Let's break this down step by step.

Step 1: Determine the Expected Frequencies

Since the hypothesis claims that all 5 categories are equally likely to be selected, the expected frequency for each category is the same. To calculate it, we need to know the total number of observations (sum of the observed frequencies).

Given the observed frequencies for each category:

• A = 9
• B = 18
• C = 21
• D = 10
• E = 21

Total observed frequency (O):
$\text{Total} = 9 + 18 + 21 + 10 + 21 = 79$

Since we have 5 categories, the expected frequency for each category (E) is: $E = \frac{79}{5} = 15.8$

So, the expected frequency for each category is 15.8.

Step 2: Calculate the Chi-Square Statistic

Now, let's calculate the value of $\chi^2$ using the formula for the test statistic: $\chi^2 = \sum \frac{(O - E)^2}{E}$

Where:

• $O$ is the observed frequency,
• $E$ is the expected frequency.

We’ll calculate the values for each category:

1. Category A: $\frac{(9 - 15.8)^2}{15.8} = \frac{(-6.8)^2}{15.8} = \frac{46.24}{15.8} = 2.931$

2. Category B: $\frac{(18 - 15.8)^2}{15.8} = \frac{(2.2)^2}{15.8} = \frac{4.84}{15.8} = 0.307$

3. Category C: $\frac{(21 - 15.8)^2}{15.8} = \frac{(5.2)^2}{15.8} = \frac{27.04}{15.8} = 1.712$

4. Category D: $\frac{(10 - 15.8)^2}{15.8} = \frac{(-5.8)^2}{15.8} = \frac{33.64}{15.8} = 2.130$

5. Category E: $\frac{(21 - 15.8)^2}{15.8} = \frac{(5.2)^2}{15.8} = \frac{27.04}{15.8} = 1.712$

Now, sum all the calculated values: $\chi^2 = 2.931 + 0.307 + 1.712 + 2.130 + 1.712 = 8.792$

Thus, the chi-square statistic is approximately $\chi^2 = 8.792$.

Step 3: Degrees of Freedom

The degrees of freedom for a Goodness of Fit test is calculated as: $\text{d.f.} = k - 1$ Where $k$ is the number of categories. In this case, $k = 5$, so: $\text{d.f.} = 5 - 1 = 4$

Thus, the degrees of freedom are 4.

Step 4: Find the p-value

To find the p-value, we look up the chi-square statistic ($\chi^2 = 8.792$) in the chi-square distribution table for 4 degrees of freedom.

Using a chi-square distribution table or calculator, the p-value for $\chi^2 = 8.792$ with 4 degrees of freedom is approximately 0.067.

Step 5: Decision and Conclusion

• p-value = 0.067.
• The significance level is $\alpha = 0.05$.

Since the p-value is greater than $\alpha$ (0.067 > 0.05), we fail to reject the null hypothesis.

Final Conclusion

There is not sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.

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Recap of Answers:

• Chi-square test statistic: $\chi^2 = 8.792$
• Degrees of freedom: 4
• p-value: 0.067
• Decision: Fail to reject the null hypothesis
• Conclusion: There is not sufficient evidence to reject the claim that all 5 categories are equally likely to be selected.

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Would you like further clarification on any of the steps? Here are some related questions you could explore next:

1. What are the assumptions required for performing a chi-square goodness of fit test?
2. How would the decision change if the p-value was less than 0.05?
3. How does the number of categories affect the chi-square statistic and the degrees of freedom?
4. What is the role of expected frequency in the chi-square test?
5. How do we handle situations where any expected frequency is below 5?

Tip: Always check that your expected frequencies are at least 5 to ensure the validity of the chi-square test.