# 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.

# 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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