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:

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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 $\alpha = 0.005$.

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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.

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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! 😊

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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.