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You are conducting a multinomial Goodness of Fit hypothesis test for the claim that the 4 categories occur with the following frequencies:
[tex]\[ H_o: p_A=0.5; \ p_B=0.1; \ p_C=0.1; \ p_D=0.3 \][/tex]

Complete the table. Report all answers accurate to three decimal places.

[tex]\[
\begin{tabular}{|c|c|c|}
\hline
Category & Observed Frequency & Expected Frequency \\
\hline
A & 40 & $\square$ \\
\hline
B & 18 & $\square$ \\
\hline
C & 10 & $\square$ \\
\hline
D & 20 & $\square$ \\
\hline
\end{tabular}
\][/tex]

What is the chi-square test-statistic for this data? (2 decimal places)

[tex]\[ \chi^2 = \square \][/tex]

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

[tex]\[ \text{P-Value} = \square \][/tex]

For significance level [tex]\(\alpha = 0.005\)[/tex],

What would be the conclusion of this hypothesis test?

A. Fail to reject the Null Hypothesis
B. Reject the Null Hypothesis

Sagot :

GinnyAnsweridn
To solve this hypothesis test and complete the required table, we need to determine the expected frequency for each category, calculate the chi-square test statistic, determine the p-value, and finally, make a conclusion based on the significance level. Here is the detailed, step-by-step solution:

### Step 1: Observed Frequencies
- Category A: 40
- Category B: 18
- Category C: 10
- Category D: 20

### Step 2: Total Number of Observations
Summing up all the observed frequencies gives:
[tex]\[ \text{Total Observations} = 40 + 18 + 10 + 20 = 88 \][/tex]

### Step 3: Expected Frequencies
The expected frequency for each category is calculated using the formula:
[tex]\[ \text{Expected Frequency} = (\text{Total Observations}) \times (\text{Expected Probability}) \][/tex]

So for each category:
- Category A: [tex]\( 88 \times 0.5 = 44.0 \)[/tex]
- Category B: [tex]\( 88 \times 0.1 = 8.8 \)[/tex]
- Category C: [tex]\( 88 \times 0.1 = 8.8 \)[/tex]
- Category D: [tex]\( 88 \times 0.3 = 26.4 \)[/tex]

### Completed Table

\begin{tabular}{|c|l|l|}
\hline
Category & Observed Frequency & Expected Frequency \\
\hline
A & 40 & 44.0 \\
\hline
B & 18 & 8.8 \\
\hline
C & 10 & 8.8 \\
\hline
D & 20 & 26.4 \\
\hline
\end{tabular}

### Step 4: Chi-Square Test Statistic Calculation
The chi-square test statistic is calculated using the formula:
[tex]\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \][/tex]
where [tex]\( O_i \)[/tex] are the observed frequencies and [tex]\( E_i \)[/tex] are the expected frequencies.

Thus,
[tex]\[ \chi^2 = \frac{(40 - 44.0)^2}{44.0} + \frac{(18 - 8.8)^2}{8.8} + \frac{(10 - 8.8)^2}{8.8} + \frac{(20 - 26.4)^2}{26.4} \approx 11.7 \][/tex]

So, the chi-square test statistic is [tex]\( 11.7 \)[/tex].

### Step 5: P-Value Calculation
The p-value corresponding to the chi-square test statistic with appropriate degrees of freedom (df) is given as:
[tex]\[ \text{P-Value} = 0.008 \][/tex]

### Step 6: Conclusion
Given the significance level alpha [tex]\( \alpha = 0.005 \)[/tex]:
- If the p-value is greater than [tex]\( \alpha \)[/tex], we fail to reject the null hypothesis.
- If the p-value is less than or equal to [tex]\( \alpha \)[/tex], we reject the null hypothesis.

Since [tex]\( 0.008 > 0.005 \)[/tex], we Fail to reject the Null Hypothesis.

### Final Summary

- Expected Frequencies:
\begin{itemize}
\item Category A: 44.0
\item Category B: 8.8
\item Category C: 8.8
\item Category D: 26.4
\end{itemize}

- Chi-Square Test Statistic: [tex]\( \chi^2 = 11.7 \)[/tex]
- P-Value: [tex]\( 0.008 \)[/tex]
- Conclusion: Fail to reject the Null Hypothesis
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