Question: Given the 5 categories in the table below, test the claim that the categories are equally likely to be selected at a $\alpha = 0.05$ significance level.
Category Observed Frequency Expected Frequency
A 24
B 9
C 14
D 23
E 13
a. Complete the table by calculating the expected frequencies.
b. What is the chi-square test statistic? Round to three decimal places.
$\chi^2 =$
c. What are the degrees of freedom?
d.f. =
d. What is the p-value for this sample? Round to four decimal places.
p-value =
e. Is the p-value less than $\alpha$?
f. Make a decision.
g. Make a conclusion.
There is sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.
There is not sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.
The sample data support the claim that all 5 categories are equally likely to be selected.
There is not sufficient sample evidence to support the claim that all 5 categories are equally likely to be selected.
1. **State the problem:** We want to test if the 5 categories are equally likely to be selected using a chi-square goodness-of-fit test at significance level $\alpha = 0.05$.
2. **Calculate total observed frequency:**
$$\text{Total} = 24 + 9 + 14 + 23 + 13 = 83$$
3. **Calculate expected frequencies:** Since categories are equally likely, expected frequency for each category is:
$$E = \frac{\text{Total}}{5} = \frac{83}{5} = 16.6$$
4. **Complete the table with expected frequencies:**
| Category | Observed ($O$) | Expected ($E$) |
|----------|----------------|---------------|
| A | 24 | 16.6 |
| B | 9 | 16.6 |
| C | 14 | 16.6 |
| D | 23 | 16.6 |
| E | 13 | 16.6 |
5. **Calculate chi-square test statistic:**
$$\chi^2 = \sum \frac{(O - E)^2}{E}$$
Calculate each term:
$$\frac{(24 - 16.6)^2}{16.6} = \frac{7.4^2}{16.6} = \frac{54.76}{16.6} \approx 3.300$$
$$\frac{(9 - 16.6)^2}{16.6} = \frac{(-7.6)^2}{16.6} = \frac{57.76}{16.6} \approx 3.481$$
$$\frac{(14 - 16.6)^2}{16.6} = \frac{(-2.6)^2}{16.6} = \frac{6.76}{16.6} \approx 0.407$$
$$\frac{(23 - 16.6)^2}{16.6} = \frac{6.4^2}{16.6} = \frac{40.96}{16.6} \approx 2.468$$
$$\frac{(13 - 16.6)^2}{16.6} = \frac{(-3.6)^2}{16.6} = \frac{12.96}{16.6} \approx 0.781$$
Sum these:
$$\chi^2 \approx 3.300 + 3.481 + 0.407 + 2.468 + 0.781 = 10.437$$
6. **Degrees of freedom:**
$$d.f. = \text{number of categories} - 1 = 5 - 1 = 4$$
7. **Find p-value:** Using chi-square distribution table or calculator for $\chi^2 = 10.437$ with $4$ degrees of freedom,
$$p\text{-value} \approx 0.0349$$
8. **Compare p-value with $\alpha$:**
$$0.0349 < 0.05$$ so the p-value is less than $\alpha$.
9. **Decision:** Since $p$-value $< \alpha$, reject the null hypothesis.
10. **Conclusion:** There is sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.