Lesson 11: Chi-Square Tests

Lesson 11: Chi-Square Tests

Objectives

Upon successful completion of this lesson, you should be able to:

  • Construct a chi-square probability distribution plot in StatKey or Minitab Express
  • Determine when a chi-square goodness-of-fit test and chi-square test of independence should be conducted
  • Compute and interpret expected counts
  • Conduct chi-square tests by hand and using Minitab Express
  • Calculate and interpret relative risk

In this lesson we will learn how to compare the proportions of more than two independent groups and how to test for a relationship between two categorical variables. In this lesson you will also review risk and odds and learn about relative risk.

Before we begin, let's review the various classifications of variables from Lesson 1:

Categorical
Names or labels (i.e., categories) with no logical order or with a logical order but inconsistent differences between groups, also known as qualitative.
Quantitative
Numerical values with magnitudes that can be placed in a meaningful order with consistent intervals, also known as numerical.

In this lesson we will be examining methods for analyzing categorical variables.


11.1 - Reviews

11.1 - Reviews

In this lesson you may need to use Minitab Express to construct frequency tables or two-way contingency tables. We'll start by reviewing these procedures. You will need to construct probability distribution plots for chi-square distributions. In earlier lessons we constructed probability distribution plots for z, t, and F distributions; the procedure is similar for a chi-square distribution. We will also review conditional probabilities and the term independence in this section.


11.1.1 - Frequency Table

11.1.1 - Frequency Table

The following example was first presented in Lesson 2.1.1.2.1.

It uses following data set (from College Board):

MinitabExpress  – Frequency Tables

To create a frequency table in Minitab Express:

  1. Open the data set
  2. On a PC: In the menu bar select STATISTICS > Describe > Tally
  3. On a Mac: In the menu bar select Statistics > Summary Statistics > Tally
  4. Double click the variable Region in the box on the left to insert the variable into the Variable box
  5. Under Statistics, check Counts and Percents
  6. Click OK

This should result in the following frequency table:

Tally
Region Count Percent
ENC 5 9.8039%
ESC 4 7.8431%
MA 3 5.8824%
MTN 8 15.6863%
NE 6 11.7647%
PAC 5 9.8039%
SA 9 17.6471%
WNC 7 13.7255%
WSC 4 7.8431%
N= 51  
Video Walkthrough

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


11.1.2 - Two-Way Contingency Table

11.1.2 - Two-Way Contingency Table

Recall from Lesson 2.1.2 that a two-way contingency table is a display of counts for two categorical variables in which the rows represented one variable and the columns represent a second variable. The starting point for analyzing the relationship between two categorical variables is to create a two-way contingency table. When one variable is obviously the explanatory variable, the convention is to use the explanatory variable to define the rows and the response variable to define the columns; this is not a hard and fast rule though.

MinitabExpress  – Constructing a Two-Way Contingency Table

  1. Open the data set:
  2. On a PC: Select STATISTICS > Cross Tabulation and Chi-square
    On a Mac: Select Statistics > Tables > Cross Tabulation and Chi-Square
  3. Select Raw data (categorical variable) from the drop down menu
  4. Double click the variable Smoke Cigarettes in the box on the left to insert the variable into the Rows box
  5. Double click the variable Biological Sex in the box on the left to insert the variable into the Columns box
  6. Click OK

This should result in the two-way table below:

Tabulated Statistics: Smoke Cigarettes, Biological Sex
Rows: Smokes Cigaretes | Columns: Biological Sex
  Female Male All
No 120 89 209
Yes 7 10 17
All 127 99 226
Cell Contents: Count
Video Walkthrough

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


11.1.3 - Probability Distribution Plots

11.1.3 - Probability Distribution Plots

In previous lessons you have constructed probabilities distribution plots for normal distributions, binomial distributions, and \(t\) distributions. This week you will use the same procedure to construct a probability distribution plot for the chi-square distribution.

MinitabExpress  – Constructing a Probability Distribution Plot

Chi-square tests of independence are always right-tailed tests. Let's find the area of a chi-square distribution with 1 degree of freedom to the right of \(\chi^2 = 1.75\). In other words, we're looking up the \(p\) value associated with a chi-square test statistic of 1.75.

