null hypothesis with proportions

9.1 Null and Alternative Hypotheses

The actual test begins by considering two hypotheses . They are called the null hypothesis and the alternative hypothesis . These hypotheses contain opposing viewpoints.

H 0 , the — null hypothesis: a statement of no difference between sample means or proportions or no difference between a sample mean or proportion and a population mean or proportion. In other words, the difference equals 0.

H a —, the alternative hypothesis: a claim about the population that is contradictory to H 0 and what we conclude when we reject H 0 .

Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data.

After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are reject H 0 if the sample information favors the alternative hypothesis or do not reject H 0 or decline to reject H 0 if the sample information is insufficient to reject the null hypothesis.

Mathematical Symbols Used in H 0 and H a :

H 0 always has a symbol with an equal in it. H a never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers use = in the null hypothesis, even with > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis.

Example 9.1

H 0 : No more than 30 percent of the registered voters in Santa Clara County voted in the primary election. p ≤ 30 H a : More than 30 percent of the registered voters in Santa Clara County voted in the primary election. p > 30

A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25 percent. State the null and alternative hypotheses.

Example 9.2

We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are the following: H 0 : μ = 2.0 H a : μ ≠ 2.0

We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : μ __ 66
• H a : μ __ 66

Example 9.3

We want to test if college students take fewer than five years to graduate from college, on the average. The null and alternative hypotheses are the following: H 0 : μ ≥ 5 H a : μ < 5

We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : μ __ 45
• H a : μ __ 45

Example 9.4

An article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third of the students pass. The same article stated that 6.6 percent of U.S. students take advanced placement exams and 4.4 percent pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6 percent. State the null and alternative hypotheses. H 0 : p ≤ 0.066 H a : p > 0.066

On a state driver’s test, about 40 percent pass the test on the first try. We want to test if more than 40 percent pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : p __ 0.40
• H a : p __ 0.40

Collaborative Exercise

Bring to class a newspaper, some news magazines, and some internet articles. In groups, find articles from which your group can write null and alternative hypotheses. Discuss your hypotheses with the rest of the class.

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5.5 - hypothesis testing for two-sample proportions.

We are now going to develop the hypothesis test for the difference of two proportions for independent samples. The hypothesis test follows the same steps as one group.

These notes are going to go into a little bit of math and formulas to help demonstrate the logic behind hypothesis testing for two groups. If this starts to get a little confusion, just skim over it for a general understanding! Remember we can rely on the software to do the calculations for us, but it is good to have a basic understanding of the logic!

We will use the sampling distribution of \(\hat{p}_1-\hat{p}_2\) as we did for the confidence interval.

For a test for two proportions, we are interested in the difference between two groups. If the difference is zero, then they are not different (i.e., they are equal). Therefore, the null hypothesis will always be:

\(H_0\colon p_1-p_2=0\)

Another way to look at it is \(H_0\colon p_1=p_2\). This is worth stopping to think about. Remember, in hypothesis testing, we assume the null hypothesis is true. In this case, it means that \(p_1\) and \(p_2\) are equal. Under this assumption, then \(\hat{p}_1\) and \(\hat{p}_2\) are both estimating the same proportion. Think of this proportion as \(p^*\).

Therefore, the sampling distribution of both proportions, \(\hat{p}_1\) and \(\hat{p}_2\), will, under certain conditions, be approximately normal centered around \(p^*\), with standard error \(\sqrt{\dfrac{p^*(1-p^*)}{n_i}}\), for \(i=1, 2\).

We take this into account by finding an estimate for this \(p^*\) using the two-sample proportions. We can calculate an estimate of \(p^*\) using the following formula:

\(\hat{p}^*=\dfrac{x_1+x_2}{n_1+n_2}\)

This value is the total number in the desired categories \((x_1+x_2)\) from both samples over the total number of sampling units in the combined sample \((n_1+n_2)\).

Putting everything together, if we assume \(p_1=p_2\), then the sampling distribution of \(\hat{p}_1-\hat{p}_2\) will be approximately normal with mean 0 and standard error of \(\sqrt{p^*(1-p^*)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\), under certain conditions.

\(z^*=\dfrac{(\hat{p}_1-\hat{p}_2)-0}{\sqrt{\hat{p}^*(1-\hat{p}^*)\left(\dfrac{1}{n_1}+\dfrac{1}{n_2}\right)}}\)

...will follow a standard normal distribution.

