Chi-square test for goodness-of-fit
When to use it
Use the chi-square test of goodness-of-fit when you have one nominal variable with two or more values (such as red, pink and white flowers). The observed counts of numbers of observations in each category are compared with the expected counts, which are some kind of theoretical expectation (such as a 1:1 sex ratio or a 1:2:1 ratio in a genetic cross).
If the expected number of observations in any category is too small, the chi-square test may give inaccurate results, and an exact test or a randomization test should be used instead. See the web page on small sample sizes for further discussion.
Null hypothesis
The statistical null hypothesis is that the number of observations in each category is equal to that predicted by a biological theory, and the alternative hypothesis is that the observed numbers are different from the expected. The null hypothesis is usually an extrinsic hypothesis, one for which the expected proportions are determined before doing the experiment. Examples include a 1:1 sex ratio or a 1:2:1 ratio in a genetic cross. Another example would be looking at an area of shore that had 59% of the area covered in sand, 28% mud and 13% rocks; if seagulls were standing in random places, your null hypothesis would be that 59% of the seagulls were standing on sand, 28% on mud and 13% on rocks.
In some situations, an intrinsic hypothesis is used. This is a null hypothesis in which the expected proportions are calculated after the experiment is done, using some of the information from the data. The best-known example of an intrinsic hypothesis is the Hardy-Weinberg proportions of population genetics: if the frequency of one allele in a population is p and the other allele is q, the null hypothesis is that expected frequencies of the three genotypes are p2, 2pq, and q2. This is an intrinsic hypothesis, because p and q are estimated from the data after the experiment is done, not predicted by theory before the experiment.
How the test works
The test statistic is calculated by taking an observed number (O), subtracting the expected number (E), then squaring this difference. The larger the deviation from the null hypothesis, the larger the difference between observed and expected is. Squaring the differences makes them all positive. Each difference is divided by the expected number, and these standardized differences are summed. The test statistic is conventionally called a "chi-square" statistic, although this is somewhat confusing (it's just one of many test statistics that follows the chi-square distribution). The equation is
chi2 = ∑(O−E)2/E
As with most test statistics, the larger the difference between observed and expected, the larger the test statistic becomes.
The distribution of the test statistic under the null hypothesis fits the theoretical chi-square distribution. This means that once you know the chi-square test statistic, you can calculate the probability of getting that value of the chi-square statistic.
The shape of the chi-square distribution depends on the number of degrees of freedom. For an extrinsic null hypothesis (the much more common situation, where you know the proportions predicted by the null hypothesis before collecting the data), the number of degrees of freedom is simply the number of values of the variable, minus one. Thus if you are testing a null hypothesis of a 1:1 sex ratio, there are two possible values (male and female), and therefore one degree of freedom. This is because once you know how many of the total are females (a number which is "free" to vary from 0 to the sample size), the number of males is determined. If there are three values of the variable (such as red, pink, and white), there are two degrees of freedom, and so on.
An intrinsic null hypothesis is one in which you estimate one or more parameters from the data in order to get the numbers for your null hypothesis. As described above, one example is Hardy-Weinberg proportions. For an intrinsic null hypothesis, the number of degrees of freedom is calculated by taking the number of values of the variable, subtracting 1 for each parameter estimated from the data, then subtracting 1 more. Thus for Hardy-Weinberg proportions with two alleles and three genotypes, there are three values of the variable (the three genotypes); you subtract one for the parameter estimated from the data (the allele frequency, p); and then you subtract one more, yielding one degree of freedom.
Example: extrinsic hypothesis
Mendel crossed peas that were heterozygotes for Smooth/wrinkled, where Smooth is dominant. The expected ratio in the offspring is 3 Smooth: 1 wrinkled. He observed 423 Smooth and 133 wrinkled.
The expected frequency of Smooth is calculated by multiplying the sample size (556) by the expected proportion (0.75) to yield 417. The same is done for green to yield 139. The number of degrees of freedom when an extrinsic hypothesis is used is the number of values of the nominal variable minus one. In this case, there are two values (Smooth and wrinkled), so there is one degree of freedom.
