A Population Proportion, Confidence Intervals | Statistics

Learning objectives

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 (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 the following 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 $\sqrt{\frac{p q}{n}}$ , 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' = \frac{x}{n}

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

$E B P = (z_{\frac{α}{2}}) (\sqrt{\frac{p^{'} q^{'}}{n}}),$ 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 $\frac{σ}{\sqrt{n}}$ . For a proportion, the appropriate standard deviation is $\sqrt{\frac{p q}{n}}$ .

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

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 $n p^{'} \geq 5$ and $n q^{'} \geq 5$ .

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).

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

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

n = 500

x = the number of successes = 421

p^{'} = \frac{x}{n} = \frac{421}{500} = 0.842

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

q ′ = 1 – p ′ = 1 – 0 .842 = 0 .158

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

Then, $z_{\frac{α}{2}} = z_{0.025} = 1.96 .$

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_0.025is 0.025, and the area to the left of z_0.025is 0.975. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table.

E B P = (z_{\frac{α}{2}}) \sqrt{\frac{p^{'} q^{'}}{n}} = (1.96) \sqrt{\frac{(0.842) (0.158)}{500}} = 0.032

p' - E B P = 0.842 - 0.032 = 0.81

p^{'} + E B P = 0.842 + 0.032 = 0.874

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

Interpretation

We 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 Level

Ninety-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.

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

x = 300 a n d n = 500

p^{'} = \frac{x}{n} = \frac{300}{500} = 0.600

q^{'} = 1 - p^{'} = 1 - 0.600 = 0.400

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

z_{\frac{α}{2}} = z_{0 .05} = 1 .645

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.

E B P = (z_{\frac{α}{2}}) \sqrt{\frac{p^{'} q^{'}}{n}} = (1.645) \sqrt{\frac{(0.60) (0.40)}{500}} = 0.036

p^{'} - E B P = 0.60 - 0.036 = 0.564

p^{'} + E B P = 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 Level

Ninety 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

Try It

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.

(0.7731, 0.8269); We estimate with 90 percent confidence that the true percent of all students in the district who are against the new legislation is between 77.31 percent and 82.69 percent.

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 of students who own an iPod and a smartphone.

Solution A

Sixty-eight percent (68 percent) of students own an iPod and a smartphone.

$p^{'} = 0.68$

$q^{'} = 1 - p^{'} = 1 - 0.68 = 0.32$

Since CL = 0.97, we know α = 1 – 0.97 = 0.03 and $\frac{α}{2}$ = 0.015.

The area to the left of z_0.015 is 0.015, and the area to the right of z_0.015 is 1 – 0.015 = 0.985.

Using the TI 83, 83+, or 84+ calculator function InvNorm(.985,0,1),

$z_{0.015} = 2.17$

$E P B = (z_{\frac{α}{2}}) \sqrt{\frac{p^{'} q^{'}}{n}} = 2.17 \sqrt{\frac{0.68 (0.32)}{300}} \approx 0.0269$

p′ – EPB = 0.68 – 0.0269 = 0.6531

p′ + EPB = 0.68 + 0.0269 = 0.7069

We are 97 percent confident that the true proportion of all students who own an iPod and a smartphone is between 0.6531 and 0.7069.

Solution B

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.

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 A

Six 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^{'} = \frac{x}{n} = \frac{8}{29} \approx 0.276

q^{'} = 1 - p^{'} = 1 - 0.276 = 0.724

Because CL = 0.95, we know α = 1 – 0.95 = 0.05, and $\frac{α}{2}$ = 0.025.

z_{0.025} = 1.96

E P B = (z_{\frac{α}{2}}) \sqrt{\frac{p^{'} q^{'}}{n}} = (1.96) \sqrt{\frac{0.276 (0.724)}{29}} \approx 0.163

\begin{matrix} p ′ – E P B = 0 .276 – 0 .163 = 0 .113 \\ p ′ + E P B = 0 .276 + 0 .163 = 0 .439 \end{matrix}

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

Try It

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.

