A **chi-square goodness of fit test** is used when we want to formally test whether or not a categorical variable follows a hypothesized distribution.

The following example show how to conduct a chi-square goodness of fit test.

**Example: Chi-Square Goodness of Fit Test**

An owner of a shop claims that 30% of all his weekend customers visit on Friday, 50% on Saturday, and 20% on Sunday. An independent researcher visits the shop on a random weekend and finds that 91 customers visit on Friday, 104 visit on Saturday, and 65 visit on Sunday.

** Is this data consistent with the shop owner’s claim? Use a 0.05 level of significance.**

**Step 1. State the hypotheses. **

The null hypothesis (H_{0}): The shop owner’s claim is correct: 30% of customers visit on Friday, 50% on Saturday, and 20% on Sunday.

The alternative hypothesis: (Ha): At least one of the proportions in the null hypothesis is not correct.

**Step 2. Determine a significance level to use.**

The problem tells us that we are to use a .05 level of significance.

**Step 3. Find the test statistic.**

The test statistic is X^{2} = Σ [ (O_{i} – E_{i})^{2} / E_{i} ]

Where Σ is just a fancy symbol that means “sum”, O_{i} is the observed frequency at level *i* of the variable, and E_{i} is the expected frequency at level *i* of the variable.

There were 260 customers who visited the shop on this particular weekend (91 on Friday + 104 on Saturday + 65 on Sunday).

According to the shop owner, we should *expect* 30% * 260 = 78 of the total customers to visit on Friday. The *observed *number of people who visited on Friday was 91. So for Friday we have:

(O – E)^{2} / E = (91 – 78)^{2} / 78 = 2.167

According to the shop owner, we should *expect* 50% * 260 = 130 of the total customers to visit on Saturday. The *observed *number of people who visited on Saturday was 104. So for Saturday we have:

(O – E)^{2} / E = (104 – 130)^{2} / 130 = 5.2

According to the shop owner, we should *expect* 20% * 260 = 52 of the total customers to visit on Sunday. The *observed *number of people who visited on Sunday was 65. So for Sunday we have:

(O – E)^{2} / E = (65 – 52)^{2} / 52 = 3.25

To find the test statistic, we simply sum up these numbers: 2.167 + 5.2 + 3.25 = **10.617**

Use the Chi-Square Calculator with a degrees of freedom = k-1 (k is the number of levels of the variable) = 3-1 = 2, Chi-square critical value = 10.617, and click “Calculate p-value” to find that the p-value = .99505. Then 1 – .99505 = **.00495**.

**Step 4. Reject or fail to reject the null hypothesis.**

Since the p-value (.00495) is less than our significance level of .05, we reject the null hypothesis.

**Step 5. Interpret the results. **

Since we rejected the null hypothesis, we have sufficient evidence to say the true distribution of customers who come in to this shop on weekends is not equal to 30% on Friday, 50% on Saturday, and 20% on Sunday.