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 	23 		
B 	5 		
C 	14 		
D 	15 		
E 	20 		

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=


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

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

### 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.

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    23  
B    5  
C    14  
D    15  
E    20

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=

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

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

Perform a chi-square goodness of fit test to evaluate the hypothesis that all 5 categories are equally likely. Includes detailed calculations of expected frequencies, Pearson residuals, chi-square test statistic (χ² = 12.287), degrees of freedom (d.f. = 4), and p-value (p = 0.015). Suitable for college-level statistics.

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