Solution A

Using “plus four,” we have x = 31 + 2 = 33 and n = 65 + 4 = 69.

$p^{'} = \frac{33}{69} \approx 0.478$

$q^{'} = 1 - p^{'} = 1 - 0.478 = 0.522$

Since CL = 0.96, we know α = 1 – 0.96 = 0.04 and $\frac{α}{2}$ = 0.02.

$z_{0.02} = 2.054$

$E P B = (z_{\frac{α}{2}}) \sqrt{\frac{p^{'} q^{'}}{n}} = (2.054) (\sqrt{\frac{(0.478) (0.522)}{69}}) ~ 0.124$

p′ – EPB = 0.478 – 0.124 = 0.354

p′ + EPB = 0.478 + 0.124 = 0.602

We are 96 percent confident that between 35.4 percent and 60.2 percent of all freshmen at State University have declared a major.

Solution B

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.

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

p^{'} = \frac{15}{54} \approx 0.278

q^{'} = 1 - p^{'} = 1 - 0.278 = 0.722

Because CL = 0.90, we know α = 1 – 0.90 = 0.10, and $\frac{α}{2}$ = 0.05.

z_{0.05} = 1.645

E P B = (z_{\frac{α}{2}}) (\sqrt{\frac{p^{'} q^{'}}{n}}) = (1.645) (\sqrt{\frac{(0.278) (0.722)}{54}}) \approx 0.100

\begin{matrix} p ′ – E P B = 0 .278 – 0 .100 = 0 .178 \\ p ′ + E P B = 0 .278 + 0 .160 = 0 .378 \end{matrix}

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.

Try It

The research group referenced in eip-251 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 eip-251.

Solution A

Using “plus-four,” we have x = 159 + 2 = 161 and n = 588 + 4 = 592.

$p^{'} = \frac{161}{592} \approx 0.272$

$q^{'} = 1 - p^{'} = 1 - 0.272 = 0.728$

Since CL = 0.90, we know α = 1 – 0.90 = 0.10 and $\frac{α}{2}$ = 0.05.

$E P B = (z_{\frac{α}{2}}) (\sqrt{\frac{p^{'} q^{'}}{n}}) = (1.645) (\sqrt{\frac{(0.272) (0.728)}{592}}) \approx 0.030$

p′ – EPB = 0.272 – 0.030 = 0.242

p′ + EPB = 0.272 + 0.030 = 0.302

We are 90 percent confident that between 24.2 percent and 30.2 percent of all teens would report having more than 500 friends on Facebook.

Solution B

Conclusion: The confidence interval for the larger sample is narrower than the interval from eip-251. Larger samples will always yield more precise confidence intervals than smaller samples. The “plus four” method has a greater impact on the smaller sample. It shifts the point estimate from 0.26 (13/50) to 0.278 (15/54). It has a smaller impact on the EPB, changing it from 0.102 to 0.100. In the larger sample, the point estimate undergoes a smaller shift: from 0.270 (159/588) to 0.272 (161/592). It is easy to see that the plus-four method has the greatest impact on smaller samples.

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

$E B P = z_{\frac{α}{2}} \times \sqrt{\frac{p' q'}{n}}$ , 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.
$n = \frac{{(z_{\frac{α}{2}})}^{2} (p^{'} q^{'})}{E B P^{2}}$ . This formula tells us that we can compute the sample size n required for a confidence level of $C l = 1 - α$ by taking the square of the critical value $z_{\frac{a}{2}}$ , 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.

Suppose 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.

From the problem, we know that EBP = 0.03 (3 percent=0.03), and $z_{\frac{α}{2}}$ z_0.05 = 1.645 because the confidence level is 90 percent.

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

n = \frac{z^{2} p^{'} q^{'}}{E B P^{2}} g i v e s n = \frac{{1.645}^{2} (0.5) (0.5)}{{0.03}^{2}} = 751.7

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.

