Chi-Square Test Chi-square is a statistical test commonly used to compare observed data with data we would expect to obtain according to a specific hypothesis. For example, if, according to Mendel’s laws, you expected 10 of 20 offspring from a cross to be male and the actual observed number was 8 males, then you might want to know about the “goodness to fit” between the observed and expected. Were the deviations (differences between observed and expected) the result of chance, or were they due to other factors.

How much deviation can occur before you, the investigator, must conclude that something other than chance is at work, causing the observed to differ from the expected. The chi-square test is always testing what scientists call the null hypothesis, which states that there is no significant difference between the expected and observed result. The formula for calculating chi-square ( [pic]2) is: [pic]2= [pic](o-e)2/e That is, chi-square is the sum of the squared difference between observed (o) and the expected (e) data (or the deviation, d), divided by the expected data in all possible categories.

For example, suppose that a cross between two pea plants yields a population of 880 plants, 639 with green seeds and 241 with yellow seeds. You are asked to propose the genotypes of the parents. Your hypothesis is that the allele for green is dominant to the allele for yellow and that the parent plants were both heterozygous for this trait. If your hypothesis is true, then the predicted ratio of offspring from this cross would be 3:1 (based on Mendel’s laws) as predicted from the results of the Punnett square (Figure B. ). Figure B. 1 – Punnett Square. Predicted offspring from cross between green and yellow-seeded plants. Green (G) is dominant (3/4 green; 1/4 yellow). To calculate [pic]2 , first determine the number expected in each category. If the ratio is 3:1 and the total number of observed individuals is 880, then the expected numerical values should be 660 green and 220 yellow. [pic] Chi-square requires that you use numerical values, not percentages or ratios. [pic] Then calculate [pic]2 using this formula, as shown in Table B. . Note that we get a value of 2. 668 for [pic]

But what does this number mean? Here’s how to interpret the [pic]2 value: 1. Determine degrees of freedom (df). Degrees of freedom can be calculated as the number of categories in the problem minus 1. In our example, there are two categories (green and yellow); therefore, there is I degree of freedom.

Determine a relative standard to serve as the basis for accepting or rejecting the hypothesis. The relative standard commonly used in biological research is p ; 0. 05.

The p value is the probability that the deviation of the observed from that expected is due to chance alone (no other forces acting). In this case, using p ; 0. 05, you would expect any deviation to be due to chance alone 5% of the time or less.

Refer to a chi-square distribution table (Table B. 2). Using the appropriate degrees of ‘freedom, locate the value closest to your calculated chi-square in the table. Determine the closestp (probability) value associated with your chi-square and degrees of freedom. In this case ([pic]2=2. 68), the p value is about 0. 10, which means that there is a 10% probability that any deviation from expected results is due to chance only. Based on our standard p > 0. 05, this is within the range of acceptable deviation. In terms of your hypothesis for this example, the observed chi-squareis not significantly different from expected. The observed numbers are consistent with those expected under Mendel’s law. Step-by-Step Procedure for Testing Your Hypothesis and Calculating Chi-Square 1. State the hypothesis being tested and the predicted results.

Gather the data by conducting the proper experiment (or, if working genetics problems, use the data provided in the problem).

Determine the expected numbers for each observational class. Remember to use numbers, not percentages. [pic] Chi-square should not be calculated if the expected value in any category is less than 5. [pic]

Calculate [pic]2 using the formula. Complete all calculations to three significant digits. Round off your answer to two significant digits.

Use the chi-square distribution table to determine significance of the value. .

Determine degrees of freedom and locate the value in the appropriate column.

Locate the value closest to your calculated 2 on that degrees of freedom df row.

Move up the column to determine the p value.

State your conclusion in terms of your hypothesis.

If the p value for the calculated [pic]2 is p ; 0. 05, accept your hypothesis. ‘The deviation is small enough that chance alone accounts for it. A p value of 0. 6, for example, means that there is a 60% probability that any deviation from expected is due to chance only.

This is within the range of acceptable deviation.

If the p value for the calculated [pic]2 is p < 0. 05, reject your hypothesis, and conclude that some factor other than chance is operating for the deviation to be so great. For example, a p value of 0. 01 means that there is only a 1% chance that this deviation is due to chance alone. Therefore, other factors must be involved. The chi-square test will be used to test for the “goodness to fit” between observed and expected data from several laboratory investigations in this lab manual.

Source: R. A. Fisher and F. Yates, Statistical Tables for Biological Agricultural and Medical Research, 6th ed. , Table IV, Oliver & Boyd, Ltd. , Edinburgh, by permission of the authors and publishers.

Chi-square is a statistical test commonly used to compare observed data with data we would expect to obtain according to a specific hypothesis. For example, if, according to Mendel’s laws, you expected 10 of 20 offspring from a cross to be male and the actual observed number was 8 males, then you might want to know about the “goodness to fit” between the observed and expected. Were the deviations (differences between observed and expected) the result of chance, or were they due to other factors.

How much deviation can occur before you, the investigator, must conclude that something other than chance is at work, causing the observed to differ from the expected. The chi-square test is always testing what scientists call the null hypothesis, which states that there is no significant difference between the expected and observed result. The formula for calculating chi-square ( [pic]2) is: [pic]2= [pic](o-e)2/e That is, chi-square is the sum of the squared difference between observed (o) and the expected (e) data (or the deviation, d), divided by the expected data in all possible categories. For example, suppose that a cross between two pea plants yields a population of 880 plants, 639 with green seeds and 241 with yellow seeds. You are asked to propose the genotypes of the parents.

