Step 1: Compute Expected Frequencies

Since the claim is that all 5 categories are equally likely to be selected, the expected frequency for each category can be calculated as follows:

$\text{Expected Frequency (E)} = \frac{\text{Total Observed Frequencies}}{\text{Number of Categories}} = \frac{23 + 5 + 14 + 15 + 20}{5} = \frac{77}{5} = 15.40$

So, the expected frequency for each category is 15.40.

Step 2: Compute Pearson Residuals $(O - E)^2 / E$

Now, compute the residuals for each category using the formula:

$\left(\frac{(O - E)^2}{E}\right)$

Category A:

$\left(\frac{(23 - 15.40)^2}{15.40}\right) = \frac{(7.60)^2}{15.40} = \frac{57.76}{15.40} = 3.751$

Category B:

$\left(\frac{(5 - 15.40)^2}{15.40}\right) = \frac{(-10.40)^2}{15.40} = \frac{108.16}{15.40} = 7.025$

Category C:

$\left(\frac{(14 - 15.40)^2}{15.40}\right) = \frac{(-1.40)^2}{15.40} = \frac{1.96}{15.40} = 0.127$

Category D:

$\left(\frac{(15 - 15.40)^2}{15.40}\right) = \frac{(-0.40)^2}{15.40} = \frac{0.16}{15.40} = 0.010$

Category E:

$\left(\frac{(20 - 15.40)^2}{15.40}\right) = \frac{(4.60)^2}{15.40} = \frac{21.16}{15.40} = 1.374$

Step 3: Chi-Square Test Statistic $(\chi^2)$

The chi-square test statistic is the sum of all the Pearson residuals:

$\chi^2 = 3.751 + 7.025 + 0.127 + 0.010 + 1.374 = 12.287$

Thus, the chi-square test statistic is:

$\boxed{\chi^2 = 12.287}$

Step 4: Degrees of Freedom

The degrees of freedom for a chi-square goodness of fit test is calculated as:

$d.f. = \text{Number of Categories} - 1 = 5 - 1 = 4$

So, the degrees of freedom is:

$\boxed{d.f. = 4}$

Step 5: Find the p-value

Using a chi-square distribution table or calculator with $\chi^2 = 12.287$ and $d.f. = 4$, the p-value is found to be:

$\boxed{p\text{-value} = 0.015}$

Summary of Results:

• Chi-square test statistic: $\chi^2 = 12.287$
• Degrees of freedom: $d.f. = 4$
• p-value: $0.015$

Since the p-value is less than the significance level $\alpha = 0.05$, you would reject the null hypothesis and conclude that the categories are not equally likely.

Would you like more details on any part of the process?

---

Here are 5 related questions to deepen your understanding:

1. How is the chi-square test used to assess goodness of fit in other contexts?
2. What assumptions are necessary for the chi-square goodness-of-fit test to be valid?
3. How do you interpret a chi-square test statistic that is close to zero?
4. Can the chi-square goodness-of-fit test be applied to continuous data?
5. What is the effect of increasing the number of categories on the degrees of freedom?

Tip: Always check the expected frequencies; if any are less than 5, consider using a different test like Fisher's exact test.