The Chi Square Statistic

Types of Data:

There are basically two types of random variables and they yield two types of data: numerical and categorical. A chi square (X2) statistic is used to investigate whether distributions of categorical variables differ from one another. Basically categorical variable yield data in the categories and numerical variables yield data in numerical form. Responses to such questions as "What is your major?" or Do you own a car?" are categorical because they yield data such as "biology" or "no." In contrast, responses to such questions as "How tall are you?" or "What is your G.P.A.?" are numerical. Numerical data can be either discrete or continuous. The table below may help you see the differences between these two variables.

 Data Type  Question Type Possible Responses
 Categorical  What is your sex? male or female
 Numerical Disrete- How many cars do you own? two or three
 Numerical Continuous - How tall are you?  72 inches

Notice that discrete data arise fom a counting process, while continuous data arise from a measuring process.

The Chi Square statistic compares the tallies or counts of categorical responses between two (or more) independent groups. (note: Chi square tests can only be used on actual numbers and not on percentages, proportions, means, etc.)

2 x 2 Contingency Table

There are several types of chi square tests depending on the way the data was collected and the hypothesis being tested. We'll begin with the simplest case: a 2 x 2 contingency table. If we set the 2 x 2 table to the general notation shown below in Table 1, using the letters a, b, c, and d to denote the contents of the cells, then we would have the following table:

Table 1. General notation for a 2 x 2 contingency table.

Variable 1

 Variable 2

 Data type 1

 Data type 2

 Totals
 Category 1

 a

b

a + b
 Category 2

 c

d

c + d
 Total

a + c

b + d

a + b + c + d = N

For a 2 x 2 contingency table the Chi Square statistic is calculated by the formula:

Note: notice that the four components of the denominator are the four totals from the table columns and rows.

Suppose you conducted a drug trial on a group of animals and you hypothesized that the animals receiving the drug would show increased heart rates compared to those that did not receive the drug. You conduct the study and collect the following data:

Ho: The proportion of animals whose heart rate increased is independent of drug treatment.

Ha: The proportion of animals whose heart rate increased is associated with drug treatment.

 

Table 2. Hypothetical drug trial results.

   Heart Rate
 Increased
 No Heart Rate
 Increase
Total
 Treated  36  14  50
 Not treated  30  25  55
 Total  66  39  105

Applying the formula above we get:

Chi square = 105[(36)(25) - (14)(30)]2 / (50)(55)(39)(66) = 3.418

Before we can proceed we eed to know how many degrees of freedom we have. When a comparison is made between one sample and another, a simple rule is that the degrees of freedom equal (number of columns minus one) x (number of rows minus one) not counting the totals for rows or columns. For our data this gives (2-1) x (2-1) = 1.

We now have our chi square statistic (x2 = 3.418), our predetermined alpha level of significance (0.05), and our degrees of freedom (df = 1). Entering the Chi square distribution table with 1 degree of freedom and reading along the row we find our value of x2 (3.418) lies between 2.706 and 3.841. The corresponding probability is between the 0.10 and 0.05 probability levels. That means that the p-value is above 0.05 (it is actually 0.065). Since a p-value of 0.65 is greater than the conventionally accepted significance level of 0.05 (i.e. p > 0.05) we fail to reject the null hypothesis. In other words, there is no statistically significant difference in the proportion of animals whose heart rate increased.

What would happen if the number of control animals whose heart rate increased dropped to 29 instead of 30 and, consequently, the number of controls whose hear rate did not increase changed from 25 to 26? Try it. Notice that the new x2 value is 4.125 and this value exceeds the table value of 3.841 (at 1 degree of freedom and an alpha level of 0.05). This means that p < 0.05 (it is now0.04) and we reject the null hypothesis in favor of the alternative hypothesis - the heart rate of animals is different between the treatment groups. When p < 0.05 we generally refer to this as a significant difference.

Table 3. Chi Square distribution table.

probability level (alpha)

Df 0.5 0.10

0.05
0.02 0.01 0.001
1 0.455 2.706 3.841 5.412 6.635 10.827
2 1.386 4.605 5.991 7.824 9.210 13.815
3 2.366 6.251 7.815 9.837 11.345 16.268
4 3.357 7.779 9.488 11.668 13.277 18.465
5 4.351 9.236 11.070 13.388 15.086 20.517

To make the chi square calculations a bit easier, plug your observed and expected values into the following applet. Click on the cell and then enter the value. Click the compute button on the lower right corner to see the chi square value printed in the lower left hand coner.




 

Chi Square Goodness of Fit (One Sample Test)

This test allows us to compae a collection of categorical data with some theoretical expected distribution. This test is often used in genetics to compare the results of a cross with the theoretical distribution based on genetic theory. Suppose you preformed a simpe monohybrid cross between two individuals that were heterozygous for the trait of interest.

Aa x Aa

The results of your cross are shown in Table 4.

 

Table 4. Results of a monohybrid coss between two heterozygotes for the 'a' gene.

 

 A

 a

 Totals
 A

 10

 42

 52
 a

 33

 15

 48
 Totals

 43

 57

 100

The penotypic ratio 85 of the A type and 15 of the a-type (homozygous recessive). In a monohybrid cross between two heterozygotes, however, we would have predicted a 3:1 ratio of phenotypes. In other words, we would have expected to get 75 A-type and 25 a-type. Are or resuls different?