References

Jensen, T. (2013). Democrats, Republicans divided on opinion of music icons. Retrieved from http://www.publicpolicypolling.com/Day2MusicPoll.pdf

Madden, M., et al. (2013). Teens, social media, and privacy. Retrieved from http://www.pewinternet.org/Reports/2013/Teens-Social-Media-And-Privacy.aspx

Princeton Survey Research Associates International. (2012). 2012 teens and privacy management survey. Retrieved from http://www.pewinternet.org/~/media//Files/Questionnaire/2013/Methods%20and%20Questions_Teens%20and%20Social%20Media.pdf

Rasmussen Reports. (2013). 52% say big-time college athletics corrupt education process. Retrieved from http://www.rasmussenreports.com/public_content/lifestyle/sports/may_2013/52_say_big_time_college_athletics_corrupt_education_process

Saad, L. (2013). Three in four U.S. workers plan to work past retirement age. Retrieved from http://www.gallup.com/poll/162758/three-fourworkers-plan-work-past-retirement-age.aspx

The Field Poll. (n.d.). Retrieved from http://field.com/fieldpollonline/subscribers/

Zogby Analytics. (2013). New SUNYIT/Zogby analytics poll: Few Americans worry about emergency situations occurring in their community; Only one in three have an emergency plan; 70% support infrastructure investment for national security. Retrieved from http://www.zogbyanalytics.com/news/299-americans-neither-worried-norprepared-in-case-of-a-disaster-sunyit-zogby-analytics-poll

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 by 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) $= (p^{'} - E B P, p^{'} + E B P) = (p^{'} - z \sqrt{\frac{p^{'} q^{'}}{n}}, p^{'} + z \sqrt{\frac{p^{'} q^{'}}{n}}) .$

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 $p^{'} = \frac{x + 2}{n + 4}$ , 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 ′

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

E B P = the error bound for a proportion = z_{\frac{α}{2}} \sqrt{\frac{p^{'} q^{'}}{n}} .

Confidence interval for a proportion:

$(lower bound, upper bound) = (p^{'} - E B P, p^{'} + E B P) = (p^{'} - z \sqrt{\frac{p^{'} q^{'}}{n}}, p^{'} + z \sqrt{\frac{p^{'} q^{'}}{n}})$

$n = \frac{z_{\frac{α}{2}}^{2} p^{'} q^{'}}{E B P^{2}}$ 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 $p' = \frac{x}{n} .$

$E B P = (z_{\frac{α}{2}}) \sqrt{\frac{p' q'}{n}} p' + q' = 1$

The confidence interval has the format (p′ – EBP, p′ + EBP).

$\bar{x}$ is a point estimate for μ.

p′ is a point estimate for ρ.

s is a point estimate for σ.

Use the following information to answer the next two exercises: Marketing companies are interested in knowing the population percentage of women who make the majority of household purchasing decisions.

When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 90 percent confident that the population proportion is estimated to within 0.05?

If it were later determined that it was important to be more than 90 percent confident and a new survey were commissioned, how would it affect the minimum number you need to survey? Why?

It would decrease, because the z-score would decrease, which would reduce the numerator and lower the number.

Use the following information to answer the next five exercises: Suppose a marketing company conducted a survey. It randomly surveyed 200 households and found that in 120 of them, the women made the majority of the purchasing decisions. We are interested in the population proportion of households where women make the majority of the purchasing decisions.

Identify the following:

x = ______
n = ______
p′ = ______

Define the random variables X and P′ in words.

X is the number of successes where the woman makes the majority of the purchasing decisions for the household. P′ is the percentage of households sampled where the woman makes the majority of the purchasing decisions for the household.

Which distribution should you use for this problem?

Construct a 95 percent confidence interval for the population proportion of households where the women make the majority of the purchasing decisions. State the confidence interval, sketch the graph, and calculate the error bound.