Finally, we can develop our hypothesis test for \(p_1-p_2\).

Hypothesis Testing for Two-Sample Proportions

Conditions :

\(n_1\hat{p}_1\), \(n_1(1-\hat{p}_1)\), \(n_2\hat{p}_2\), and \(n_2(1-\hat{p}_2)\) are all greater than five

Test Statistic:

\(z^*=\dfrac{\hat{p}_1-\hat{p}_2-0}{\sqrt{\hat{p}^*(1-\hat{p}^*)\left(\dfrac{1}{n_1}+\dfrac{1}{n_2}\right)}}\)

...where \(\hat{p}^*=\dfrac{x_1+x_2}{n_1+n_2}\).

The critical values, p-values, and decisions will all follow the same steps as those from a hypothesis test for a one-sample proportion.

Lecture 26: Inference about a population proportion

Corinne Riddell (modifications by Alan Hubbard and Tomer Altman)

November 1, 2024

• So far we’ve learned the z-test and the t-test that apply when the variable of interest is continuous
• We applied these tests to one-sample (e.g., \(H_0: \mu=8\) ) and two-sample settings (e.g., \(H_0: \mu_1 = \mu_2\) )
• Today, we will generalize these procedures to binary data, for which we estimate a proportion \(\hat{p}\) from a sample and use that as our best guess of the underlying population parameter \(p\)
• Notation: \(\bar{x}\) is to \(\mu\) as \(\hat{p}\) is to \(p\)
• Confidence interval for a proportion
• Sample size estimates for a proportion
• Hypothesis tests for a proportion

Recall the sampling distribution for \(\hat{p}\)

The sampling distribution for \(\hat{p}\) is centered on \(p\) with a standard error of \(\sqrt{\frac{p(1-p)}{n}}\)

If we follow the same format for the CI from previous chapters we would get:

\[\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\] This is what is known as the large sample confidence interval for a population proportion

• overly conservative (e.g., coverage significantly above 95%), or
• anti-conservative (less than 95 of the 95% confidence intervals would contain the true value for the proportion \(p\) )
• We typically say that such a method has poor coverage because it contains the true probability \(p\) less than the specified amount - (typically less than 95%).
• To overcome this, we will modify how we calculate the confidence interval slightly using what is known as the “plus four” method

Introducing: the “Plus Four” Method

If you add two imaginary successes and two failures to the data set (increasing the sample size by four imaginary trials), the interval can have better performance

Let \(\tilde{p} = \frac{\text{number of successes + 2}}{n+4}\)

• Let \(SE = \sqrt{\frac{\tilde{p}(1-\tilde{p})}{n+4}}\)
• Then the CI is: \[\tilde{p} \pm z^* \sqrt{\frac{\tilde{p}(1-\tilde{p})}{n+4}}\]

This is called the “Plus Four” Method

Note we use \(z^*\) rather than \(t^*\) . This is because the standard error of the sampling distribution is completely determined by \(p\) and \(n\) , we don’t need to estimate a second parameter.

In addition, for smaller samples (when one can rely on the CLT), one can rely on alternative methods to get inference that do not rely on normality of the \(\hat{p}\)

Use this method when \(n \geq 10\)

Why does the “Plus Four” Method work?

It is a simplification of a more complex method known as the Wilson Score Interval

You don’t need to know why it works, just that it is better to use this “plus four” trick if you’re making a confidence interval for a proportion by hand

Note: if the number of successes and failures is relatively large, then the “plus four” method will converge with the large sample CI method

Two methods so far…

We have so far introduced the large sample method to calculate the CI for p:

\[\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\] And the plus four method to calculate the CI for p:

• \(\tilde{p} = \frac{\text{number of successes + 2}}{n+4}\)

We are going to talk about two more methods.

What does R use?

R has two functions to calculate confidence intervals for proportions

The first function is prop.test (analogous to t.test ) to calculate confidence intervals and hypothesis tests for binomial proportions

This function uses the “Wilson score interval with a continuity correction”. Thus, when you use the prop.test function, you don’t need to “plus 4”, it will do it for you (and does an even better job because of the continuity correction) .

prop.test() corresponds to the Wilson score method. You do not need to know how to calculate the Wilson score method by hand. You only need to know how to use R to perform this method.