  1. On a PC: from the menu select STATISTICS > Distribution Plot
    On a Mac: from the menu select Statistics > Probability Distributions > Distribution Plot
  2. Select Display Probability 
  3. For Distribution select Chi-Square
  4. For Degrees of freedom enter 1
  5. Select A specified X value
  6. Select Right tail
  7. For X value enter 1.75

This should result in the following output:

Chi-square distribution plot made using Minitab Express; degrees of freedom equal 1; the area to the right of a chi-square value of 1.75 is 0.185877

Video Walkthrough

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

Example: Area to the Right of Chi-Sq = 6.25, df=3


11.1.4 - Conditional Probabilities and Independence

11.1.4 - Conditional Probabilities and Independence

In Lesson 2 you were introduced to conditional probabilities and independent events. These definitions are reviewed below along with some examples.

Recall that if events A and B are independent then \(P(A) = P(A \mid B)\). In other words, whether or not event B occurs does not change the probability of event A occurring.

Conditional Probability

The probability of one event occurring given that it is known that a second event has occurred. This is communicated using the symbol \(\mid\) which is read as "given."

For example, \(P(A\mid B)\) is read as "Probability of A given B."

Independent Events
Unrelated events. The outcome of one event does not impact the outcome of the other event.

Example: Queens & Hearts

If a card is randomly drawn from a standard 52-card deck, the probability of the card being a queen is independent from the probability of the card being a heart.  If I tell you that a randomly selected card is a queen, that does not change the likelihood of it being a heart, diamond, club, or spade. 

Using a conditional probability to prove this:

\(P(Queen) = \dfrac{4}{52}=0.077\)

\(P(Queen \mid Heart) = \dfrac {1}{13} = 0.077\)

Example: Gender and Pass Rate

Data concerning two categorical variables can be displayed in a contingency table. 

Contingency Table of Gender and Pass
  Pass Did Not Pass Total
Men 6 9 15
Women 10 15 25
Total 16 24 40

If gender and passing are independent, then the probability of passing will not change if a case's gender is known. This could be written as \(P(Pass) = P(Pass \mid Man)\).

\(P(Pass) = \dfrac{16}{40} = 0.4\)

\(P(Pass \mid Man) = \dfrac{6}{15}=0.4\)

In this sample, gender and passing are independent. 


11.2 - Goodness of Fit Test

11.2 - Goodness of Fit Test

A chi-square goodness-of-fit test can be conducted when there is one categorical variable with more than two levels. If there are exactly two categories, then a one proportion z test may be conducted. The levels of that categorical variable must be mutually exclusive. In other words, each case must fit into one and only one category.

We can test that the proportions are all equal to one another or we can test any specific set of proportions.

If the expected counts, which we'll learn how to compute shortly, are all at least five, then the chi-square distribution may be used to approximate the sampling distribution. If any expected count is less than five, then a randomization test should be conducted. 

Possible Research Questions
  • When randomly selecting a card from a deck with replacement, are we equally likely to select a heart, diamond, spade, and club?
  • According to one research study, in the United States 2% of adults identify as homosexual, 2% as bisexual, and 96% as heterosexual. Are these proportions different in the population of Penn State students?
  • A concessions stand sells blue, red, purple, and green freezer pops. They survey a sample of children and ask which of the four colors is their favorite. They want to know if the colors differ in popularity. 
Test Statistic

In conducting a goodness-of-fit test, we compare observed counts to expected counts. Observed counts are the number of cases in the sample in each group. Expected counts are computed given that the null hypothesis is true; this is the number of cases we would expect to see in each cell if the null hypothesis were true.

Expected Cell Value
\(E=n (p_i)\)

\(n\) is the total sample size
\(p_i\) is the hypothesized proportion of the "ith" group

The observed and expected values are then used to compute the chi-square (\(\chi^2\)) test statistic.

Chi-Square (\(\chi^2\)) Test Statistic

\(\chi^2=\sum \dfrac{(Observed-Expected)^2}{Expected}\)

Approximating the Sampling Distribution

StatKey has the ability to conduct a randomization test for a goodness-of-fit test. There is an example of this in Section 7.1 of the Lock5 textbook.  If all expected values are at least five, then the sampling distribution can be approximated using a chi-square distribution.

Like the t distribution, the chi-square distribution varies depending on the degrees of freedom. Degrees of freedom for a chi-square goodness-of-fit test are equal to the number of groups minus 1. The distribution plot below compares the chi-square distributions with 2, 4, and 6 degrees of freedom.