Calculate the chi square statistic x2 by completing the following steps:

  1. For each observed number in the table subtract the corresponding expected number (O — E).
  2. Square the difference [ (O —E)2 ].
  3. Divide the squares obtained for each cell in the table by the expected number for that cell [ (O - E)2 / E ].
  4. Sum all the values for (O - E)2 / E. This is the chi square statistic.

For our example, the calculation would be:

  Observed Expected

(O — E)
(O — E)2 (O — E)2/ E
A-type 85 75 10 100 1.33
a-type 15 25 10 100 4.0
Total 100 100      5.33

We now have our chi square statistic (x2 = 5.33), our predetermined alpha level of significalnce (0.05), and our degrees of freedom (df =1). Entering the Chi square distribution table with 1 degree of freedom and reading along the row we find our value of x2 5.33) lies between 3.841 and 5.412. The corresponding probability is 0.05<P<0.02. This is smaller than the conventionally accepted significance level of 0.05 or 5%, so the null hypothesis that the two distributions are the same is rejected. In other words, when the computed x2 statistic exceeds the critical value in the table for a 0.05 probability level, then we can reject the null hypothesis of equal distributions. Since our x2 statistic (5.33) exceeded the critical value for 0.05 probability level (3.841) we can reject the null hypothesis that the observed values of our cross are the same as the theoretical distribution of a 3:1 ratio.

Table 3. Chi Square distribution table.

probability level (alpha)

Df 0.5 0.10

0.05
0.02 0.01 0.001
1 0.455 2.706 3.841 5.412 6.635 10.827
2 1.386 4.605 5.991 7.824 9.210 13.815
3 2.366 6.251 7.815 9.837 11.345 16.268
4 3.357 7.779 9.488 11.668 13.277 18.465
5 4.351 9.236 11.070 13.388 15.086 20.517

To put this into context, it means that we do not have a 3:1 ratio of A_ to aa offspring.

To make the chi square calculations a bit easier, plug your observed and expected values into the following java applet.

Click on the cell and then enter the value. Click the compute button on the lower right corner to see the chi square value printed in the lower left hand coner.




 

Chi Square Test of Independence

For a contingency table that has r rows and c columns, the chi square test can be thought of as a test of independence. In a test ofindependence the null and alternative hypotheses are:

Ho: The two categorical variables are independent.

Ha: The two categorical variables are related.

We can use the equation Chi Square = the sum of all the (fo - fe)2 / fe

Here fo denotes the frequency of the observed data and fe is the frequency of the expected values. The general table would look something like the one below:

   Category I Category II Category III

Row Totals
 Sample A

 a

b

c

a+b+c
 Sample B

 d

e

f

d+e+f
 Sample C

 g

h

i

g+h+i
 Column Totals

 a+d+g

b+e+h

c+f+i

 a+b+c+d+e+f+g+h+i=N

Now we need to calculate the expected values for each cell in the table and we can do that using the the row total times the column total divided by the grand total (N). For example, for cell a the expected value would be (a+b+c)(a+d+g)/N.

Once the expected values have been calculated for each cell, we can use the same procedure are before for a simple 2 x 2 table.

 Observed Expected |O - E| (O — E)2  (O — E)2/ E
         

Suppose you have the following categorical data set.

Table . Incidence of three types of malaria in three tropical regions.

   Asia Africa

South America

Totals
 Malaria A

31

14

45

90
 Malaria B

2

5

53

60
 Malaria C

53

45

2

100
 Totals

 86

64

100

250

 

We could now set up the following table:

 Observed

Expected

|O -E|

 (O — E)2

 (O — E)2/ E
 31  30.96  0.04  0.0016  0.0000516
 14  23.04  9.04 81.72 3.546
 45  36.00  9.00 81.00 2.25
 2  20.64  18.64 347.45 16.83
 5  15.36  10.36 107.33 6.99
 53  24.00  29.00 841.00 35.04
 53  34.40  18.60 345.96 10.06
 45  25.60  19.40 376.36 14.70
 2  40.00  38.00  1444.00 36.10

Chi Square = 125.516

Degrees of Freedom = (c - 1)(r - 1) = 2(2) = 4

Table 3. Chi Square distribution table.

probability level (alpha)

Df 0.5 0.10

0.05
0.02 0.01 0.001
1 0.455 2.706 3.841 5.412 6.635 10.827
2 1.386 4.605 5.991 7.824 9.210 13.815
3 2.366 6.251 7.815 9.837 11.345 16.268
4 3.357 7.779 9.488 11.668 13.277 18.465
5 4.351 9.236 11.070 13.388 15.086 20.517

Reject Ho because 125.516 is greater than 9.488 (for alpha = 0.05)

Thus, we would reject the null hypothesis that there is no relationship between location and type of malaria. Our data tell us there is a relationship between type of malaria and location, but that's all it says.

Follow the link below to access a java-based program for calculating Chi Square statistics for contingency tables of up to 9 rows by 9 columns. Enter the number of row and colums in the spaces provided on the page and click the submit button. A new form will appear asking you to enter your actual data into the cells of the contingency table. When finished entering your data, click the "calculate now" button to see the results of your Chi Square analysis. You may wish to print this last page to keep as a record.

Chi Square,

This page was created as part of the Mathbeans Project. The java applets were created by David Eck and modified by Jim Ryan. The Mathbeans Project is funded by a grant from the National Science Foundation DUE-9950473.