There is a fourth method to compute confidence intervals for proportions that is often used called the Clopper Pearson method , also known as the “Exact method” . It is implemented with the second R function, binom.test() .

The exact method is statistically conservative, meaning that it gives better coverage than it suggests. That is, a 95% CI computed under this method includes the true proportion in the interval more than 95% of the time.

It can be thought of as an inversion of hypothesis testing

• It finds the set of all values of the parameter \(p_{null}\) , where we consider the null value in a two-sided hypothesis test, would lead to a p-value \(p \ge 0.05\)
• It calculates the p-value using the binomial distribution function
• As a reminder, this was the cumulative probability function for binomial (Lec 16):

\[ P(X \leq x)=\sum_{i=0}^x {n \choose i}p_{null}^i(1-p_{null})^{n-i}\]

• Generally, exact methods are considered the gold-standard for making confidence intervals as they provide the stated coverage with no asymptotic assumptions (that is, do not rely on normal sampling distributions for the estimator, in this case, \(\hat{p}\) )

Example applying all the methods

Suppose that 500 elderly individuals suffered hip fractures, of which 100 died within a year of their fracture. Compute the 95% CI for the proportion who died using:

• the large sample method
• the “plus four” method (by hand)
• the Wilson Score method (using prop.test )
• the Clopper Pearson Exact method (using binom.test )

Example of large sample method to calculate the CI for a proportion (by hand)

• Our estimate for the proportion is \(\hat{p}=20\%\)
• Using the large sample method, the 95% confidence interval is 16.5% to 23.5%
• Remember, this method has poor coverage, meaning that fewer than 95 of the 100 intervals we could make would contain the true value \(p\)

Example using the “plus four” method to calculate the CI for a proportion (by hand)

Using the plus 4 method, the confidence interval is 16.7% to 23.7%.

Example using the Wilson Score method to calculate the CI for a proportion (using R)

• The 95% confidence interval using the Wilson Score method is 16.6% to 23.8%
• Note that the prop.test function is also conducting a two-sided hypothesis test (where \(H_0: p_0=0.5\) , unless otherwise specified)
• You can ignore the testing-related output and focus on the CI output when using the function to make a CI

Example using the Clopper Pearson “Exact” method to calculate the CI for a proportion (using R)

• The 95% confidence interval using the exact binomial test is 16.6% to 23.8%
• It has larger coverage (contains the true value more often) which necessitates a wider interval
• Note that the binom.test function is also conducting a two-sided hypothesis test (where \(H_0: p_0=0.5\) , unless otherwise specified)

Summary of the confidence intervals across the methods

*with continuity correction

**also known as the exact method

• Only the large sample method is symmetric around \(\hat{p} = 20\%\) . This is okay. Symmetric confidence intervals are applicable to Normal sampling distributions, which might or might not describe the distribution of \(\hat{p}\) .
• Non-symmetric CIs make sense because \(p\) is bounded between 0 and 1. For example, if \(p\) is very small, say 0.012, you would not want a CI that has a lower bound which is negative, this would not make sense.
• When the Normal approximation assumptions are satisfied, the methods give very similar results.

Another example of the “plus four” method (by hand)

[We are including another example for you to read so you can practice working out the calculations by hand.]

A study examined a random sample of 75 SARS patients, of which 64 developed recurrent fever.

Therefore \(\hat{p}=64/75=85.33\%\)

Using the plus 4 method: \(\tilde{p}=\frac{64 + 2}{75+4} = 83.54\%\)

\[SE=\sqrt{\frac{\tilde{p}(1-\tilde{p})}{75+4}}=\sqrt\frac{{.8354\times {(1- 0.8354)}}}{79} = 0.04172\] Thus the plus four 95% CI is: \(\tilde{p} \pm 1.96 \times SE = 0.8354 \pm 0.04172 = 79.37\% \text{ to } 87.71\%\)

How big should the sample be to estimate a proportion?

Suppose that you want to estimate a sample size for a proportion within a given margin of error. That is, you want to put a maximum bound on the width of the corresponding confidence interval.

Let \(m\) denote the desired margin of error. Then \(m = z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)

We can solve this equation for \(n\) , but we also need to plug in a value for \(p\) . To do that we make a guess for \(p\) denoted by \(p^*\) .