Probability distribution plot made using Minitab Express; 3 chi-square distributions are overlaid with degrees of freedom of 2, 4, and 6

To find the p-value we find the area under the chi-square distribution to the right of our test statistic. A chi-square test is always right-tailed. 


11.2.1 - Five Step Hypothesis Testing Procedure

11.2.1 - Five Step Hypothesis Testing Procedure

The examples on the following pages use the five step hypothesis testing procedure outlined below. This is the same procedure that we used to conduct a hypothesis test for a single mean, single proportion, difference in two means, and difference in two proportions.

Step 1: Check assumptions and write hypotheses

When conducting a chi-square goodness-of-fit test, it makes the most sense to write the hypotheses first. The hypotheses will depend on the research question. The null hypothesis will always contain the equalities and the alternative hypothesis will be that at least one population proportion is not as specified in the null.

In order to use the chi-square distribution to approximate the sampling distribution, all expected counts must be at least five.

Expected Count
\(E=np_i\)

Where \(n\) is the total sample size and \(p_i\) is the hypothesized population proportion in the "ith" group.

To check this assumption, compute all expected counts and confirm that each is at least five.

Step 2: Compute the test statistic

In Step 1 you already computed the expected counts. Use this formula to compute the chi-square test statistic:

Chi-Square Test Statistic
\(\chi^2=\sum \dfrac{(O-E)^2}{E}\)
Where \(O\) is the observed count for each cell and \(E\) is the expected count for each cell.
Step 3: Determine the p-value

Construct a chi-square distribution with degrees of freedom equal to the number of groups minus one. The p-value is the area under that distribution to the right of the test statistic that was computed in Step 2. You can find this area by constructing a probability distribution plot in Minitab Express. 

Step 4: Make a decision

Unless otherwise stated, use the standard 0.05 alpha level.

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

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

Step 5: State a real-world conclusion

Go back to the original research question and address it directly. If you rejected the null hypothesis, then there is evidence that at least one of the population proportions is not as stated in the null hypothesis. If you failed to reject the null hypothesis, then there is not evidence that any of the population proportions are different from what is stated in the null hypothesis. 


11.2.1.1 - Video: Cupcakes (Equal Proportions)

11.2.1.1 - Video: Cupcakes (Equal Proportions)

11.2.1.2- Cards (Equal Proportions)

11.2.1.2- Cards (Equal Proportions)

Research question

When randomly selecting a card from a deck with replacement, are we equally likely to select a heart, diamond, spade, and club?

I randomly selected a card from a standard deck 40 times with replacement. I pulled 13 hearts, 8 diamonds, 8 spades, and 11 clubs.

Let's use the five-step hypothesis testing procedure:

Step 1: Check assumptions and write hypotheses

\(H_0: p_h=p_d=p_s=p_c=0.25\)
\(H_a:\) at least one \(p_i\) is not as specified in the null

We can use the null hypothesis to check the assumption that all expected counts are at least 5.

\(Expected\;count=n (p_i)\)

All \(p_i\) are 0.25. \(40(0.25)=10\), thus this assumption is met and we can approximate the sampling distribution using the chi-square distribution.

Step 2: Compute the test statistic

 \(\chi^2=\sum \dfrac{(Observed-Expected)^2}{Expected} \)

All expected values are 10. Our observed values were 13, 8, 8, and 11.

\(\chi^2=\dfrac{(13-10)^2}{10}+\dfrac{(8-10)^2}{10}+\dfrac{(8-10)^2}{10}+\dfrac{(11-10)^2}{10}\)
\(\chi^2=\dfrac{9}{10}+\dfrac{4}{10}+\dfrac{4}{10}+\dfrac{1}{10}\)
\(\chi^2=1.8\)

Step 3: Determine the p-value

Our sampling distribution will be a chi-square distribution.

\(df=k-1=4-1=3\)

We can find the p-value by constructing a chi-square distribution with 3 degrees of freedom to find the area to the right of \(\chi^2=1.8\)

Chi-squared distribution plot made using Minitab Express; degrees of freedom equal 3; area to the right of chi-squared value of 1.8 is 0.614935

The p-value is 0.614935

Step 4: Make a decision

\(p>0.05\) therefore we fail to reject the null hypothesis.

