Chi-Square Goodness of Fit Test on a TI-84 Calculator


Chi-Square Goodness of Fit Test is used to determine whether or not a categorical variable follows a hypothesized distribution.

This tutorial explains how to perform a Chi-Square Goodness of Fit Test on a TI-84 calculator.

Example: Chi-Square Goodness of Fit Test on a TI-84 Calculator

A shop owner claims that an equal number of customers come into his shop during each weekday. To test this hypothesis, an independent researcher records the number of customers that come into the shop on a given week and finds the following:

  • Monday: 50 customers
  • Tuesday: 60 customers
  • Wednesday: 40 customers
  • Thursday: 47 customers
  • Friday: 53 customers

We will use the following steps to perform a Chi-Square goodness of fit test to determine if the data is consistent with the shop owner’s claim.

Step 1: Input the data.

First, we will input the data values for the expected number of customers each day and the observed number of customers each day. Press Stat  and then press EDIT . Enter the following values for the observed number of customers in column L1 and the values for the expected number of customers in column L2:

Raw values in TI-84 calculator

Note: There were 250 customers total. Thus, if the shop owner expects an equal number to come into the shop each day then that would be 50 customers per day.

Step 2: Perform the Chi-Square goodness of fit test.

Next, we will perform the Chi-Square goodness of fit test. Press Stat and then scroll over to TESTS. Then scroll down to X2GOF-Test and press Enter.

Chi-Square goodness of fit test on a TI-84 calculator

For Observed, choose list L1. For Expected, choose list L2. For df (degrees of freedom), enter # categories – 1. In our case, we have 5-1 = 4. Then highlight Calculate and press Enter.

Chi-square goodness of fit test on a TI-84 calculator

The following output will automatically appear:

Chi-Square goodness of fit test output on TI-84 calculator

Step 3: Interpret the results.

The X2 test statistic for the test is 4.36 and the corresponding p-value is 0.3595. Since this p-value is not less than 0.05, we fail to reject the null hypothesis. This means we do not have sufficient evidence to say that the true distribution of customers is different from the distribution that the shop owner claimed.

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