The result is chi-square=0.35, 1 d.f., P=0.557, indicating that the null hypothesis cannot be rejected; there is no significant difference between the observed and expected frequencies.
Example: intrinsic hypothesis
McDonald et al. (1996) examined variation at the CVJ5 locus in the American oyster, Crassostrea virginica. There were two alleles, L and S, and the genotype frequencies in Panacea, Florida were 14 LL, 21 LS, and 25 SS. The estimate of the L allele proportion from the data is 49/120=0.408. Using the Hardy-Weinberg formula and this estimated allele proportion, the expected genotype proportions are 0.167 LL, 0.483 LS, and 0.350 SS. There are three classes (LL, LS and SS) and one parameter estimated from the data (the L allele proportion), so there is one degree of freedom. The result is chi-square=4.54, 1 d.f., P=0.033, which is significant at the 0.05 level. We can reject the null hypothesis that the data fit the expected Hardy-Weinberg proportions.
Graphing the results
If there are just two values of the nominal variable, you wouldn't display the result in a graph, as that would be a bar graph with just one bar. Instead, you just report the proportion; for example, Mendel found 23.9% wrinkled peas in his cross.
With more than two values of the nominal variable, you'd usually present the results of a goodness-of-fit test in a table of observed and expected proportions. If the expected values are obvious (such as 50%) or easily calculated from the data (such as Hardy–Weinberg proportions), you can omit the expected numbers from your table. For a presentation you'll probably want a graph showing both the observed and expected proportions, to give a visual impression of how far apart they are. You should use a bar graph for the observed proportions; the expected can be shown with a horizontal dashed line, or with bars of a different pattern.
 |
| Genotype proportions at the CVJ5 locus in the American oyster. Horizontal dashed lines indicate the expected proportions under Hardy–Weinberg equilibrium; error bars indicate 95% confidence intervals. |
 |
| Genotype proportions at the CVJ5 locus in the American oyster. Gray bars are observed proportions, with 95% confidence intervals; white bars are expected proportions under Hardy–Weinberg equilibrium. |
One way to get the horizontal lines on the graph is to set up the graph with the observed proportions and error bars, set the scale for the Y-axis to be fixed for the minimum and maximum you want, and get everything formatted (fonts, patterns, etc.). Then replace the observed proportions with the expected proportions in the spreadsheet; this should make the columns change to represent the expected values. Using the spreadsheet drawing tools, draw horizontal lines at the top of the columns. Then put the observed proportions back into the spreadsheet. Of course, if the expected proportion is something simple like 25%, you can just draw the horizontal line all the way across the graph.
Similar tests
There are several tests that use chi-square statistics. The one described here is formally known as Pearson's chi-square. It is by far the most common chi-square test, so it is usually just called the chi-square test.
You have a choice of four goodness-of-fit tests: the exact binomial test or exact multinomial test, the G-test of goodness-of-fit,, the chi-square test of goodness-of-fit, or the randomization test. For small values of the expected numbers, the chi-square and G-tests are inaccurate, because the distribution of the test statistics do not fit the chi-square distribution very well.
The usual rule of thumb is that you should use the exact test or randomization test when the smallest expected value is less than 5, and the chi-square and G-tests are accurate enough for larger expected values. This rule of thumb dates from the olden days when statistics were done by hand, and the calculations for the exact test were very tedious and to be avoided if at all possible. Nowadays, computers make it just as easy to do the exact test or randomization test as the computationally simpler chi-square or G-test. I recommend that you use the exact test or randomization test when the total sample size is less than 1000. See the web page on small sample sizes for further discussion.
The chi-square test gives approximately the same results as the G-test. Unlike the chi-square test, the G-values are additive, which means they can be used for more elaborate statistical designs, such as repeated G-tests of goodness-of-fit. The G-test is therefore preferred by many, even for simpler designs where the additivity is not important. On the other hand, the chi-square test is more familiar to more people, and it's always a good idea to use statistics that your readers are familiar with when possible.
