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Chi-square goodness of fit formula, To see if a categorical variable follows a hypothesized distribution, a Chi-Square Goodness of Fit Test is utilized.

This lesson will show you how to use R to run a Chi-Square Goodness of Fit Test.

Chi-square goodness of fit formula in R

Every day, an equal number of clients enter a business, according to a vendor. To test this theory, a corporate executive records the number of customers who visit the shop in a given week and discovers the following.

Monday: 250 customers, Tuesday: 230 customers, Wednesday: 265 customers, Thursday: 235 customers, and Friday: 223 customers

To evaluate if the data is consistent with the vendor claim, do the Chi-Square goodness of fit test in R using the instructions below.

Gather information.

First, we’ll make two arrays to store our observed frequencies and expected customer proportions for each day.

observedfreq <- c(250, 230, 265, 235, 223)
expectedprop <- c(0.2, 0.2, 0.2, 0.2, 0.2)

The expected frequency sum should be 1.

Use the Chi-Square Goodness of Fit Test to see if you’re a good fit.

Let’s see the null and alternative hypotheses for a Chi-Square Goodness of Fit Test are as follows.

H0: A variable follows a hypothesized distribution.
H1: A variable does not follow a hypothesized distribution.

The Chi-Square Goodness of Fit Test can then be performed using the chisq.test() function, which has the following syntax.

chisq.test(x, p)

where:
x: The observed frequencies are represented numerically as a vector.
p: a numerical vector of proportions to be expected.

In our example, the following code demonstrates how to utilize this function.

conduct a Chi-Square Goodness-of-Fit Test

chisq.test(x= observedfreq, p= expectedprop)
Chi-squared test for given probabilities
data: observedfreq
X-squared = 4.7265, df = 4, p-value = 0.3165

The p-value for the Chi-Square test is 0.3165, and the Chi-Square test statistic is 4.7.

The p-value is equivalent to a Chi-Square value with n-1 degrees of freedom, where n is the number of categories. degrees of freedom= 5-1 = 4 in this situation.

The Chi-Square to P-Value Calculator can be used to establish that the p-value for X2 = 4.7 with degrees of freedom= 4 is 0.3165.

Conclusion

We cannot reject the null hypothesis since the p-value (0.3165) is not less than 0.05.

This means we don’t have enough evidence to conclude that the genuine customer distribution differs from the vendor’s claimed distribution.

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