CI: (0.5321, 0.6679)

View original figure (external; image not reproduced in this text edition)

Original figure description: This is a normal distribution curve. The peak of the curve coincides with the point 0.6 on the horizontal axis. A central region is shaded between points 0.5321 and 0.6679.

EBM: 0.0679

List two difficulties the company might have in obtaining random results if this survey were done by email.

Use the following information to answer the next five exercises: Of 1,050 randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage earners, 250 identified themselves as mid-level managers, and 160 identified themselves as executives. In the survey, 82 percent of manual laborers preferred trucks, 62 percent of non-manual wage earners preferred trucks, 54 percent of mid-level managers preferred trucks, and 26 percent of executives preferred trucks.

We are interested in finding the 95 percent confidence interval for the percentage of executives who prefer trucks. Define random variables X and P′ in words.

X is the number of successes where an executive prefers a truck. P′ is the percentage of executives sampled who prefer a truck.

Which distribution should you use for this problem?

Construct a 95 percent confidence interval. State the confidence interval, sketch the graph, and calculate the error bound.

CI: (0.19432, 0.33068)

View original figure (external; image not reproduced in this text edition)

Original figure description: This is a normal distribution curve. The peak of the curve coincides with the point 0.26 on the horizontal axis. A central region is shaded between points 0.1943 and 0.3307.

EBM: 0.0707

Suppose we want to lower the sampling error. What is one way to accomplish that?

The sampling error given in the survey is ±2 percent. Explain what the ±2 percent means.

The sampling error means that the true mean can be 2 percent above or below the sample mean.

Use the following information to answer the next five exercises: A poll of 1,200 voters asked what the most significant issue was in the upcoming election. Sixty-five percent answered "the economy." We are interested in the population proportion of voters who believe the economy is the most important.

Define the random variable X in words.

Define the random variable P′ in words.

P′ is the proportion of voters sampled who said the economy is the most important issue in the upcoming election.

Which distribution should you use for this problem?

Construct a 90 percent confidence interval, and state the confidence interval and the error bound.

CI: (0.62735, 0.67265);

EBM: 0.02265

What would happen to the confidence interval if the level of confidence were 95 percent?

Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 p.m., Monday night, ages 8 to 12, beginning ice-skating class is picked. In that class are 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.

What is being counted?

the number of girls, ages 8 to 12, in the 5 p.m. Monday night beginning ice-skating class

In words, define the random variable X.

Calculate the following:

x = _______
n = _______
p′ = _______

x = 64
n = 80
p′ = 0.8

State the estimated distribution of X. X~ ________

Define a new random variable P′. What is p′ estimating?

In words, define the random variable P′.

State the estimated distribution of P′. Construct a 92 percent confidence interval for the true proportion of girls in the ages 8 to 12 beginning ice-skating classes at the Ice Chalet.

$P^{'} ~ N (0.8, \sqrt{\frac{(0.8) (0.2)}{80}})$

CI = (0.72171, 0.87829).

How much area is in both tails (combined)?

How much area is in each tail?

0.04

Calculate the following:

lower limit
upper limit
error bound

The 92 percent confidence interval is _______.

(0.72; 0.88)

Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample proportion.

View original figure (external; image not reproduced in this text edition)

Original figure description: Normal distribution curve with two vertical upward lines from the x-axis to the curve. The confidence interval is between these two lines. The residual areas are on either side.

In one complete sentence, explain what the interval means.

With 92 percent confidence, we estimate the proportion of girls, ages 8 to 12, in a beginning ice-skating class at the Ice Chalet to be between 72 percent and 88 percent.

Using the same p′ and level of confidence, suppose that n were increased to 100. Would the error bound become larger or smaller? How do you know?

Using the same p′ and n = 80, how would the error bound change if the confidence level were increased to 98 percent? Why?