Your hypothesis is that the allele for green is dominant to the allele for yellow and that the parent plants were both heterozygous for this trait. If your hypothesis is true, then the predicted ratio of offspring from this cross would be 3:1 (based on Mendel’s laws) as predicted from the results of the Punnett square (Figure B. 1). Figure B. 1 – Punnett Square. Predicted offspring from cross between green and yellow-seeded plants. Green (G) is dominant (3/4 green; 1/4 yellow). To calculate [pic]2 , first determine the number expected in each category. If the ratio is 3:1 and the total number of observed individuals is 880, then the expected numerical values should be 660 green and 220 yellow. [pic]

Chi-square requires that you use numerical values, not percentages or ratios. [pic] Then calculate [pic]2 using this formula, as shown in Table B. 1. Note that we get a value of 2. 668 for [pic]2. But what does this number mean? Here’s how to interpret the [pic]2 value: 1. Determine degrees of freedom (df). Degrees of freedom can be calculated as the number of categories in the problem minus 1. In our example, there are two categories (green and yellow); therefore, there is I degree of freedom. 2. Determine a relative standard to serve as the basis for accepting or rejecting the hypothesis. The relative standard commonly used in biological research is p ; 0. 05.

The p value is the probability that the deviation of the observed from that expected is due to chance alone (no other forces acting). In this case, using p ; 0. 05, you would expect any deviation to be due to chance alone 5% of the time or less. 3. Refer to a chi-square distribution table (Table B. 2). Using the appropriate degrees of ‘freedom, locate the value closest to your calculated chi-square in the table. Determine the closestp (probability) value associated with your chi-square and degrees of freedom. In this case ([pic]2=2. 668), the p value is about 0. 10, which means that there is a 10% probability that any deviation from expected results is due to chance only. Based on our standard p > 0. 05, this is within the range of acceptable deviation.

In terms of your hypothesis for this example, the observed chi-squareis not significantly different from expected. The observed numbers are consistent with those expected under Mendel’s law.

Step-by-Step Procedure for Testing Your Hypothesis and Calculating Chi-Square

State the hypothesis being tested and the predicted results. Gather the data by conducting the proper experiment (or, if working genetics problems, use the data provided in the problem).

Determine the expected numbers for each observational class. Remember to use numbers, not percentages. [pic] Chi-square should not be calculated if the expected value in any category is less than 5.

Calculate [pic]2 using the formula. Complete all calculations to three significant digits. Round off your answer to two significant digits.

Use the chi-square distribution table to determine significance of the value.

Determine degrees of freedom and locate the value in the appropriate column.

Locate the value closest to your calculated [pic]2 on that degrees of freedom df row.

Move up the column to determine the p value.

State your conclusion in terms of your hypothesis.

If the p value for the calculated [pic]2 is p ; 0. 05, accept your hypothesis. ‘The deviation is small enough that chance alone accounts for it. A p value of 0. , for example, means that there is a 60% probability that any deviation from expected is due to chance only. This is within the range of acceptable deviation. b. If the p value for the calculated [pic]2 is p < 0. 05, reject your hypothesis, and conclude that some factor other than chance is operating for the deviation to be so great. For example, a p value of 0. 01 means that there is only a 1% chance that this deviation is due to chance alone. Therefore, other factors must be involved. The chi-square test will be used to test for the “goodness to fit” between observed and expected data from several laboratory investigations in this lab manual.

Frequency Distributions

One important set of statistical tests allows us to test for deviations of observed frequencies from expected frequencies. To introduce these tests, we will start with a simple, non-biological example. We want to determine if a coin is fair. In other words, are the odds of flipping the coin heads-up the same as tails-up. We collect data by flipping the coin 200 times. The coin landed heads-up 108 times and tails-up 92 times. At first glance, we might suspect that the coin is biased because heads resulted more often than than tails. However, we have a more quantitative way to analyze our results, a chi-squared test. To perform a chi-square test (or any other statistical test), we first must establish our null hypothesis.

In this example, our null hypothesis is that the coin should be equally likely to land head-up or tails-up every time. The null hypothesis allows us to state expected frequencies. For 200 tosses, we would expect 100 heads and 100 tails. The next step is to prepare a table as follows.

The Observed values are those we gather ourselves. The expected values are the frequencies expected, based on our null hypothesis. We total the rows and columns as indicated. It’s a good idea to make sure that the row totals equal the column totals (both total to 400 in this example). Using probability theory, statisticians have devised a way to determine if a frequency distribution differs from the expected distribution. To use this chi-square test, we first have to calculate chi-squared. Chi-squared = ? (observed-expected)2/(expected) We have two classes to consider in this example, heads and tails. Chi-squared = (100-108)2/100 + (100-92)2/100 = (-8)2/100 + (8)2/100 = 0. 4 + 0. 64 = 1. 28 Now we have to consult a table of critical values of the chi-squared distribution.

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. For a 2 x 2 contingency table the Chi Square statistic is calculated by the formula: [pic] 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.

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. 25 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.

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? [pic] 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.

Probability level (alpha) 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.