Step 5: State a "real world" conclusion

There is not evidence that the proportion of hearts, diamonds, spades, and clubs that are randomly drawn from this deck are different.


11.2.1.3 - Roulette Wheel (Different Proportions)

11.2.1.3 - Roulette Wheel (Different Proportions)

Research Question

An American roulette wheel contains 38 slots: 18 red, 18 black, and 2 green.  A casino has purchased a new wheel and they want to know if there is any evidence that the wheel is unfair. They spin the wheel 100 times and it lands on red 44 times, black 49 times, and green 7 times.

Step 1: Check assumptions and write hypotheses

If the wheel is fair then \(p_{red}=\dfrac{18}{38}\), \(p_{black}=\dfrac{18}{38}\), and \(p_{green}=\dfrac{2}{38}\).

All of these proportions combined equal 1.

\(H_0: p_{red}=\dfrac{18}{38},\;p_{black}=\dfrac{18}{38}\;and\;p_{green}=\dfrac{2}{38}\)

\(H_a: At\;least\;one\;p_i\;is \;not\;as\;specified\;in\;the\;null\)

In order to conduct a chi-square goodness of fit test all expected values must be at least 5. 

For both red and black: \(Expected \;count=100(\dfrac{18}{38})=47.368\)

For green: \(Expected\;count=100(\dfrac{2}{38})=5.263\)

All expected counts are at least 5 so we can conduct a chi-square goodness of fit test. 

Step 2: Compute the test statistic

 \(\chi^2=\sum \dfrac{(Observed-Expected)^2}{Expected} \)

In the first step we computed the expected values for red and black to be 47.368 and for green to be 5.263.

 \(\chi^2= \dfrac{(44-47.368)^2}{47.368}+\dfrac{(49-47.368)^2}{47.368}+\dfrac{(7-5.263)^2}{5.263} \)

 \(\chi^2=0.239+0.056+0.573=0.868\)

Step 3: Determine the p-value

Our sampling distribution will be a chi-square distribution.

\(df=k-1=3-1=2\)

We can find the p-value by constructing a chi-square distribution with 2 degrees of freedom to find the area to the right of \(\chi^2=0.868\)

Chi-squared distribution plot made using Minitab Express; degrees of freedom equal 2; area to the right of chi-squared value of 0.868 is shaded with a proportion of 0.647912

The p-value is 0.647912

Step 4: Make a decision

\(p>0.05\) therefore we should fail to reject the null hypothesis. 

Step 5: State a "real world" conclusion

There is not evidence that this roulette wheel is unfair.


11.2.2 - Minitab Express: Goodness-of-Fit Test

11.2.2 - Minitab Express: Goodness-of-Fit Test

Research Question:

When randomly selecting a card from a deck with replacement, are we equally likely to select a heart, diamond, spade, and club?

I randomly selected a card from a standard deck 40 times with replacement. I pulled 13 hearts, 8 diamonds, 8 spades, and 11 clubs.

MinitabExpress  – Conducting a Chi-Square Goodness-of-Fit Test

Summarized Data, Equal Proportions

To perform a chi-square goodness-of-fit test in Minitab Express using summarized data we first need to enter the data into the worksheet. Below you can see that we have one column with the names of each group and one column with the observed counts for each group.

  Suit Count
1 Hearts 13
2 Diamonds 8
3 Spades 8
4 Clubs 11
  1. On a PC: Select STATISTICS > Chi-Square Goodness-of-Fit
    On a Mac: Select Statistics > Tables > Chi-Square Goodness-of-Fit
  2. From the drop-down box select Summarized data in a column
  3. Double-click Count to enter it into the Observed Counts box
  4. Double-click Suit to enter it into the Category names box
  5. Click OK

This should result in the following output:

Chi-Square Goodness-of-Fit Test: Count
Observed and Expected Counts
Category Observed Test
Proportion
Expected Contribution
to Chi-Sq
Hearts 13 0.250000 10 0.90
Diamonds 8 0.250000 10 0.40
Spades 8 0.250000 10 0.40
Clubs 11 0.250000 10 0.10
Chi-Square Test
N DF Chi-Sq P-Value
40 3 1.80 0.6149
Video Walkthrough

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

All expected values are at least 5 so we can use the chi-square distribution to approximate the sampling distribution. Our results are \(\chi^2 (3) = 1.80\). \(p = 0.6149\). Because our p-value is greater than the standard alpha level of 0.05, we fail to reject the null hypothesis. There is not evidence that the proportions are different in the population.

