To solve this problem, we will compute each required value step-by-step.

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Step 1: Pooled Variance ($s_p^2$)

The pooled variance is computed using the formula:
$s_p^2 = \frac{SS_1 + SS_2}{df_1 + df_2}$
where $df_1 = n_1 - 1$ and $df_2 = n_2 - 1$.

Calculations:

• For each sample, $n_1 = n_2 = 9$, so $df_1 = df_2 = 9 - 1 = 8$.
• Total degrees of freedom $df_{\text{total}} = df_1 + df_2 = 8 + 8 = 16$.
• Sum of squares: $SS_1 = 546$, $SS_2 = 606$.

$s_p^2 = \frac{SS_1 + SS_2}{df_{\text{total}}} = \frac{546 + 606}{16} = \frac{1152}{16} = 72$

Thus, the pooled variance is $s_p^2 = 72$.

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Step 2: Estimated Standard Error ($s(M_1 - M_2)$)

The estimated standard error for the difference between two sample means is: $s(M_1 - M_2) = \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2}}$

Calculations:

$s(M_1 - M_2) = \sqrt{\frac{72}{9} + \frac{72}{9}} = \sqrt{8 + 8} = \sqrt{16} = 4$

Thus, the estimated standard error is $s(M_1 - M_2) = 4$.

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Step 3: Test Statistic ($t$)

The $t$-statistic is computed as: $t = \frac{M_1 - M_2}{s(M_1 - M_2)}$

Given:

• Mean difference $M_1 - M_2 = 8$,
• $s(M_1 - M_2) = 4$.

$t = \frac{8}{4} = 2$

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Step 4: Critical Value ($t_{\text{critical}}$)

For a two-tailed test at $\alpha = 0.05$ with $df = 16$, we consult a $t$-distribution table or use software.
From the table: $t_{\text{critical}} = \pm 2.120$

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Step 5: Decision

• If $|t| > t_{\text{critical}}$, we reject the null hypothesis.
• Here, $|t| = 2$ and $t_{\text{critical}} = \pm 2.120$.

Since $|t| = 2$ is not greater than $t_{\text{critical}} = 2.120$, we fail to reject the null hypothesis.

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Conclusion

There is not enough evidence to conclude a significant difference between the two treatments.
Final Answer:

• $t_{\text{critical}} = \pm 2.120$
• $t = 2$
• Fail to reject the null hypothesis; there is no significant difference.

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Let me know if you'd like further explanation or have any questions! Here are some related questions to explore:

1. How do degrees of freedom influence the critical $t$-value?
2. What assumptions are made when pooling variances?
3. How would the conclusion change if $n = 20$ per sample?
4. What is the effect size, and how is it computed here?
5. Why is the test two-tailed instead of one-tailed?

Tip: Always check the assumptions of homogeneity of variance before pooling variances!