\(p^*\) is your best estimate for the underlying proportion. You might gather this estimate from a completed pilot study or based on previous studies published by someone else. If you have no best guess, you can use \(p^*=0.5\) . This will produce the most conservative estimate of \(n\) . However if the true \(p\) is less than 0.3 or greater than 0.7, the sample size estimated may be much larger than you need.

Rearranging the formula on the last slide for \(n\) , we get:

\(m = z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)

\(\sqrt{n}m=z^*\sqrt{p(1-p)}\)

\(\sqrt{n}=\frac{z^*}{m}\sqrt{p(1-p)}\)

\(n = (\frac{z^*}{m})^2p^*(1-p^*)\)

This last formula is the one we will use to estimate the required sample size.

Example of estimating sample size

Suppose after the general election, you were interested in estimating the number of STEM undergraduate students who voted. So you want to do a study to estimate this proportion. How many students should you include in your sample?

First you need to decide what margin of error you desire. Suppose it is 4 percentage points or \(m=0.04\) for a 95% CI.

If you had no idea what proportion of STEM students voted then you let \(p^*=0.5\) and solve for \(n\) :

\(n = (\frac{z^*}{m})^2p^*(1-p^*) = (\frac{1.96}{0.04})^2\times 0.5 \times 0.5 = 600.25 = 601\)

This implies you would need to sample 601 students to get an estimate with a 95% confidence interval that is +/- 4 percentage points.

However, suppose you found a previous study that estimated the number of STEM students who voted to be 25%. Then what sample size would you need to detect this proportion?

\(n = (\frac{z^*}{m})^2p^*(1-p^*) = (\frac{1.96}{0.04})^2\times 0.25 \times 0.75 = 450.19 = 451\)

What if you want the width of the 95% confidence interval to be 6 percentage points. What would \(m\) be in this case?

The width of the 95% CI is equal to twice the margin of error. So if you want the width to be 0.06, then this is equivalent to saying you want a margin of error of 0.03.

Hypothesis tests of a proportion

When you only have one sample what is the null hypothesis? You’re interested in knowing whether there is evidence against the null hypothesis that the population proportion \(p\) is equal to some specified value \(p_0\) . That is:

\[H_0: p = p_0\]

For example, you may want to test whether there is evidence against the null hypothesis that \(p = 0.25\) .

Recall the sampling distribution for the proportion:

• Normally distributed
• Centered at \(p_0\) under the null distribution
• Has a standard error of \(\sqrt{\frac{p_0(1-p_0)}{n}}\)

The test statistic for the null hypothesis is:

\[z = \frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\] This is a z-test (not a t-test) so we compared to the standard Normal distribution and ask what is the probability of observing a \(z\) value of this magnitude (or more extreme).

One sided alternatives:

• \(H_a: p > p_0\)
• \(H_a: p < p_0\)

Two-sided alternative:

• \(H_a: p \ne p_0\)

When to use this test? Use this test when the expected number of successes and failures is \(\geq\) 10. That is, when \(np_0\geq10\) and \(n(1-p_0)\geq 10\) .

Example of a hypothesis test for a proportion

Consider a SRS of 200 patients undergoing treatment to alleviate side effects from a rigorous drug regimen at a particular hospital, where 33 patients experienced reduced or no side effects.

\(\hat{p}=33/200=0.165=16.5\%\)

Suppose that historically, the rate of patients with little or no side effects is 10%. Does the new treatment increase the rate? That is:

\(H_0: p = 0.10\)

\(H_a: p > 0.10\)

Step 1: Calculate \(\hat{p} = 16.5\%\) from previous slide.

Step 2: Calculate the standard error of the sampling distribution for \(p\) under the null hypothesis: \(SE = \sqrt{\frac{p_0(1-p_0)}{n}} = \sqrt{\frac{0.1(1-0.1)}{200}} = 0.0212132\)

Step 3: Calculate the z-test for the proportion:

\(z = \frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} = \frac{0.165-0.10}{0.0212132} = 3.06413\)

Step 4: Calculate the probability of seeing a z-value of this magnitude or larger :

Step 5: Evaluate the evidence against the null hypothesis. Because the p-value is so small (0.1%), there is little chance of seeing a proportion equal to 16.5% or larger if the true proportion is actually 10%. Thus, there is evidence in favor of the alternative hypothesis, that the underlying proportion is larger than 10%.

