8.1.2 - Hypothesis Testing

8.1.2 - Hypothesis Testing

A hypothesis test for a proportion is used when you are comparing one group to a known or hypothesized population proportion value. In other words, you have one sample with one categorical variable. The hypothesized value of the population proportion is symbolized by \(p_0\) because this is the value in the null hypothesis (\(H_0\)).

If \(np_0 \ge 10\) and \(n(1-p_0) \ge 10\) then the distribution of sample proportions is approximately normal and can be estimated using the normal distribution. That sampling distribution will have a mean of \(p_0\) and a standard deviation (i.e., standard error) of \(\sqrt{\frac{p_0 (1-p_0)}{n}}\)

Recall that the standard normal distribution is also known as the z distribution. Thus, this is known as a "single sample proportion z test" or "one sample proportion z test." 

If \(np_0 < 10\) or \(n(1-p_0) < 10\) then the distribution of sample proportions follows a binomial distribution. We will not be conducting this test by hand in this course, however you will learn how this can be conducted using Minitab Express using the exact method.


8.1.2.1 - Normal Approximation Method Formulas

8.1.2.1 - Normal Approximation Method Formulas

Here we will be using the five step hypothesis testing procedure to compare the proportion in one random sample to a specified population proportion using the normal approximation method.

1. Check assumptions and write hypotheses

In order to use the normal approximation method, the assumption is that both \(n p_0 \geq 10\) and \(n (1-p_0) \geq 10\). Recall that \(p_0\) is the population proportion in the null hypothesis.

Research Question Is the proportion different from \(p_0\)? Is the proportion greater than \(p_0\)? Is the proportion less than \(p_0\)?
Null Hypothesis, \(H_{0}\) \(p=p_0\) \(p= p_0\) \(p= p_0\)
Alternative Hypothesis, \(H_{a}\) \(p\neq p_0\) \(p> p_0\) \(p< p_0\)
Type of Hypothesis Test Two-tailed, non-directional Right-tailed, directional Left-tailed, directional

Where \(p_0\) is the hypothesized population proportion that you are comparing your sample to.

2. Calculate the test statistic

When using the normal approximation method we will be using a z test statistic. The z test statistic tells us how far our sample proportion is from the hypothesized population proportion in standard error units. Note that this formula follows the basic structure of a test statistic that you learned last week: \(test\;statistic=\frac{sample\;statistic-null\;parameter}{standard\;error}\)

Test statistic: One Group Proportion
\(z=\frac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}}\)

\(\widehat{p}\) = sample proportion
\(p_{0}\) = hypothesize population proportion
\(n\) = sample size

3. Determine the p-value

Given that the null hypothesis is true, the p value is the probability that a randomly selected sample of n would have a sample proportion as different, or more different, than the one in our sample, in the direction of the alternative hypothesis. We can find the p value by mapping the test statistic from step 2 onto the z distribution. 

Note that p-values are also symbolized by \(p\). Do not confuse this with the population proportion which shares the same symbol.

We can look up the \(p\)-value using Minitab Express by constructing the sampling distribution.  Because we are using the normal approximation here, we have a \(z\) test statistic that we can map onto the \(z\) distribution. Recall, the z distribution is a normal distribution with a mean of 0 and standard deviation of 1. If we are conducting a one-tailed (i.e., right- or left-tailed) test, we look up the area of the sampling distribution that is beyond our test statistic. If we are conducting a two-tailed (i.e., non-directional) test there is one additional step: we need to multiple the area by two to take into account the possibility of being in the right or left tail. 

4. Make a decision

We can decide between the null and alternative hypotheses by examining our p-value. If \(p \leq \alpha\) reject the null hypothesis. If \(p>\alpha\) fail to reject the null hypothesis. Unless stated otherwise, assume that \(\alpha=.05\).

When we reject the null hypothesis our results are said to be statistically significant.

5. State a "real world" conclusion

Based on our decision in step 4, we will write a sentence or two concerning our decision in relation to the original research question.


