The Chi-Squared Distribution

When sample sizes are large enough, the shape of the null distribution for the chi-squared statistic is approximately a chi-squared distribution. Here are some important facts about this family of distributions.

Comparing the test statistic to a Chi-squred distribution

Golf balls, revisited

In our golf balls example, the null distribution is approximately a chi-squared distribution with 3 (one less than the number of cells in our table) degrees of freedom.

set.seed(1234)
golfballs <- as.table(c(137, 138, 107, 104)); 
dimnames(golfballs) <- list(1:4)
golfballs
##   1   2   3   4 
## 137 138 107 104
chisq(golfballs)
## X.squared 
##  8.469136
Golf_rand <- do(2000) * chisq(tally( ~ resample(1:4, size = 486)))
head(Golf_rand, 3)
##   X.squared
## 1 8.7818930
## 2 2.3292181
## 3 0.3539095
gf_dhistogram( ~ X.squared, data = Golf_rand, boundary = 0) %>%
  gf_vline(xintercept = ~ 8.469, color = "navy") %>%
  gf_dist('chisq', df = 3, color = "red")

prop( ~(X.squared >= 8.469), data = Golf_rand)
## prop_TRUE 
##     0.036
1 - pchisq(8.469, df = 3)
## [1] 0.03725096
# illustrated version
1 - xpchisq(8.469, df = 3)
## [1] 0.03725096

As we see, we get essentially the same p-value either way.

Rule of Thumb

The results should be quite similar to the randomization results as long as the sample is large enough. The usual rule of thumb is that the expected counts should be at least 5 in each cell.1 Our expected counts are much larger than this.

Using chisq.test()

The chisq.test() function can do all this work for us.

# randomization method
chisq.test(c(137, 138, 107, 104), p = c(0.25, 0.25, 0.25, 0.25),
           simulate.p.value = TRUE)
## 
##  Chi-squared test for given probabilities with simulated p-value (based
##  on 2000 replicates)
## 
## data:  c(137, 138, 107, 104)
## X-squared = 8.4691, df = NA, p-value = 0.03848
# theory method
chisq.test(c(137, 138, 107, 104), p = c(0.25, 0.25, 0.25, 0.25))
## 
##  Chi-squared test for given probabilities
## 
## data:  c(137, 138, 107, 104)
## X-squared = 8.4691, df = 3, p-value = 0.03725

You can get even more information if you use xchisq.test()

xchisq.test(c(137, 138, 107, 104), p = c(0.25, 0.25, 0.25, 0.25))
## 
##  Chi-squared test for given probabilities
## 
## data:  x
## X-squared = 8.4691, df = 3, p-value = 0.03725
## 
##  137.00   138.00   107.00   104.00 
## (121.50) (121.50) (121.50) (121.50)
##  [1.98]   [2.24]   [1.73]   [2.52] 
## < 1.41>  < 1.50>  <-1.32>  <-1.59> 
##        
## key:
##  observed
##  (expected)
##  [contribution to X-squared]
##  <Pearson residual>

Why 3 degrees of freedom?

If we know the sample size (the number of golf balls we collect) and we know any 3 of the cell counts in the table, then the last cell count is completely determined, it is not free. We have only 3 degrees of freedom.

Example:

Suppose we collect a new data set. This time there are only 100 golf balls. In each table below, determine the missing observed count.

number on golf ball 1 2 3 4
observed count 20 30 25 ?
number on golf ball 1 2 3 4
observed count 31 28 ? 20
number on golf ball 1 2 3 4
observed count ? 28 24 20

Goodness of Fit Practice

Skill building

  1. In each example below, use the test statistic and degrees of freedom to compute a p-value using the chi-squared distribution.

    1. \(X^2 = 3.5\); df = 3.
    2. \(X^2 = 7.2\); df = 3.
    3. \(X^2 = 7.2\); df = 2.
    4. \(X^2 = 15.52\); df = 5.
    5. \(X^2 = 15.52\); df = 10.