Another Example

Suppose that there were 100 elderly individuals with falls observed, and 2 died. Here are the 95% CIs applying the four different methods:

Elderly falls example

We can graphically compare the CIs from the previous slide:

• Notice how different the intervals are, especially large sample vs. others.
• Notice that the large sample lower bound is non-sensical (i.e., we can’t have negative proportions!)
• The large sample CI differs from the others because the Normal approximation assumptions are not satisfied.
• Here is the code to arrive at the estimates for the previous table:

Teach yourself statistics

Hypothesis Test for a Proportion

This lesson explains how to conduct a hypothesis test of a proportion, when the following conditions are met:

• The sampling method is simple random sampling .
• Each sample point can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure.
• The sample includes at least 10 successes and 10 failures.
• The population size is at least 20 times as big as the sample size.

This approach consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results.

State the Hypotheses

Every hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis . The hypotheses are stated in such a way that they are mutually exclusive. That is, if one is true, the other must be false; and vice versa.

Formulate an Analysis Plan

The analysis plan describes how to use sample data to accept or reject the null hypothesis. It should specify the following elements.

• Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used.
• Test method. Use the one-sample z-test to determine whether the hypothesized population proportion differs significantly from the observed sample proportion.

Analyze Sample Data

Using sample data, find the test statistic and its associated P-Value.

σ = sqrt[ P * ( 1 - P ) / n ]

z = (p - P) / σ

• P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a z-score, use the Normal Distribution Calculator to assess the probability associated with the z-score. (See sample problems at the end of this lesson for examples of how this is done.)

Interpret Results

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P-value to the significance level , and rejecting the null hypothesis when the P-value is less than the significance level.

Test Your Understanding

In this section, two hypothesis testing examples illustrate how to conduct a hypothesis test of a proportion. The first problem involves a a two-tailed test; the second problem, a one-tailed test.

Sample Size Calculator

As you probably noticed, the process of testing a hypothesis about a proportion can be complex. Stat Trek's Sample Size Calculator can do the same job quickly and easily. When you need to test a hypothesis, consider using the Sample Size Calculator. The calculator is free. It can found in the Stat Trek main menu under the Stat Tools tab. Or you can tap the button below.

Problem 1: Two-Tailed Test

The CEO of a large electric utility claims that 80 percent of his 1,000,000 customers are very satisfied with the service they receive. To test this claim, the local newspaper surveyed 100 customers, using simple random sampling. Among the sampled customers, 73 percent say they are very satisified. Based on these findings, can we reject the CEO's hypothesis that 80% of the customers are very satisfied? Use a 0.05 level of significance.

Solution: The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P = 0.80

Alternative hypothesis: P ≠ 0.80

• Formulate an analysis plan . For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test .

σ = sqrt [(0.8 * 0.2) / 100]

σ = sqrt(0.0016) = 0.04

z = (p - P) / σ = (.73 - .80)/0.04 = -1.75

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a two-tailed test , the P-value is the probability that the z-score is less than -1.75 or greater than 1.75. We use the Normal Distribution Calculator to find P(z < -1.75) = 0.04. Since the standard normal distribution is symmetric with a mean of zero, we know that P(z > 1.75) = 0.04. Thus, the P-value = 0.04 + 0.04 = 0.08.

• Interpret results . Since the P-value (0.08) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the sample included at least 10 successes and 10 failures, and the population size was at least 10 times the sample size.

Problem 2: One-Tailed Test Suppose the previous example is stated a little bit differently. Suppose the CEO claims that at least 80 percent of the company's 1,000,000 customers are very satisfied. Again, 100 customers are surveyed using simple random sampling. The result: 73 percent are very satisfied. Based on these results, should we accept or reject the CEO's hypothesis? Assume a significance level of 0.05.

Null hypothesis: P >= 0.80

Alternative hypothesis: P < 0.80

σ = sqrt[ P * ( 1 - P ) / n ] = sqrt [(0.8 * 0.2) / 100]

• Interpret results . Since the P-value (0.04) is less than the significance level (0.05), we cannot accept the null hypothesis.