Question 1181367
Here's the solution to the hypothesis test:

1. **Hypotheses:**

* H₀: There is no significant difference in the performance of the two groups. (μ₁ = μ₂)
* H₁: There is a significant difference in the performance of the two groups. (μ₁ ≠ μ₂)

2. **Level of Significance:** α = 0.05

3. **Test:** Since we have two independent samples, small sample sizes (both less than 30), and unknown population standard deviations, we use a two-sample t-test.  Because the sample standard deviations are different, we'll use the version of the t-test that *does not* assume equal variances.

4. **Critical Region:** This is a two-tailed test (because H₁ is μ₁ ≠ μ₂).  We need to find the degrees of freedom (df). For unequal variances, the calculation for df is complex. A conservative approach is to use the smaller of n₁-1 and n₂-1 which is min(15-1, 17-1)=14.  For df = 14 and α = 0.05 (two-tailed), the critical t-value from the t-table is approximately ±2.145.  Our critical region is t < -2.145 or t > 2.145.

5. **Solution (Calculations):**

* **Calculate the pooled variance (Sp²):**  Since we're not assuming equal variances, we *don't* pool the variances.

* **Calculate the test statistic (t):**

   t = (x̄₁ - x̄₂) / sqrt(s₁²/n₁ + s₂²/n₂)
   t = (85 - 81) / sqrt(4²/15 + 5²/17)
   t = 4 / sqrt(1.067 + 1.471)
   t = 4 / sqrt(2.538)
   t ≈ 4 / 1.593
   t ≈ 2.51

6. **Decision:**  Our calculated t-value (2.51) falls *within* the critical region (t > 2.145).  Therefore, we *reject* the null hypothesis.

7. **Conclusion:** There *is* sufficient evidence at the 5% level of significance to conclude that there is a significant difference in the performance of the two groups of students. Method A appears to be more effective than Method B.

**Summary:**

* **t-computed:** ≈ 2.51
* **t-tabular (critical value):** ≈ ±2.145
* **Decision:** Reject H₀
* **Conclusion:**  There is a statistically significant difference between the two methods.