The chi-square test of independence is used for two nominal variables, not one.
Power analysis
If your nominal variable has just two values, you can use this power analysis for proportions web page to do your power analysis. This web page is set up for one-tailed tests, rather than the more common two-tailed tests, so enter alpha = 2.5 instead of alpha = 5 percent. Note that you'll get slightly different results, depending on whether you make the observed proportion smaller or larger than the expected; use whichever gives you a larger sample size.
If your nominal variable has more than two values, use this power and sample size page. Choose "Generic chi-square test" from the box on the left side of the page (if you don't see the list of tests, make sure your web browser has Java turned on). Under "Prototype data," enter the chi-square value and sample size for some fake data. For example, if you're doing a genetic cross with an expected 1:2:1 ratio, and your minimum effect size is 10 percent more heterozygotes than expected, use the chi-square spreadsheet to do a chi-square test on observed numbers of 20:60:20 compared to expected proportions of 1:2:1. The spreadsheet gives you a chi-square value of 4.00 and an n of 100, which you enter under "Prototype data". Then set d (the degrees of freedom) equal to 2, and leave alpha at 0.05. The sliders can then be slid back and forth to yield the desired result. For example, if you slide the Power to 0.90, n is equal to 316. Note that the absolute values of the prototype data don't matter, only their relative relationship; you could have used 200:600:200, which would give you a chi-square value of 40.0 and an n of 1000, and gotten the exact same result.
How to do the test
Spreadsheet
I have set up aspreadsheet for the chi-square test of goodness-of-fit. It is largely self-explanatory. It will calculate the degrees of freedom for you if you're using an extrinsic null hypothesis; if you are using an intrinsic hypothesis, you must enter the degrees of freedom into the spreadsheet.
Web pages
There are also web pages that will perform this test here, here, or here. None of these web pages lets you set the degrees of freedom to the appropriate value for testing an intrinsic null hypothesis.
SAS
Here is a SAS program that uses PROC FREQ for a chi-square test. It uses the Mendel pea data from above, and it assumes you've already counted the number of smooth and wrinkled peas. The weight count; command tells SAS that the 'count' variable is the number of times each value of 'texture' was observed. chisq tells SAS to do a chi-square test, and testp=(75 25); tells it the expected percentages. The expected percentages must add up to 100. The expected percentages are given for the values of 'texture' in alphabetical order: 75 percent 'smooth', 25 percent 'wrinkled'.
data peas; input texture $ count; cards; smooth 423 wrinkled 133 ; proc freq data=peas; weight count; tables texture / chisq testp=(75 25); run;
Here's a SAS program that uses PROC FREQ for a chi-square test on raw data. I've used three dots to indicate that I haven't shown the complete data set.
data peas; input texture $; cards; smooth wrinkled smooth smooth wrinkled smooth . . . smooth smooth ; proc freq data=peas; tables texture / chisq testp=(75 25); run;
The output includes the following:
Chi-Square Test
for Specified Proportions
-------------------------
Chi-Square 0.3453
DF 1
Pr > ChiSq 0.5568
You would report this as "chi-square=0.3453, 1 d.f., P=0.5568."
Further reading
Sokal and Rohlf, p. 701.
Zar, pp. 462-466.
References
McDonald, J.H., B.C. Verrelli and L.B. Geyer. 1996. Lack of geographic variation in anonymous nuclear polymorphisms in the American oyster, Crassostrea virginica. Molecular Biology and Evolution 13: 1114-1118.
⇐ Previous topic | Next topic ⇒
This page was last revised January 2, 2008. Its address is http://udel.edu/~mcdonald/statchigof.html.
©2007-2008 by John H. McDonald. You can probably do what you want with this content; see the permissions page at http://udel.edu/~mcdonald/statpermissions.html for details.