<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 8.2: Testing a Proportion Hypothesis

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Test a hypothesis about a population proportion by applying the binomial distribution approximation.
• Test a hypothesis about a population proportion using the P\begin{align*}P-\end{align*}value.

## Introduction

In the previous section we studied the test statistic that is used when you are testing hypotheses about the mean of a population and you have a large sample (>30)\begin{align*}(>30)\end{align*}.

Often statisticians are interest in making inferences about a population proportion. For example, when we look at election results we often look at the proportion of people that vote and who this proportion of voters choose. Typically, we call these proportions percentages and we would say something like “Approximately 68 percent of the population voted in this election and 48 percent of these voters voted for Barack Obama.”

So how do we test hypotheses about proportions? We use the same process as we did when testing hypotheses about populations but we must include sample proportions as part of the analysis. This lesson will address how we investigate hypotheses around population proportions and how to construct confidence intervals around our results.

### Hypothesis Testing about Population Proportions by Applying the Binomial Distribution Approximation

We could perform tests of population proportions to answer the following questions:

• What percentage of graduating seniors will attend a 4-year college?
• What proportion of voters will vote for John McCain?
• What percentage of people will choose Diet Pepsi over Diet Coke?

To test questions like these, we make hypotheses about population proportions. For example,

H0:35%\begin{align*}H_0: 35\%\end{align*} of graduating seniors will attend a 4-year college.

H0:42%\begin{align*}H_0:42\%\end{align*} of voters will vote for John McCain.

H0:26%\begin{align*}H_0:26\%\end{align*} of people will choose Diet Pepsi over Diet Coke.

To test these hypotheses we follow a series of steps:

• Hypothesize a value for the population proportion P\begin{align*}P\end{align*} like we did above.
• Randomly select a sample.
• Use the sample proportion p^\begin{align*}\hat{p}\end{align*} to test the stated hypothesis.

To determine the test statistic we need to know the sampling distribution of the sample proportion. We use the binomial distribution which illustrates situations in which two outcomes are possible (for example, voted for a candidate, didn’t vote for a candidate), remembering that when the sample size is relatively large, we can use the normal distribution to approximate the binomial distribution. The test statistic is

zz=sample estimatevalue under the null hypothesisstandard error under the null hypothesis=p^p0p0(1p0)n\begin{align*}z &= \frac{\text{sample estimate}-\text{value under the null hypothesis}}{\text{standard error under the null hypothesis}}\\ z &= \frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\end{align*}

where:

p0\begin{align*}p_0\end{align*} is the hypothesized value of the proportion under the null hypothesis

n\begin{align*}n\end{align*} is the sample size

Example: We want to test a hypothesis that 60 percent of the 400 seniors graduating from a certain California high school will enroll in a two or four-year college upon graduation. What would be our hypotheses and the test statistic?

Since we want to test the proportion of graduating seniors and we think that proportion is around 60 percent, our hypotheses are:

H0:pHa:p=.6.6\begin{align*}H_0: p &= .6\\ H_a: p & \neq .6\end{align*}

The test statistic would be z=p^.6.6(1.6)n\begin{align*}z=\frac{\hat{p}-.6}{\sqrt{\frac{.6(1-.6)}{n}}}\end{align*}. To complete this calculation we would have to have a value for the sample size (n).

### Testing a Proportion Hypothesis

Similar to testing hypotheses dealing with population means, we use a similar set of steps when testing proportion hypotheses.

• Determine and state the null and alternative hypotheses.
• Set the criterion for rejecting the null hypothesis.
• Calculate the test statistic.
• Decide whether to reject or fail to reject the null hypothesis.
• Interpret your decision within the context of the problem.

Example: A congressman is trying to decide on whether to vote for a bill that would legalize gay marriage. He will decide to vote for the bill only if 70 percent of his constituents favor the bill. In a survey of 300 randomly selected voters, 224 (74.6%) indicated that they would favor the bill. Should he or should he not vote for the bill?

First, we develop our null and alternative hypotheses.

H0:pHa:p=.7>.7\begin{align*}H_0: p &=.7\\ H_a: p &> .7\end{align*}

Next, we should set the criterion for rejecting the null hypothesis. Choose α=.05\begin{align*}\alpha=.05\end{align*} and since the null hypothesis is considering p>.7\begin{align*}p > .7\end{align*}, this is a one tailed test. Using a standard z\begin{align*}z\end{align*} table or the TI 83/84 calculator we find the critical value for a one tailed test at an alpha level of .05 to be 1.645.

