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In this image, a chi-square goodness-of-fit test is being performed. Let's break down each part of the question:

### a. Completing the table:
The table already shows the observed frequencies for each category (A, B, C, D, E) and the expected frequency for each category (16.4). There’s no further action required here since the expected values have already been filled in.

### b. Chi-square test statistic:
The chi-square test statistic has been calculated as:
\[
\chi^2 = 14.887
\]
This value was rounded to three decimal places as requested.

### c. Degrees of freedom (d.f.):
The degrees of freedom for a chi-square test is calculated using the formula:
\[
\text{d.f.} = k - 1
\]
Where \( k \) is the number of categories. Since there are 5 categories (A, B, C, D, E), the degrees of freedom is:
\[
d.f. = 5 - 1 = 4
\]
This matches the given value of \( d.f. = 4 \).

### d. p-value for the sample:
The p-value is given as \( 0.005 \), which has been rounded to four decimal places.

### e. Is the p-value less than \( \alpha \)?
Assuming \( \alpha = 0.05 \) (the typical significance level), the p-value of 0.005 is indeed less than \( \alpha \), so the answer is **Yes**.

### f. Decision:
Since the p-value is less than \( \alpha \), the null hypothesis is rejected. The correct decision is to "Reject the null."

### g. Conclusion:
By rejecting the null hypothesis, the conclusion is:
"There is sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected."

### Summary:
- The chi-square test statistic is 14.887.
- Degrees of freedom are 4.
- The p-value is 0.005.
- Since the p-value is less than \( \alpha \), reject the null hypothesis.
- Conclusion: There is sufficient evidence to reject the claim that all 5 categories are equally likely to be selected.

Would you like any more details or clarifications on this?

Here are 5 relative questions for further understanding:
1. How is the chi-square test statistic calculated from the observed and expected frequencies?
2. What does rejecting the null hypothesis mean in a chi-square test?
3. How does changing the significance level \( \alpha \) affect the decision-making process?
4. What are some assumptions of the chi-square goodness-of-fit test?
5. How are expected frequencies determined in a chi-square test?

**Tip**: In hypothesis testing, always compare the p-value to your significance level \( \alpha \) to make your decision about rejecting or failing to reject the null hypothesis.

A chi-square goodness-of-fit test is performed using observed and expected frequencies for five categories. The chi-square test statistic is 14.887, degrees of freedom is 4, and the p-value is 0.005. Is there sufficient evidence to reject the claim that all 5 categories are equally likely to be selected?

Learn how to perform a chi-square goodness-of-fit test using observed and expected frequencies. This example includes a test statistic of 14.887, degrees of freedom of 4, and a p-value of 0.005. We explain how to reject the null hypothesis at a significance level of 0.05. Suitable for students in Grades 10-12.

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