The example above tested equal population proportions. Minitab Express also has the ability to conduct a chi-square goodness-of-fit test when the hypothesized population proportions are not all equal. To do this, you can choose to test specified proportions or to use proportions based on historical counts:

Screen shot from Minitab Express showing where to change the null proportions


11.2.2.1 - Video Example: Tulips (Summarized Data, Equal Proportions)

11.2.2.1 - Video Example: Tulips (Summarized Data, Equal Proportions)

A company selling tulip bulbs claims they have equal proportions of white, pink, and purple bulbs and that they fill customer orders by randomly selecting bulbs from the population of all of their bulbs.

You ordered 30 bulbs and received 16 white, 8 pink, and 6 purple.

Is there evidence the bulbs you received were not randomly selected from a population with an equal proportion of each color?

 

Minitab Express is used to conduct a hypothesis test to address this research question. 


11.2.2.2 - Video Example: Roulette (Summarized Data, Different Proportions)

11.2.2.2 - Video Example: Roulette (Summarized Data, Different Proportions)

11.3 - Chi-Square Test of Independence

11.3 - Chi-Square Test of Independence

The chi-square (\(\chi^2\)) test of independence is used to test for a relationship between two categorical variables. Recall that if two categorical variables are independent, then \(P(A) = P(A \mid B)\). The chi-square test of independence uses this fact to compute expected values for the cells in a two-way contingency table under the assumption that the two variables are independent (i.e., the null hypothesis is true).

Even if two variables are independent in the population, samples will vary due to random sampling variation. The chi-square test is used to determine if there is evidence that the two variables are not independent in the population using the same hypothesis testing logic that we used with one mean, one proportion, etc.

Again, we will be using the five step hypothesis testing procedure:

1. Check assumptions and write hypotheses.

The assumptions are that the sample is randomly drawn from the population and that all expected values are at least 5 (we will see what expected values are later).

Our hypotheses are:

     \(H_0:\) There is not a relationship between the two variables in the population (they are independent)

     \(H_a:\) There is a relationship between the two variables in the population (they are dependent)

Note: When you're writing the hypotheses for a given scenario, use the names of the variables, not the generic "two variables."

2. Calculate an appropriate test statistic
Chi-Square Test Statistic
\(\chi^2=\sum \dfrac{(Observed-Expected)^2}{Expected}\)
Expected Cell Value
\(E=\dfrac{row\;total \; \times \; column\;total}{n}\)
3. Determine the p-value

The p-value can be found using Minitab Express. Look up the area to the right of your chi-square test statistic on a chi-square distribution with the correct degrees of freedom. Chi-square tests are always right-tailed tests. 

Degrees of Freedom: Chi-Square Test of Independence
\(df=(number\;of\;rows-1)(number\;of\;columns-1)\)
4. Make a decision

If \(p \leq \alpha\) reject the null hypothesis.

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

5. State a "real world" conclusion

Write a conclusion in terms of the original research question.


11.3.1 - Example: Gender and Online Learning

11.3.1 - Example: Gender and Online Learning

A sample of 314 Penn State students was asked if they have ever taken an online course. Their genders were also recorded. The contingency table below was constructed. Use a chi-square test of independence to determine if there is a relationship between gender and whether or not someone has taken an online course.

  Have you taken an online course?
  Yes No
Men 43 63
Women 95 113
1. Check assumptions and write hypotheses

\(H_0:\) There is not a relationship between gender and whether or not someone has taken an online course (they are independent)

\(H_a:\) There is a relationship between gender and whether or not someone has taken an online course (they are dependent)

Looking ahead to our calculations of the expected values, we can see that all expected values are at least 5. This means that the sampling distribution can be approximated using the \(\chi^2\) distribution. 

2. Calculate the test statistic

In order to compute the chi-square test statistic we must know the observed and expected values for each cell. We are given the observed values in the table above. We must compute the expected values. The table below includes the row and column totals.