The error bound would increase. Assuming all other variables are kept constant, as the confidence level increases, the area under the curve corresponding to the confidence level becomes larger, which creates a wider interval and thus a larger error.

If you decreased the allowable error bound, why would the minimum sample size increase (keeping the same level of confidence)?

Homework

Insurance companies are interested in knowing the population percentage of drivers who always buckle up before riding in a car.

When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 95 percent confident that the population proportion is estimated to within 0.03?
If it were later determined that it was important to be more than 95 percent confident and a new survey was commissioned, how would that affect the minimum number you would need to survey? Why?

1,068
The sample size would need to be increased, because the critical value increases as the confidence level increases.

Suppose that the insurance companies did conduct a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up.

1. x = __________
2. n = __________
3. p′ = __________
Define the random variables X and P′ in words.
Which distribution should you use for this problem? Explain your choice.
Construct a 95 percent confidence interval for the population proportion who claim they always buckle up.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
If this survey were done by telephone, list three difficulties the companies might have in obtaining random results.

According to a recent survey of 1,200 people, 61 percent believe that the president is doing an acceptable job. We are interested in the population proportion of people who believe the president is doing an acceptable job.

Define the random variables X and P′ in words.
Which distribution should you use for this problem? Explain your choice.
Construct a 90 percent confidence interval for the population proportion of people who believe the president is doing an acceptable job.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.

X = the number of people who believe that the president is doing an acceptable job;

P′ = the proportion of people in a sample who believe that the president is doing an acceptable job.
$N (0.61, \sqrt{\frac{(0.61) (0.39)}{1200}})$
1. CI: (0.59, 0.63)
2. Check student’s solution.
3. EBM: 0.02

An article regarding dating and marriage recently appeared in a major newspaper. Of the 1,709 randomly selected adults, 315 identified themselves as ethnicity A, 323 identified themselves as ethnicity B, 254 identified themselves as ethnicity C, and 779 identified themselves as ethnicity D. In this survey, 86 percent of ethnicity B said that they would welcome a person of ethnicity A into their families. Among ethnicity C, 77 percent would welcome a person of ethnicity D into their families, 71 percent would welcome a person of ethnicity A, and 66 percent would welcome a person of ethnicity B.

We are interested in finding the 95 percent confidence interval for the percent of all ethnicity B adults who would welcome a person of ethnicity D into their families. Define the random variables X and P′ in words.
Which distribution should you use for this problem? Explain your choice.
Construct a 95 percent confidence interval.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.

Refer to the information in eip-idp49345824.

Construct three 95 percent confidence intervals:
1. percentage of all ethnicity C who would welcome a person of ethnicity D into their families
2. percentage of all ethnicity C who would welcome a person of ethnicity A into their families
3. percentage of all ethnicity C who would welcome a person of ethnicity B into their families
Even though the three point estimates are different, do any of the confidence intervals overlap? Which?
For any intervals that do overlap, in words, what does this imply about the significance of the differences in the true proportions?
For any intervals that do not overlap, in words, what does this imply about the significance of the differences in the true proportions?

1. (0.72, 0.82)
2. (0.65, 0.76)
3. (0.60, 0.72)
Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap.
We can say that there does not appear to be a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a Latino person into their families.
We can say that there is a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a black person into their families.

Stanford University conducted a study of whether running is healthy for men and women over age 50. During the first eight years of the study, 1.5 percent of the 451 members of the 50-Plus Fitness Association died. We are interested in the proportion of people over 50 who ran and died in the same eight year period.

Define the random variables X and P′ in words.
Which distribution should you use for this problem? Explain your choice.
Construct a 97 percent confidence interval for the population proportion of people over 50 who ran and died in the same 8-year period.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
Explain what a 97 percent confidence interval means for this study.

A telephone poll of 1,000 adult Americans was reported in an issue of a national magazine. One of the questions asked, “What is the main problem facing the country?” Twenty percent responded "crime". We are interested in the population proportion of adult Americans who believe that crime is the main problem.

