You are conducting a Goodness of Fit Chi-Squared hypothesis test (α = 0.05) for the claim that all 5 categories are equally likely to be selected. Complete the table.
Category 	Observed Frequency	Expected Frequency	((O-E)^2/E)
A 	9 	Correct
	Correct
B 	18 	Correct
	Correct
C 	21 	Correct
	Correct
D 	10 	Correct
	Correct
E 	21 	Correct
	Correct

Round Expected values accurate to exactly 2 decimal places. Retain unrounded numbers for future calculations.
Round the ((O-E)^2/E) values to 3 decimals.

What is the chi-square test-statistic for this data? (Round answer to 3 decimal places, and remember to use the unrounded Pearson residuals in your calculations.)
χ2=
Incorrect

What are the degrees of freedom for this test?
d.f.= Correct



What is the p-value for this sample? (Round answer to 3 decimal places.)
p-value =


The p-value is...

    less than (or equal to) α

greater than α

This test statistic leads to a decision to...

    reject the null
    fail to reject the null

As such, the final conclusion is that...

    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.



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.

---

### 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.

---

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.

You are conducting a Goodness of Fit Chi-Squared hypothesis test (α = 0.05) for the claim that all 5 categories are equally likely to be selected. Complete the table. What is the chi-square test-statistic for this data? What are the degrees of freedom for this test? What is the p-value for this sample? This test statistic leads to a decision to reject the null or fail to reject the null. As such, the final conclusion is that there is sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.

Explore the Chi-Squared Goodness of Fit Test for determining if five categories are equally likely. This analysis includes calculating the chi-square statistic, degrees of freedom, and p-value, leading to the conclusion on the null hypothesis. Key statistics concepts are covered, suitable for Grades 11-12.

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