9.12: Estimating a Population Proportion (1 of 3)

Last updated
Save as PDF

Page ID: 251397

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$

Learning Objectives

Construct a confidence interval to estimate a population proportion when conditions are met. Interpret the confidence interval in context.

Introduction

In “Estimating a Population Proportion,” we continue our discussion of estimating a population proportion with a confidence interval. Recall that the purpose of a confidence interval is to use a sample proportion to construct an interval of values that we can be reasonably confident contains the true population proportion.

The basic idea is summarized here:

When we select a random sample from the population of interest, we expect the sample proportion to be a good estimate of the population proportion. But we also know that sample proportions vary, so we expect some error. (Remember that the error here is due to chance. It is not due to a mistake that anyone made.)
For a given sample proportion, we will not know the amount of error, so we use the standard error as an estimate for the average amount of error we expect in sample proportions. (Recall that the standard error is the expected standard deviation of sample proportions when we take many, many random samples.)
If a normal model is a good fit for the sampling distribution, then about 95% of sample proportions estimate the population proportion within 2 standard errors. We say that we are 95% confident that the following interval contains the population proportion.

$\begin{array}{l}p\text{}&PlusMinus;\text{}\mathrm{margin}\text{}\mathrm{of}\text{}\mathrm{error}\\ p\text{}&PlusMinus;\text{}2(\mathrm{standard}\text{}\mathrm{error})\\ p\text{}&PlusMinus;\text{}2\sqrt{\frac{p(1-p)}{n}}\end{array}$

You may realize that this formula for the confidence interval is a bit odd, since our goal in calculating the confidence interval is to estimate the population proportion p. Yet the formula requires that we know p. In the section “Introduction to Statistical Inference,” we used an estimate for p from a previous study when calculating the confidence interval. This is not the usual way statisticians estimate the standard error, but it captured the main idea and allowed us to practice finding and interpreting confidence intervals. Now, we develop a different way to estimate standard error that is commonly used in statistical practice.

Example

Community College Students and Gender

According to a 2010 report from the American Council on Education, females make up 57% of the college population in the United States. Students in a statistics class at Tallahassee Community College want to determine the proportion of female students at TCC. They select a random sample of 135 TCC students and find that 72 are female, which is a sample proportion of 72 / 135 ≈ 0.533. So 53.3% of the students in the sample are female.

What can they conclude about the proportion of females at the college? How confident can they be in their estimate?

To answer these questions, we need to find a confidence interval.

Checking conditions:

We learned in Linking Probability to Statistical Inference that a confidence interval comes from a normal model of the sampling distribution. Let’s first make sure that a normal model is appropriate here. Recall the two conditions for using a normal model for sample proportions:

The sample must be random.
The expected number of successes in the sample, np, and the expected number of failures, n(1 – p), are both greater than or equal to 10. In symbols, this is np ≥ 10 and n(1 − p) ≥ 10. Recall that success doesn’t mean good and failure doesn’t mean bad. A success is just what we are counting.

When we try to check these conditions, we have a problem. We do not know p, the population proportion. In fact, p is what we are trying to estimate! So we cannot determine the expected number of successes and failures. Our solution to this problem is to adjust these conditions. Advanced theory tells us that if the actual number of successes and failures in the sample are greater than or equal to 10, then a normal model is still a good fit.

This sample contains 72 successes (female students) and 63 failures (male students). Both are greater than 10. We therefore use the normal model for the sampling distribution.

Finding the margin of error:

We know that a sample proportion is only an estimate for the population proportion. We do not expect the sample proportion to equal the population proportion, so there is some error due to random chance. We use the standard deviation of the sample proportions to describe the amount of error we can expect in random samples. We call this the standard error.

In Linking Probability to Statistical Inference, we learned that the standard error of the sample proportion depends on the population proportion and sample size. Here is the formula for the standard error:

$\sqrt{\frac{p(1-p)}{n}}$

When we use a normal model for the sampling distribution, 95% of sample proportions estimate the population proportion within approximately 2 standard errors. So the margin of error is the following:

$2\text{}\sqrt{\frac{p(1-p)}{n}}$

Now let’s calculate the margin of error for the TCC estimate of 53.3%. Notice that we have the same problem we had earlier. We don’t know p, the population proportion. So we can’t calculate the margin of error! Our solution to this problem is to estimate the standard error using the sample proportion in place of p. We call this the estimated standard error, and the formula is:

$\sqrt{\frac{\stackrel{ˆ}{p}(1-\stackrel{ˆ}{p})}{n}}$

For this example, the estimated standard error is

$\sqrt{\frac{0.533(1-0.533)}{135}}\text{}\approx \text{}0.043$

So the margin of error for the 95% confidence interval is:

$2\text{}\sqrt{\frac{0.533(1-0.533)}{135}}\text{}\approx \text{}2(0.043)\text{}=\text{}0.086$

Finding the confidence interval:

We can interpret the margin of error by saying we are 95% confident that the proportion of all students at TCC who are female is within 0.086 of our sample proportion of 0.533. We can then write the interval in the following form:

$\stackrel{ˆ}{p}\text{}&PlusMinus;\text{}\mathrm{margin}\text{}\mathrm{of}\text{}\mathrm{error}=0.533\text{}&PlusMinus;\text{}0.086$

When we add and subtract the margin of error from the sample proportion, the confidence interval is 0.447 to 0.619.

Conclusion:

We are 95% confident that the proportion of all TCC students who are female is between 0.447 and 0.619. We can also make this statement using percentages. We are 95% confident that the percentage of all TCC students who are female is between 44.7% and 61.9%.

Recall from Linking Probability to Statistical Inference that 95% confidence means this method captures the population proportion about 95% of the time.

Summary

Conditions for using the normal model of the sampling distribution:

In Linking Probability to Statistical Inference, we saw that a normal model describes the behavior of sample proportions if np ≥ 10 and n(1 − p) ≥ 10. These formulas say that the expected number of successes and failures in the sample must be 10 or greater. In Inference for One Proportion, we will never know the value of the population proportion p, so we estimate p with a sample proportion. Now we will assume that we can use a normal model if $n\stackrel{ˆ}{p}≥10\text{}\mathrm{and}\text{}n(1-\stackrel{ˆ}{p})≥10$ .

These formulas say that the actual number of successes and failures in the sample are 10 or greater.

The 95% confidence interval for estimating population proportion p :

In Linking Probability to Statistical Inference, we learned that the error in an estimate is related to the spread in the sampling distribution. We saw that the standard error of the sampling distribution of sample proportions is given by this formula:

$\sqrt{\frac{p(1-p)}{n}}$

In Inference for One Proportion, we are estimating the population proportion p. So we estimate the standard error by replacing p with the sample proportion, which affects the margin of error in the confidence interval. We have the following adjustment to the confidence interval formula:

$\stackrel{ˆ}{p}\text{}&PlusMinus;\text{}\mathrm{margin}\text{}\mathrm{of}\text{}\mathrm{error}=\stackrel{ˆ}{p}\text{}&PlusMinus;\text{}2\text{}\sqrt{\frac{\stackrel{ˆ}{p}(1-\stackrel{ˆ}{p})}{n}}$

Learn By Doing

Foothill College’s athletic department wants to calculate the proportion of students who have attended a women’s basketball game at the college. They use student email addresses, randomly choose 220 students, and email them. Of the 145 who responded, 22 had attended a women’s basketball game.

https://assessments.lumenlearning.co...sessments/3592

https://assessments.lumenlearning.co...sessments/3593

https://assessments.lumenlearning.co...sessments/3903