0.03 (3 percent = 0.03), and $zα2zα2$z0.05 = 1.645 because the confidence level is 90 percent.

To calculate the sample size n, use the formula and make the substitutions.

Round the answer to the next higher value. The sample size should be 752 cell phone customers ages 50+ in order to be 90 percent confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion of all customers ages 50+ who use text messaging on their cell phones.

Try It 8.14

An internet marketing company wants to determine the current percentage of customers who click on ads on their smartphones. How many customers should the company survey in order to be 90 percent confident that the estimated proportion is within 5 percentage points of the true population proportion of customers who click on ads on their smartphones? Assume that the sample proportion p′ is 0.50.