  Have you taken an online course?  
  Yes No  
Men 43 63 106
Women 95 113 208
  138 176 314
\(E=\dfrac{row\;total \times column\;total}{n}\)
\(E_{Men,\;Yes}=\dfrac{106\times138}{314}=46.586\)
\(E_{Men,\;No}=\dfrac{106\times176}{314}=59.414\)
\(E_{Women,\;Yes}=\dfrac{208\times138}{314}=91.414\)
\(E_{Women,\;No}=\dfrac{208 \times 176}{314}=116.586\)

Note that all expected values are at least 5, thus this assumption of the \(\chi^2\) test of independence has been met. 

Observed and expected counts are often presented together in a contingency table. In the table below, expected values are presented in parentheses.

  Have you taken an online course?  
  Yes No  
Men 43 (46.586) 63 (59.414) 106
Women 95 (91.414) 113 (116.586) 208
  138 176 314

\(\chi^2=\sum \dfrac{(O-E)^2}{E} \)

\(\chi^2=\dfrac{(43-46.586)^2}{46.586}+\dfrac{(63-59.414)^2}{59.414}+\dfrac{(95-91.414)^2}{91.414}+\dfrac{(113-116.586)^2}{116.586}=0.276+0.216+0.141+0.110=0.743\)

The chi-square test statistic is 0.743

\(df=(number\;of\;rows-1)(number\;of\;columns-1)=(2-1)(2-1)=1\)

3. Determine the p-value

We can determine the p-value by constructing a chi-square distribution plot with 1 degree of freedom and finding the area to the right of 0.743.

Distribution Plot - Chi-Square, DF=1

\(p = 0.388702\)

4. Make a decision

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

5. State a "real world" conclusion

There is not evidence that gender and whether or not an individual has completed an online course are related.

Note that we cannot say for sure that these two categorical variables are independent, we can only say that we do not have evidence that they are dependent.


11.3.2 - Minitab Express: Test of Independence

11.3.2 - Minitab Express: Test of Independence

If you have a data file with the responses for individual cases then you have "raw data" and can follow the directions below. If you have a table filled with data, then you have "summarized data." There is an example of conducting a chi-square test of independence using summarized data on a later page. After data entry the procedure is the same for both data entry methods.

MinitabExpress  – Chi-square Test Using Raw Data

Research question: Is there a relationship between where a student sits in class and whether they have ever cheated?

Null hypothesis: Seat location and cheating are not related in the population. 
Alternative hypothesis: Seat location and cheating are related in the population.

To perform a chi-square test of independence in Minitab Express using raw data:

  1. Open Minitab data set
  2. On a PC: Select STATISTICS > Cross Tabulation and Chi-Square
    On a Mac: Select Statistics > Tables > Cross Tabulation and Chi-Square
  3. Double-click on the variable Seating to insert it into the Rows box
  4. Double-click on the variable Ever Cheat to insert it into the Columns box
  5. Click the Display tab and check the boxes Chi-square test for association and Expected cell counts
  6. Click OK

This should result in the following output:

Cell contents grouped by Back, Front, Middle, All; First row: count, Next row: expected count

Rows: Seating | Columns: Ever Cheat
  No Yes All
Back 24 8 32
  24.21 7.79  
Front 38 8 46
  34.81 11.19  
Middle 109 39 148
  111.98 36.02  
All 1714 55 226
Chi-Square Test
  Chi-Square DF P-Value
Pearson 1.54 2 0.4633
Likelihood Ratio 1.63 2 0.4435
Video Walkthrough

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

All expected values are at least 5 so we can use the Pearson chi-square test statistic. Our results are \(\chi^2 (2) = 1.54\). \(p = 0.4633\). Because our \(p\) value is greater than the standard alpha level of 0.05, we fail to reject the null hypothesis. There is not evidence of a relationship in the population between seat location and whether a student has cheated.


11.3.2.1 - Video Example: Dog & Cat Ownership (Raw Data)

11.3.2.1 - Video Example: Dog & Cat Ownership (Raw Data)

This example uses the dataset:

1. Check any necessary assumptions and write null and alternative hypotheses.

 \(H_0:\) There is not a relationship between dog ownership and cat ownership in the population of all World Campus STAT 200 students
\(H_a:\) There is a relationship between dog ownership and cat ownership in the population of all World Campus STAT 200 students

Cell contents grouped by No, Yes, Missing; First row: count, Next row: expected count

Rows: Dog | Columns: Cat
  No Yes All
BaNock 183 69 252
  176.02 75.98  
Yes 183 89 272
  189.98 82.02  
Missing 1 0  
All 366 158 524

Assumption: All expected counts are at least 5. The expected counts here are 176.02, 75.98, 189.98, and 82.02, so this assumption has been met.