Define the random variables X and P′ in words.
Which distribution should you use for this problem? Explain your choice.
Construct a 95 percent confidence interval for the population proportion of adult Americans who believe that crime is the main problem.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
Suppose we want to lower the sampling error. What is one way to accomplish that?
The sampling error given by the group of researchers who conducted the poll is ±3 percent. In one to three complete sentences, explain what the ±3 percent represents.

X = the number of adult Americans who believe that crime is the main problem; P′ = the proportion of adult Americans who believe that crime is the main problem.
Because we are estimating a proportion, that P′ = 0.2 and n = 1,000, the distribution we should use is $N (0.2, \sqrt{\frac{(0.2) (0.8)}{1000}})$ .
1. CI: (0.18, 0.22)
2. Check student’s solution.
3. EBM: 0.02
One way to lower the sampling error is to increase the sample size.
The stated ± 3 percent represents the maximum error bound. This means that those doing the study are reporting a maximum error of 3 percent. Thus, they estimate the percentage of adult Americans who the percentage of adult Americans who that crime is the main problem to be between 18 percent and 22 percent.

Refer to exercise111. Another question in the poll asked, “[How much are] you worried about the quality of education in our schools?” Sixty-three percent responded “a lot”. We are interested in the population proportion of adult Americans who are worried a lot about the quality of education in our schools.

Define the random variables X and P′ in words.
Which distribution should you use for this problem? Explain your choice.
Construct a 95 percent confidence interval for the population proportion of adult Americans who are worried a lot about the quality of education in our schools.
1. State the confidence interval.
2. Sketch the graph.
3. Calculate the error bound.
The sampling error given by the group of researchers who conducted the poll is ±3 percent. In one to three complete sentences, explain what the ±3 percent represents.

Use the following information to answer the next three exercises: According to a Field Poll, 79 percent of California adults (actual results are 400 out of 506 surveyed) believe that education and our schools is one of the top issues facing California. We wish to construct a 90 percent confidence interval for the true proportion of California adults who believe that education and the schools is one of the top issues facing California.

A point estimate for the true population proportion is _______.

0.90
1.27
0.79
400

A 90 percent confidence interval for the population proportion is _______.

(0.761, 0.820)
(0.125, 0.188)
(0.755, 0.826)
(0.130, 0.183)

The error bound is approximately _____.

1.581
0.791
0.059
0.030

Use the following information to answer the next two exercises: Five hundred eleven (511) homes in a certain southern California community are randomly surveyed to determine whether they meet minimal earthquake preparedness recommendations. One hundred seventy-three (173) of the homes surveyed meet the minimum recommendations for earthquake preparedness, and 338 do not.

Find the confidence interval at the 90 percent confidence level for the true population proportion of southern California community homes meeting at least the minimum recommendations for earthquake preparedness.

(0.2975, 0.3796)
(0.6270, 0.6959)
(0.3041, 0.3730)
(0.6204, 0.7025)

The point estimate for the population proportion of homes that do not meet the minimum recommendations for earthquake preparedness is ______.

0.6614
0.3386
173
338

On May 23, 2013, a polling group reported that of the 1,005 people surveyed, 76 percent of U.S. workers believe that they will continue working past retirement age. The confidence level for this study was reported at 95 percent with a ±3 percent margin of error.

Determine the estimated proportion from the sample.
Determine the sample size.
Identify CL and α.
Calculate the error bound based on the information provided.
Compare the error bound in Part d to the margin of error reported by the polling group. Explain any differences between the values.
Create a confidence interval for the results of this study.
A reporter is covering the release of this study for a local news station. How should she explain the confidence interval to her audience?

A national survey of 1,000 adults was conducted on May 13, 2013, by a group of researchers. It concluded with 95 percent confidence that 49 percent to 55 percent of Americans believe that big-time college sports programs corrupt the process of higher education.

