Chi-square Goodness of Fit Test in R

  • What is chi-square goodness of fit test?
  • Example data and questions
  • Statistical hypotheses
  • R function: chisq.test()
    • Answer to Q1: Are the colors equally common?
    • Answer to Q2 comparing observed to expected proportions
    • Access to the values returned by chisq.test() function
  • See also
  • Infos

What is chi-square goodness of fit test?

The chi-square goodness of fit test is used to compare the observed distribution to an expected distribution, in a situation where we have two or more categories in a discrete data. In other words, it compares multiple observed proportions to expected probabilities.


Chi-square Goodness of Fit test in R

Example data and questions

For example, we collected wild tulips and found that 81 were red, 50 were yellow and 27 were white.

  1. Question 1:

Are these colors equally common?

If these colors were equally distributed, the expected proportion would be 1/3 for each of the color.

  1. Question 2:

Suppose that, in the region where you collected the data, the ratio of red, yellow and white tulip is 3:2:1 (3+2+1 = 6). This means that the expected proportion is:

  • 3/6 (= 1/2) for red
  • 2/6 ( = 1/3) for yellow
  • 1/6 for white

We want to know, if there is any significant difference between the observed proportions and the expected proportions.

Statistical hypotheses

  • Null hypothesis (\(H_0\)): There is no significant difference between the observed and the expected value.
  • Alternative hypothesis (\(H_a\)): There is a significant difference between the observed and the expected value.

R function: chisq.test()

The R function chisq.test() can be used as follow:

                  chisq.test(x, p)

  • x: a numeric vector
  • p: a vector of probabilities of the same length of x.

Answer to Q1: Are the colors equally common?

                    tulip <- c(81, 50, 27) res <- chisq.test(tulip, p = c(1/3, 1/3, 1/3)) res                  
                                          Chi-squared test for given probabilities data:  tulip X-squared = 27.886, df = 2, p-value = 8.803e-07

The function returns: the value of chi-square test statistic ("X-squared") and a a p-value.

The p-value of the test is 8.80310^{-7}, which is less than the significance level alpha = 0.05. We can conclude that the colors are significantly not commonly distributed with a p-value = 8.80310^{-7}.

Note that, the chi-square test should be used only when all calculated expected values are greater than 5. 

                    # Access to the expected values res$expected                  
                    [1] 52.66667 52.66667 52.66667

Answer to Q2 comparing observed to expected proportions

                    tulip <- c(81, 50, 27) res <- chisq.test(tulip, p = c(1/2, 1/3, 1/6)) res                  
                                          Chi-squared test for given probabilities data:  tulip X-squared = 0.20253, df = 2, p-value = 0.9037

The p-value of the test is 0.9037, which is greater than the significance level alpha = 0.05. We can conclude that the observed proportions are not significantly different from the expected proportions.

Access to the values returned by chisq.test() function

The result of chisq.test() function is a list containing the following components:


  • statistic: the value the chi-squared test statistic.
  • parameter: the degrees of freedom
  • p.value: the p-value of the test
  • observed: the observed count
  • expected: the expected count

The format of the R code to use for getting these values is as follow:

                    # printing the p-value res$p.value                  
                    [1] 0.9036928                  
                    # printing the mean res$estimate                  
                    NULL

Infos

This analysis has been performed using R software (ver. 3.2.4).


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