  2. How large is the chi-squared statistic if df = 4 and the p-value is 0.05?

  3. How large is the chi-squared statistic if df = 2 and the p-value is 0.05?

  4. Answer the questions below about this use of xchisq.test().

    xchisq.test(c(53, 108, 39), p = c(0.25, 0.5, 0.25))
    ## 
##  Chi-squared test for given probabilities
## 
## data:  x
## X-squared = 3.24, df = 2, p-value = 0.1979
## 
##   53.00   108.00    39.00 
## ( 50.00) (100.00) ( 50.00)
##  [0.18]   [0.64]   [2.42] 
## < 0.42>  < 0.80>  <-1.56> 
##      
## key:
##  observed
##  (expected)
##  [contribution to X-squared]
##  <Pearson residual>
    1. What is the null hypothesis for this test?
    2. Why is df = 2?
    3. Explain how to compute the number 0.64 that appears in the output.
    4. Show how to calculate the p-value using pchisq() or xpchisq().
    5. Square each of the Pearson residuals. What do you observe? (Note: residuals here are defined differently from regression residuals.)

More Examples

We now know 4 ways to do each of the examples below.

  • Randomization
    • Using resample() to create a randomization distribution and using that to get your p-value.
    • Using chisq.test() or xchisq.test() with simulate.p.value = TRUE.
  • Theoretical

    • “By hand” using the chi-squared distribution – computing the expected counts and chi-squared statistic and using pchisq() or xpchisq() to get the p-value.

    • Using chisq.test() or xchisq.test() to do all the work for you. (This is the default method for chisq.test() and xchisq.test().)

Use the examples below to practice each method. If you use the theoretical method, make sure that our rule of thumb is satisfied. You don’t necessarily need to do each method on each problem, but be sure you know how to do each method. In practice, we will prefer to use chisq.test() or xchisq.test() because they are so much easier to use and provide us all the information we need.

5. Plant Genetics. According to a genetic model, the ratio of colors of seed coats should be 12:3:1. The observed colors are given in the table below. Is this data compatible with this particular genetic theory?

Seed coat color black brown pale
Count 321 77 31

6. More Plant Genetics. Below are some data from one of Mendel’s original studies of pea plants.

Phenotype yellow, round yellow, wrinkled green, round green, wrinkled
Count 315 101 108 32

According to his theory of independent assortment (Mendel’s second law), the phenotype ratios should be 9:3:3:1. Is this data compatible with this particular genetic theory?

7. Birth Days. A sample of 700 random births revealed the following data regarding the day of the week on which the birth occurred. Is there evidence that some days are more common than others or might the differences in our table simply be due to sampling variability?

Day Sun Mon Tue Wed Thr Fri Sat
Births 84 110 124 104 94 112 72

8. One proportion or Chi-squared? Fifty subjects were given two unmarked cups of coffee and asked which one they preferred. 31 preferred the fresh brewed coffee and 19 preferred the instant coffee.

Is this evidence that a majority of people prefer fresh brewed coffee? Conduct the hypothesis test two ways.

  • Using a test for one proportion
  • Using a chi-squared goodness of fit test

How do the results compare?

9. Frizzled Feathers The Frizzle fowl is a variety of chicken with curled feathers. In a 1930 experiment, Launder and Dunn crossed Frizzle fowls with with the Leghorn variety which has straight feathers. The first generation (F1) produced all slightly frizzled chicks. This made the researchers suspect a co-dominant genetic model. To test this, they interbred F1 to get F2 chicks and recorded the feather type for each chick.

frizzled slightly frizzled straight
23 50 20

The codominant model predicts a 1:2:1 ratio. Why?

Use these data to assess the genetic model.

  1. Two notes: First, if you use chisq.test(), it will warn you if it thinks the p-value might not be very accurate. In those situations you should use simulate.p.value = TRUE. Second, the rule of thumb isn’t magic. As always, bigger is better: The larger the expeted counts are, the more accurate the approximation is. But there is no hard cut-off between large enough and not large enough. If there five or ore cells and most of the expected counts are above 5, it is probably OK if one of then is a bit smaller.↩︎