Betty recorded the number of visitors at her art gallery during the week:


**Day**

**Visitor**

Tuesday

28

Wednesday

20

Thursday

18

Friday

14


She expected to have 20 visitors each day. To determine whether the number of visitors follows a uniform distribution, a chi-square test for goodness of fit should be performed (alpha = 0.10).



**Using the data above, what is the chi-square test statistic (answer choices are rounded to the hundredths place)?**

A.)

5.2

B.)

0.83

C.)

0.20

D.)

12.5

To calculate the chi-square test statistic for goodness of fit, we need to follow these steps:

1. **Determine the expected frequency**:
   Betty expected 20 visitors each day, so the expected frequency for each day is 20.

2. **Write the observed frequencies**:
   From the dataTo calculate the chi-square test statistic for goodness of fit, we need to follow these steps:

1. **Determine the expected frequency**:
   Betty expected 20 visitors each day, so the expected frequency for each day is 20.

2. **Write the observed frequencies**:
   From the 
   - Tuesday: 28
   - Wednesday: 20
   - Thursday: 18
   - Friday: 14

3. **Apply the chi-square formula**:
   The chi-square statistic is calculated using the formula:
   \[
   \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}
   \]
   where:
   - \( O_i \) is the observed frequency for each category (each day).
   - \( E_i \) is the expected frequency for each category (20 visitors per day in this case).

   We compute this for each day:

   - **Tuesday**: \( \frac{(28 - 20)^2}{20} = \frac{64}{20} = 3.2 \)
   - **Wednesday**: \( \frac{(20 - 20)^2}{20} = \frac{0}{20} = 0 \)
   - **Thursday**: \( \frac{(18 - 20)^2}{20} = \frac{4}{20} = 0.2 \)
   - **Friday**: \( \frac{(14 - 20)^2}{20} = \frac{36}{20} = 1.8 \)

4. **Sum the values**:
   \[
   \chi^2 = 3.2 + 0 + 0.2 + 1.8 = 5.2
   \]

So, the chi-square test statistic is **5.2**.

**Answer**: A.) 5.2

---

Would you like more details or have any questions? Here are some related questions to expand on this:

1. What is the critical value for this chi-square test at \(\alpha = 0.10\)?
2. How would the test result differ with an alpha of 0.05?
3. What are the degrees of freedom in this problem?
4. How do we interpret the result of a chi-square goodness of fit test?
5. Can this chi-square test be applied to larger datasets, and how does it scale?

**Tip**: For a chi-square test, the degrees of freedom are usually calculated as the number of categories minus one.

Betty recorded the number of visitors at her art gallery during the week: Tuesday: 28, Wednesday: 20, Thursday: 18, Friday: 14. She expected to have 20 visitors each day. To determine whether the number of visitors follows a uniform distribution, a chi-square test for goodness of fit should be performed (alpha = 0.10). Using the data above, what is the chi-square test statistic (answer choices are rounded to the hundredths place)? A) 5.2, B) 0.83, C) 0.20, D) 12.5

Perform a chi-square goodness of fit test to determine if Betty's art gallery visitor numbers follow a uniform distribution. This problem uses the chi-square test formula, with expected visitors being 20 per day and observed visitors ranging from 14 to 28. Find out how to calculate the test statistic, suitable for Grades 11-12.

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