Question 1207903
To address this hypothesis testing problem, we'll break down the process step-by-step. Here are the steps for parts (a), (b), and (c):

### (a) Find the Rejection Region for the Test

In hypothesis testing, the rejection region is determined based on the significance level (α). Given α = 0.01, we need to find critical values for the t-distribution with the appropriate degrees of freedom.

1. **Degrees of freedom (df)**: Since we have two samples, we can calculate the degrees of freedom using the formula:
   \[
   df = n_1 + n_2 - 2
   \]
   where \(n_1 = 18\) and \(n_2 = 13\). Therefore,
   \[
   df = 18 + 13 - 2 = 29
   \]

2. **Determine the critical t-value**: Since we're not told whether this is a one-tailed or two-tailed test, I'll assume it's a two-tailed test for the sake of this example. For α = 0.01 (which is split between the two tails), each tail would contain 0.005. You can find the critical t-value using a t-table or a calculator for \(df = 29\) and α/2 = 0.005.

   Using a t-table or calculator, you would find:
   - Critical t values for df = 29 at the 0.005 level.

   For a two-tailed test:
   - \(t < t_{\alpha/2}\) (negative critical value) and \(t > t_{(1-\alpha/2)}\) (positive critical value).
   
   Let's assume that the t-table shows critical values of:
   \[
   t_{\alpha/2} \approx \pm 2.756
   \]
   Thus, the rejection region will be:
   \[
   t < -2.756 \quad \text{and} \quad t > 2.756
   \]

3. **Final Result**:
   - For the rejection region:
     - **t > 2.756**
     - **t < -2.756**  

### (b) Find the Value of the Test Statistic

To compute the t-statistic, we can use the following formula for two independent samples:
\[
t = \frac{\bar{X_1} - \bar{X_2}}{S_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}
\]
where:
- \(\bar{X_1} = 34.6\) (mean of the first sample)
- \(\bar{X_2} = 32.1\) (mean of the second sample)
- \(S_p\) is the pooled standard deviation, calculated as:
\[
S_p = \sqrt{\frac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2}}
\]

Given:
- Sample variance \(S_1^2 = 4.5\)
- Sample variance \(S_2^2 = 5.9\)

1. Calculate \(S_p\):
   \[
   S_p = \sqrt{\frac{(18 - 1) \cdot 4.5 + (13 - 1) \cdot 5.9}{18 + 13 - 2}} = \sqrt{\frac{17 \cdot 4.5 + 12 \cdot 5.9}{29}} 
   \]
   \[
   = \sqrt{\frac{76.5 + 70.8}{29}} = \sqrt{\frac{147.3}{29}} \approx \sqrt{5.08} \approx 2.253
   \]

2. Now, substitute back into the t formula:
\[
t = \frac{34.6 - 32.1}{2.253\sqrt{\frac{1}{18} + \frac{1}{13}}}
\]
Calculate the individual components:
\[
\sqrt{\frac{1}{18} + \frac{1}{13}} = \sqrt{0.05556 + 0.07692} = \sqrt{0.13248} \approx 0.363
\]

And now find \(t\):
\[
t = \frac{2.5}{2.253 \times 0.363} \approx \frac{2.5}{0.817} \approx 3.059
\]

So:
\[
t \approx 3.059 \quad \text{(rounded to three decimal places)}
\]

### (c) Find the Approximate p-value for the Test

To find the p-value associated with the test statistic \(t \approx 3.059\) and \(df = 29\):

1. Because this is a one-tailed test, you would look up the tail probability of \(t = 3.059\) in the t-distribution table or use statistical software/calculator.

2. Generally, for values beyond \(t > 2.756\), we know that the p-value will be less than alpha = 0.01. Thus, we can check ranges. For \(t \approx 3.059\), it is likely to be in the range of < 0.01.

After checking the statistical tables or calculations, we find that:
\[
\text{p-value} < 0.010
\]

### Conclusion

So, your answers would be:
- (a) \( t > 2.756 \); \( t < -2.756 \)
- (b) \( t = 3.059 \)
- (c) p-value < 0.010

Feel free to ask if you have any further questions or need clarification on any of the steps!