You are conducting a multinomial Goodness of Fit hypothesis test for the claim that the 4 categories occur with the following frequencies:
  Ho Pa=0.1;  Pb =0.3; Pc= 0.5 ; Pd =0.1  


Complete the table. Report all answers accurate to three decimal places.
Category	Observed
Frequency	Expected
Frequency
A	33	
B	43	
C	44	
D	16	What is the chi-square test-statistic for this data? (2 decimal places) xsqared
 


What is the P-Value? (3 decimal places)
P-Value = 



For significance level alpha 0.005,

What would be the conclusion of this hypothesis test?
Fail to reject the Null Hypothesis
Reject the Null Hypothesis


Report all answers accurate to three decimal places.
Question HelpQuestion 2:  Video1



To solve this problem, we will compute the chi-square test statistic (\( \chi^2 \)) and p-value for the hypothesis test. Here are the steps:

---

### Step 1: Calculate Expected Frequencies
The expected frequency for each category is calculated using the formula:
\[
\text{Expected Frequency} = (\text{Total Observations}) \times (\text{Expected Proportion})
\]

### Step 2: Compute the Chi-Square Test Statistic
The chi-square test statistic is given by:
\[
\chi^2 = \sum \frac{(\text{Observed Frequency} - \text{Expected Frequency})^2}{\text{Expected Frequency}}
\]

### Step 3: Determine the P-Value
Using the chi-square distribution with \( k - 1 \) degrees of freedom (where \( k \) is the number of categories), compute the p-value.

### Step 4: Compare the P-Value to the Significance Level
Decide whether to reject or fail to reject \( H_0 \) based on the comparison of p-value to \( \alpha = 0.005 \).

---

### Step-by-Step Calculation

#### 1. Observed and Expected Frequencies
Let’s compute the expected frequencies for each category:

- Total Observations: \( 33 + 43 + 44 + 16 = 136 \)

| **Category** | **Observed (O)** | **Proportion (P)** | **Expected (E)** = \( 136 \times P \) |
|---------------|-------------------|---------------------|-------------------------------------|
| A             | 33               | 0.1                 | \( 136 \times 0.1 = 13.6 \)         |
| B             | 43               | 0.3                 | \( 136 \times 0.3 = 40.8 \)         |
| C             | 44               | 0.5                 | \( 136 \times 0.5 = 68.0 \)         |
| D             | 16               | 0.1                 | \( 136 \times 0.1 = 13.6 \)         |

#### 2. Chi-Square Test Statistic
\[
\chi^2 = \sum \frac{(O - E)^2}{E}
\]

| **Category** | **O** | **E** | \( (O - E)^2 \) | \( \frac{(O - E)^2}{E} \) |
|---------------|--------|-------|-----------------|--------------------------|
| A             | 33     | 13.6  | \( (33 - 13.6)^2 = 372.1 \) | \( \frac{372.1}{13.6} = 27.35 \) |
| B             | 43     | 40.8  | \( (43 - 40.8)^2 = 4.84 \)  | \( \frac{4.84}{40.8} = 0.12 \)  |
| C             | 44     | 68.0  | \( (44 - 68.0)^2 = 576.0 \) | \( \frac{576.0}{68.0} = 8.47 \) |
| D             | 16     | 13.6  | \( (16 - 13.6)^2 = 5.76 \)  | \( \frac{5.76}{13.6} = 0.42 \)  |

Summing these:
\[
\chi^2 = 27.35 + 0.12 + 8.47 + 0.42 = 36.36
\]

#### 3. Degrees of Freedom and P-Value
Degrees of freedom (\( \text{df} \)) = \( k - 1 = 4 - 1 = 3 \).

Using a chi-square table or calculator, for \( \chi^2 = 36.36 \) and \( \text{df} = 3 \), we find:

\[
\text{P-Value} < 0.001
\]

#### 4. Hypothesis Test Conclusion
At \( \alpha = 0.005 \), \( \text{P-Value} < \alpha \), so we **reject the null hypothesis**.

---

### Final Answers
1. \( \chi^2 = 36.36 \) (2 decimal places)
2. P-Value \( < 0.001 \) (3 decimal places)
3. Conclusion: **Reject the Null Hypothesis**

Would you like to see the detailed steps or calculations for p-values? Let me know! 😊 

---

Here are 5 follow-up questions:
1. What would happen if the sample size increased?
2. How does changing the significance level (\( \alpha \)) affect the hypothesis test?
3. Why is it necessary to calculate degrees of freedom?
4. How does the chi-square test differ from other goodness-of-fit tests?
5. Can this method be applied to continuous data?

**Tip:** Always double-check observed and expected frequencies for accuracy in hypothesis testing.

You are conducting a multinomial Goodness of Fit hypothesis test for the claim that the 4 categories occur with the following frequencies:
  Ho Pa=0.1; Pb=0.3; Pc=0.5; Pd=0.1

Complete the table. Report all answers accurate to three decimal places.
Category Observed Frequency Expected Frequency
A 33
B 43
C 44
D 16

What is the chi-square test-statistic for this data? (2 decimal places)
What is the P-Value? (3 decimal places)
For significance level alpha 0.005,
What would be the conclusion of this hypothesis test?
Fail to reject the Null Hypothesis
Reject the Null Hypothesis

Learn how to conduct a Chi-Square Goodness of Fit test for multinomial data. This problem evaluates observed versus expected frequencies across four categories, calculates the test statistic (χ² = 36.36), determines the p-value (<0.001), and concludes the hypothesis test. Suitable for advanced high school or college-level statistics.

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