Find the point estimate and the error bound for this confidence interval.
Can we (with 95 percent confidence) conclude that more than half of all American adults believe this?
Use the point estimate from Part a and n = 1,000 to calculate a 75 percent confidence interval for the proportion of American adults who believe that major college sports programs corrupt higher education.
Can we (with 75 percent confidence) conclude that at least half of all American adults believe this?

p′ = $\frac{(0 .55 + 0 .49)}{2}$ = 0.52; EBP = 0.55 0.52 = 0.03
No, the confidence interval includes values less than or equal to 0.50. It is possible that less than half of the population believe this.
CL = 0.75, so α = 1 – 0.75 = 0.25 and $\frac{α}{2} = 0.125 . z_{\frac{α}{2}} = 1.150$ . (The area to the right of this z is 0.125, so the area to the left is 1 – 0.125 = 0.875.)
$E B P = (1.150) \sqrt{\frac{0.52 (0.48)}{1, 000}} \approx 0.018$
(p′ - EBP, p′ + EBP) = (0.52 – 0.018, 0.52 + 0.018) = (0.502, 0.538)
Alternate Solution

STAT TESTS A: 1-PropZinterval with x = (0.52)(1,000), n = 1,000, CL = 0.75.

Answer is (0.502, 0.538).
Yes, this interval does not fall below 0.50, so we can conclude that at least half of all American adults believe that major sports programs corrupt education – but we do so with only 75 percent confidence.

A polling group recently conducted a survey asking adults across the United States about music preferences. When asked, 80 of the 571 participants download music weekly.

Create a 99 percent confidence interval for the true proportion of American adults who download music weekly.
This survey was conducted through automated telephone interviews on May 6 and 7, 2013. The error bound of the survey compensates for sampling error, or natural variability among samples. List some factors that could affect the survey’s outcome that are not covered by the margin of error.
Without performing any calculations, describe how the confidence interval would change if the confidence level decreased from 99 percent to 90 percent.

You plan to conduct a survey on your college campus to learn about the political awareness of students. You want to estimate the true proportion of college students on your campus who voted in the 2012 presidential election with 95 percent confidence and a margin of error no greater than 5 percent. How many students must you interview?

CL = 0.95; α = 1 – 0.95 = 0.05; $\frac{α}{2}$ = 0.025; $z_{\frac{α}{2}}$ = 1.96. Use p′ = q′ = 0.5.

$n = \frac{z_{\frac{α}{2}}^{2} p^{'} q^{'}}{E B P^{2}} = \frac{{1.96}^{2} (0.5) (0.5)}{{0.05}^{2}} = 384.16 .$

You need to interview at least 385 students to estimate the proportion to within 5 percent at 95 percent confidence.

In a recent poll, 9 of 48 respondents rated the likelihood of a certain event occurring in their community as likely or very likely. Use the plus-four method to create a 97 percent confidence interval for the proportion of American adults who believe that the event is likely or very likely. Explain what this confidence interval means in the context of the problem. A local poll in a New England town found that nine of 48 households think winter-proofing their cars is very important. Use the plus-four method to create a 97 percent confidence interval for the proportion of town residents who think winter-proofing their cars is very important. Explain what this confidence interval means in the context of this scenario.

Glossary

binomial distribution: a discrete random variable (RV) that arises from Bernoulli trials; there are a fixed number, n, of independent trials
Independent means that the result of any trial (for example, trial 1) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances, the binomial RV X is defined as the number of successes in n trials. The notation is X~B(n,p). The mean is μ = np, and the standard deviation is σ = $\sqrt{n p q}$ . The probability of exactly x successes in n trials is $P (X = x) = (\frac{n}{x}) p^{x} q^{n - x}$ .
error bound for a population proportion (EBP): the margin of error; depends on the confidence level, the sample size, and the estimated (from the sample) proportion of successes
plus-four confidence interval: plus-four confidence interval when you add two imaginary successes and two imaginary failures (four overall) to your sample

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