2. Calculate an appropriate test statistic.
Chi-Square Test
  Chi-Square DF P-Value
Pearson 1.77 1 0.1833
Likelihood Ratio 1.77 1 0.1828

Since the assumption was met in step 1, we can use the Pearson chi-square test statistic.

\(Pearson\;\chi^2 = 1.77\)

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

\(p = 0.1833\)

4. Decide between the null and alternative hypotheses.

Our p value is greater than the standard 0.05 alpha level, so we fail to reject the null hypothesis.

5. State a "real world" conclusion.

There is not evidence of a relationship between dog ownership and cat ownership in the population of all World Campus STAT 200 students.


11.3.2.2 - Video Example: Coffee and Tea (Summarized Data)

11.3.2.2 - Video Example: Coffee and Tea (Summarized Data)

11.3.3 - Relative Risk

11.3.3 - Relative Risk

A chi-square test of independence will give you information concerning whether or not a relationship between two categorical variables in the population is likely. As was the case with the single sample and two sample hypothesis tests that you learned earlier this semester, with a large sample size statistical power is high and the probability of rejecting the null hypothesis is high, even if the relationship is relatively weak. In addition to examining statistical significance by looking at the p value, we can also examine practical significance by computing the relative risk.

In Lesson 2 you learned that risk is often used to describe the probability of an event occurring. Risk can also be used to compare the probabilities in two different groups. First, we'll review risk, then you'll be introduced to the concept of relative risk.

The risk of an outcome can be expressed as a fraction or as the percent of a group that experiences the outcome.

Risk
\(Risk=\dfrac{n\;with\;outcome}{total\;n}\)

Examples of Risk

Asthma

60 out of 1000 teens have asthma. The risk is \(\frac{60}{1000}=.06\). This means that 6% of all teens experience asthma.

Flu

45 out of 100 children get the flu each year. The risk is \(\frac{45}{100}=.45\) or 45%

Relative Risk
Relative risk compares the risk of a particular outcome in two different groups.
Relative Risk
\(Relative\ Risk=\dfrac{Risk\ in\ Group\ 1}{Risk\ in\ Group\ 2}\)

Thus, relative risk gives the risk for group 1 as a multiple of the risk for group 2.

Examples of Relative Risk

Asthma

Suppose, 7% of teenage girls have asthma compared to 5% of teenage boys with asthma. Using the females as group 1, what is the relative risk for girls compared to boys?

  •  \(Relative\;risk=\dfrac{7 \%}{5 \%}=1.4\) 

The risk of asthma is 1.4 times as great for teenage girls as it is for teenage boys.

Flu

Suppose that the risk of a child getting the flu this year is .45 and the risk of an adult getting the flu this year is .10. What is the relative risk of children compared to adults?

  • \(Relative\;risk=\dfrac{.45}{.10}=4.5\)

Children are 4.5 times more likely than adults to get the flu this year.

Caution

Watch out for relative risk statistics where no baseline information is given about the actual risk. For instance, it doesn't mean much to say that beer drinkers have twice the risk of stomach cancer as non-drinkers unless we know the actual risks. The risk of stomach cancer might actually be very low, even for beer drinkers. For example, 2 in a million is twice the size of 1 in a million but is would still be a very low risk. This is known as the baseline with which other risks are compared.


11.4 - Lesson 11 Summary

11.4 - Lesson 11 Summary

Objectives

Upon successful completion of this lesson, you should be able to:

  • Construct a chi-square probability distribution plot in StatKey or Minitab Express.
  • Determine when a chi-square goodness-of-fit test and chi-square test of independence should be conducted.
  • Compute and interpret expected counts.
  • Conduct chi-square tests by hand and using Minitab Express.
  • Calculate and interpret relative risk.

In this lesson you added two new statistical procedures to your repertoire: the chi-square goodness-of-fit test for one categorical variable and the chi-square test of independence for two variables. In the remaining weeks we will continue to learn more procedures.


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