A dentist wondered if his appointments were distributed evenly from Monday to Friday each week. He took a random sample of 500 appointments and recorded which day of the week they were booked for. His results are:

DayMondayTuesdayWednesdayThursdayFridayAppointment Frequency

96

113

119

91

81

The dentist would like to use these results to conduct a  χ 2 χ 2 goodness-of-fit test to determine if the distribution of appointments agrees with an even distribution using a 2.5% level of significance.

Chi Square Distribution Table

**a.** Calculate the test statistic.

 χ 2= χ 2= 

Round to two decimal places if necessary

**b.** Using the Chi Square Distribution Table, determine the critical value(s) for the hypothesis test.

 +

**c.** Conclude whether to reject the null hypothesis or not based on the test statistic.

Reject

Fail to Reject

To perform the χ² goodness-of-fit test, we follow these steps:

### **a. Calculate the test statistic (χ²)**

1. **Null Hypothesis (H₀):** The appointments are distributed evenly across the days.
2. **Alternative Hypothesis (H₁):** The appointments are not distributed evenly across the days.

First, we calculate the expected frequency for each day under the null hypothesis, which assumes an even distribution:

\[
\text{Expected Frequency} = \frac{\text{Total Appointments}}{\text{Number of Days}} = \frac{500}{5} = 100
\]

Next, we calculate the test statistic (χ²) using the formula:

\[
\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}
\]

where \( O_i \) is the observed frequency and \( E_i \) is the expected frequency.

| Day       | Observed (O) | Expected (E) | (O - E)  | (O - E)²  | (O - E)² / E |
|-----------|--------------|--------------|----------|-----------|--------------|
| Monday    | 96           | 100          | -4       | 16        | 0.16         |
| Tuesday   | 113          | 100          | 13       | 169       | 1.69         |
| Wednesday | 119          | 100          | 19       | 361       | 3.61         |
| Thursday  | 91           | 100          | -9       | 81        | 0.81         |
| Friday    | 81           | 100          | -19      | 361       | 3.61         |
| **Total** | **500**      | **500**      |          |           | **9.88**     |

Thus, the test statistic is \( \chi^2 = 9.88 \).

### **b. Determine the critical value(s)**

The degrees of freedom (df) are calculated as:

\[
\text{df} = \text{Number of categories} - 1 = 5 - 1 = 4
\]

Using the Chi Square Distribution Table at a 2.5% level of significance and 4 degrees of freedom, the critical value is \( \chi^2_{0.025, 4} = 11.143 \).

### **c. Conclusion**

To determine whether to reject the null hypothesis, we compare the test statistic to the critical value:

- **Test statistic (χ²) = 9.88**
- **Critical value = 11.143**

Since the test statistic (9.88) is less than the critical value (11.143), we **fail to reject the null hypothesis**. There is insufficient evidence to conclude that the appointments are not distributed evenly across the days.

---

**Do you have any questions or would you like further details?**

### Follow-up Questions

1. What is the purpose of the χ² goodness-of-fit test in this context?
2. How do we determine the expected frequency for each category in a χ² test?
3. What does the test statistic represent in a χ² test?
4. How is the critical value for a χ² test determined?
5. What is the significance level, and how does it affect hypothesis testing?
6. What does "failing to reject the null hypothesis" imply in this scenario?
7. How would the conclusion change if the test statistic were greater than the critical value?
8. What assumptions must be met to correctly apply the χ² goodness-of-fit test?

### Tip

When conducting hypothesis tests, it's essential to define the null and alternative hypotheses clearly and understand what each hypothesis represents in the context of the problem.

Learn how to conduct a chi-square goodness-of-fit test to determine if a dentist's appointments are evenly distributed across weekdays. Includes step-by-step calculation of test statistic, critical value determination, and conclusion based on significance level.

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