8.1.2.1.1 - Video Example: Male Babies

8.1.2.1.1 - Video Example: Male Babies

8.1.2.1.2 - Example: Handedness

8.1.2.1.2 - Example: Handedness

Research Question: Are more than 80% of American's right handed?

In a sample of 100 Americans, 87 were right handed.

1. Check assumptions and write hypotheses

\(np_0 = 100(0.80)=80\)

\(n(1-p_0) = 100 (1-0.80) = 20\)

Both \(np_0\) and \(n(1-p_0)\) are at least 10 so we can use the normal approximation method. 

This is a right-tailed test because we want to know if the proportion is greater than 0.80.

\(H_{0}\colon p=0.80\)
\(H_{a}\colon p>0.80\)

2. Calculate the test statistic
Test statistic: One Group Proportion

\(z=\dfrac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}}\)

\(\widehat{p}\) = sample proportion
\(p_{0}\) = hypothesize population proportion
\(n\) = sample size

\(\widehat{p}=\dfrac{87}{100}=0.87\), \(p_{0}=0.80\), \(n=100\)

\(z= \dfrac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}}= \dfrac{0.87-0.80}{\sqrt{\frac{0.80 (1-0.80)}{100}}}=1.75\)

Our \(z\) test statistic is 1.75.

3. Determine the p-value associated with the test statistic

This is a right-tailed test so we need to find the area to the right of the test statistic, \(z=1.75\), on the z distribution.

Using Minitab Express, we find the probability \(P(z\geq1.75)=0.0400592\) which may be rounded to \(p\; value=0.0401\).

Distribution plot of Density vs X - Normal, Mean=0, StDev=1

4. Make a decision

\(p\leq .05\), therefore our decision is to reject the null hypothesis

5. State a "real world" conclusion

Yes, there is statistical evidence to state that more than 80% of all Americans are right handed.


8.1.2.1.3 - Example: Ice Cream

8.1.2.1.3 - Example: Ice Cream

Research Question: Is the percentage of Creamery customers who prefer chocolate ice cream over vanilla less than 80%?

In a sample of 50 customers 60% preferred chocolate over vanilla.

1. Check assumptions and write hypotheses

\(np_0 = 50(0.80) = 40\)

\(n(1-p_0)=50(1-0.80) = 10\)

Both \(np_0\) and \(n(1-p_0)\) are at least 10. We can use the normal approximation method.

This is a left-tailed test because we want to know if the proportion is less than 0.80.

\(H_{0}\colon p=0.80\)
\(H_{a}\colon p<0.80\)

2. Calculate the test statistic
Test statistic: One Group Proportion

\(z=\dfrac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}}\)

\(\widehat{p}\) = sample proportion
\(p_{0}\) = hypothesize population proportion
\(n\) = sample size

\(\widehat{p}=0.60\), \(p_{0}=0.80\), \(n=50\)

\(z= \dfrac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}}= \dfrac{0.60-0.80}{\sqrt{\frac{0.80 (1-0.80)}{50}}}=-3.536\)

Our \(z\) test statistic is -3.536.

3. Determine the p-value

This is a left-tailed test so we need to find the area to the right of our test statistic, \(z=-3.536\).

Distribution Plot of Density vs X - Normal, Mean=0, StDev=1

From the Minitab Express output above, the p-value is 0.0002031

4. Make a decision

\(p \leq.05\), therefore our decision is to reject the null hypothesis.

5. State a "real world" conclusion

Yes, there is evidence that the percentage of all Creamery customers who prefer chocolate ice cream over vanilla is less than 80%.


8.1.2.1.4 - Example: Overweight Citizens

8.1.2.1.4 - Example: Overweight Citizens

According to the Center for Disease Control (CDC), the percent of adults 20 years of age and over in the United States who are overweight is 69.0% (see http://www.cdc.gov/nchs/fastats/obesity-overweight.htm). One city’s council wants to know if the proportion of overweight citizens in their city is different from this known national proportion. They take a random sample of 150 adults 20 years of age or older in their city and find that 98 are classified as overweight. Let’s use the five step hypothesis testing procedure to determine if there is evidence that the proportion in this city is different from the known national proportion.

1. Check assumptions and write hypotheses

\(np_0 =150 (0.690)=103.5 \)

\(n (1-p_0) =150 (1-0.690)=46.5\)

Both \(n p_0\) and \(n (1-p_0)\) are at least 10, this assumption has been met.

Research question: Is this city’s proportion of overweight individuals different from 0.690?

This is a non-directional test because our question states that we are looking for a differences as opposed to a specific direction. This will be a two-tailed test.

\(H_{0}\colon p=0.690\)
\(H_{a}\colon p\neq 0.690\)

2. Calculate the test statistic
Test statistic: One Group Proportion

\(z=\dfrac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}}\)

\(\widehat{p}\) = sample proportion
\(p_{0}\) = hypothesize population proportion
\(n\) = sample size

\(\widehat{p}=\dfrac{98}{150}=.653\)

\( z =\dfrac{0.653- 0.690 }{\sqrt{\frac{0.690 (1- 0.690)}{150}}} = -0.980 \)

Our test statistic is \(z=-0.980\)

3. Determine the p-value

This is a non-directional (i.e., two-tailed) test, so we need to find the area under the z distribution that is more extreme than \(z=-0.980\). 

 

In Minitab Express, we find the proportion of a normal curve beyond \(\pm0.980\):

Distribution Plot of Density vs X - Normal, Mean=0, StDev=1

\(p-value=0.163543+0.163543=0.327086\)

4. Make a decision

\(p>\alpha\), therefore we fail to reject the null hypothesis

5. State a "real world" conclusion

There is not sufficient evidence to state that the proportion of citizens of this city who are overweight is different from the national proportion of 0.690.


8.1.2.2 - Minitab Express: Hypothesis Tests for One Proportion

8.1.2.2 - Minitab Express: Hypothesis Tests for One Proportion

A hypothesis test for one proportion can be conducted in Minitab Express. This can be done using raw data or summarized data.

  • If you have a data file with every individual's observation, then you have raw data.
  • If you do not have each individual observation, but rather have the sample size and number of successes in the sample, then you have summarized data.

The next two pages will show you how to use Minitab Express to conduct this analysis using either raw data or summarized data.

Note that the default method for constructing the sampling distribution in Minitab Express is to use the exact method.  If \(np_0 \geq 10\) and \(n(1-p_0) \geq 10\) then you will need to change this to the normal approximation method.  This must be done manually. Minitab Express will use the method that you select, it will not check assumptions for you!


8.1.2.2.1 - Minitab Express: 1 Proportion z Test, Raw Data

8.1.2.2.1 - Minitab Express: 1 Proportion z Test, Raw Data

If you have data in a Minitab Express worksheet, then you have what we call "raw data."  This is in contrast to "summarized data" which you'll see on the next page.

In order to use the normal approximation method both \(np_0 \geq 10\) and \(n(1-p_0) \geq 10\). Before we can conduct our hypothesis test we must check this assumption to determine if the normal approximation method or exact method should be used. This must be checked manually. Minitab Express will not check assumptions for you.

In the example below, we want to know if there is evidence that the proportion of students who are male is different from 0.50.

\(n=226\) and \(p_0=0.50\)

\(np_0 = 226(0.50)=113\) and \(n(1-p_0) = 226(1-0.50)=113\)

Both \(np_0 \geq 10\) and \(n(1-p_0) \geq 10\) so we can use the normal approximation method. 

MinitabExpress  – Conducting a One Sample Proportion z Test: Raw Data

Research question: Is the proportion of students who are male different from 0.50?

  1. Open Minitab Express file:
  2. On a PC: Select STATISTICS > One Sample > Proportion
    On a Mac: Select Statistics > 1-Sample Inference > Proportion 
  3. Double-click the variable Biological Sex to insert it into the Sample box
  4. Check the box next to Perform hypothesis test and enter 0.50 in the Hypothesized proportion box
  5. Click the Options tab
  6. Use the default Alternative hypothesis setting of Proportion ≠ hypothesized value 
  7. Use the default Confidence level of 95
  8. Select Normal approximation method
  9. Click OK

The result should be the following output:

Z-Value P-Value
-1.86 0.0625

95% CI for the Proportion

1-Sample Proportion: Biological Sex
Method
Event: Biological Sex = Male
p: proportion where Biological Sex = Male
Normal approximation is used for this analysis.
Descriptive Statistics
N Event Sample p 95% CI for p
226 99 0.438053 (0.373368, 0.502738)
Test
Null hypothesis H 0: p = 0.5
Alternative hypothesis H 1: p ≠ 0.5
Video Walkthrough

Select your operating system below to see a step-by-step guide for this example.

We could summarize these results using the five-step hypothesis testing procedure:

1. Check assumptions and write hypotheses

\(np_0 = 226(0.50)=113\) and \(n(1-p_0) = 226(1-0.50)=113\) therefore the normal approximation method will be used.

 \(H_0: p = 0.50\)

 \(H_a: p \ne 0.50\)

2. Calculate the test statistic

From the Minitab Express output, \(z\) = -1.86

3. Determine the p-value

From the Minitab Express output, \(p\) = 0.0625

4. Make a decision

\(p > \alpha\), fail to reject the null hypothesis

5. State a "real world" conclusion

There is NOT evidence that the proportion of all students in the population who are male is different from 0.50.


8.1.2.2.2 - Minitab Express: 1 Sample Proportion z test, Summary Data

8.1.2.2.2 - Minitab Express: 1 Sample Proportion z test, Summary Data

The following example uses a scenario in which we want to know if the proportion of college women who think they are overweight is less than 40%. We collect data from a random sample of 129 college women and 37 said that they think they are overweight.

First, we should check assumptions to determine if the normal approximation method or exact method should be used:

\(np_0=129(0.40)=51.6\) and \(n(1-p_0)=129(1-0.40)=77.4\) both values are at least 10 so we can use the normal approximation method.

MinitabExpress  – Performing a One Proportion z Test with Summarized Data

To perform a one sample proportion z test with summarized data in Minitab Express:

  1. Open Minitab Express without data
  2. On a PC: In the menu bar select STATISTICS > One Sample > Proportion
  3. On a Mac: In the menu bar select Statistics > 1-Sample Inference > Proportion
  4. From the drop-down menu change Sample data in a column to Summarized data
  5. For Number of events enter 37 and for Number of trials enter 129
  6. Check the box for Perform hypothesis test
  7. In the Hypothesized proportion box enter 0.40
  8. Click the Options tab
  9. Change the Alternative hypothesis to Proportion < hypothesized value
  10. Use the default Confidence level of 95
  11. Change the Method to Normal approximation
  12. Click OK

This should result in the following output:

1-Sample Proportion
Method
p: event proportion
Normal approximation is used for this analysis.
Descriptive Statistics
N Event Sample p 95% Upper Bound for p
129 37 0.286822 0.352321
Test
Null hypothesis H 0: p = 0.4
Alternative hypothesis H 1: p < 0.4
Z-Value P-Value
-2.62 0.0043

95% Upper Bound for the Proportion

Video Walkthrough

Select your operating system below to see a step-by-step guide for this example.

We could summarize these results using the five-step hypothesis testing procedure:

1. Check assumptions and write hypotheses

\(np_0=129(0.40)=51.6\) and \(n(1-p_0)=129(1-0.40)=77.4\) both values are at least 10 so we can use the normal approximation method.

\(H_0: p = 0.40\)

\(H_a: p < 0.40\)

2. Calculate the test statistic

From output, \(z\) = -2.62

3. Determine the p-value

From output, \(p\) = 0.0043

4. Make a decision

\(p \leq \alpha\), reject the null hypothesis

5. State a "real world" conclusion

There is evidence that the proportion of women in the population who think they are overweight is less than 40%.


8.1.2.2.2.1 - Video Example: Gym Members (Normal Approx. Method)

8.1.2.2.2.1 - Video Example: Gym Members (Normal Approx. Method)

Research question: Are less than 50% of all individuals with a membership at one gym female?

A simple random sample of 60 individuals with a membership at one gym was collected. Each individual's biological sex was recorded. There were 24 females. 


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