The test statistic is z=.74.7.7(1.7)3001.51\begin{align*}z=\frac{.74-.7}{\sqrt{\frac{.7(1-.7)}{300}}} \approx1.51 \end{align*}

Since our critical value is 1.645 and our test statistic is 1. 51, we cannot reject the null hypothesis. This means that we cannot conclude that the population proportion is greater than .70 with 95 percent certainty. Given this information, it is not safe to conclude that at least 70 percent of the voters would favor this bill with any degree of certainty. Even though the proportion of voters supporting the bill is over 70 percent, this could be due to chance and is not statistically significant.

Example: Admission staff from a local university is conducting a survey to determine the proportion of incoming freshman that will need financial aid. A survey on housing needs, financial aid and academic interests is collected from 400 of the incoming freshman. Staff hypothesized that 30 percent of freshman will need financial aid and the sample from the survey indicated that 101 (25.3%) would need financial aid. Is this an accurate guess?

First, we develop our null and alternative hypotheses.

H0:pHa:p=.3.3\begin{align*}H_0: p &= .3\\ H_a: p & \neq .3\end{align*}

Next, we should set the criterion for rejecting the null hypothesis. The .05 alpha level is used and for a two tailed test the critical values of the test statistic are 1.96 and -1.96.

To calculate the test statistic:

z=.25.3.3(1.3)4002.18\begin{align*}z=\frac{.25-.3}{\sqrt{\frac{.3(1-.3)}{400}}} \approx -2.18\end{align*}

Since our critical values are ±1.96\begin{align*}\pm 1.96\end{align*} and 2.18<1.96\begin{align*}-2.18 < -1.96\end{align*} we can reject the null hypothesis. This means that we can conclude that the population of freshman needing financial aid is significantly more or less than 30 percent. Since the test statistic is negative, we can conclude with 95% certainty that in the population of incoming freshman, less than 30 percent of the students will need financial aid.

## Lesson Summary

In statistics, we also make inferences about proportions of a population. We use the same process as in testing hypotheses about populations but we must include hypotheses about proportions and the proportions of the sample in the analysis. To calculate the test statistic needed to evaluate the population proportion hypothesis, we must also calculate the standard error of the proportion which is defined as sp=p0(1p0)n\begin{align*}s_p=\sqrt{\frac{p_0(1-p_0)}{n}}\end{align*}

The formula for calculating the test statistic for a population proportion is

z=p^p0p0(1p0)n\begin{align*}z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\end{align*}

where:

p^\begin{align*}\hat{p}\end{align*} is the sample proportion

p0\begin{align*}p_0\end{align*} is the hypothesized population proportion

We can construct something called the confidence interval that specifies the level of confidence that we have in our results. The formula for constructing a confidence interval for the population proportion is p^±zα2(p^(1p^)n)\begin{align*}\hat{p} \pm z_{\frac{\alpha}{2}} \left(\frac{\hat{p}(1-\hat{p})}{n}\right)\end{align*}.

For an explanation on finding the mean and standard deviation of a sampling proportion, p, and normal approximation to binomials (7.0)(9.0)(15.0)(16.0), see American Public University, Sampling Distribution of Sample Proportion (8:24)

For a calculation of the z-statistic and associated P-Value for a 1-proportion test (18.0), see kbower50, Test of 1 Proportion: Worked Example (3:51)

## Review Questions

1. The test statistic helps us determine ___.
2. True or false: In statistics, we are able to study and make inferences about proportions, or percentages, of a population.
3. A state senator cannot decide how to vote on an environmental protection bill. The senator decides to request her own survey and if the proportion of registered voters supporting the bill exceeds 0.60, she will vote for it. A random sample of 750 voters is selected and 495 are found to support the bill.
1. What are the null and alternative hypotheses for this problem?
2. What is the observed value of the sample proportion?
3. What is the standard error of the proportion?
4. What is the test statistic for this scenario?
5. What decision would you make about the null hypothesis if you had an alpha level of .01?

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

Show Hide Details
Description
Tags:
Subjects: