Question 1181367: - Solve for the t - computed and determine the t - tabular
A course in mathematics is taught to 15 students by explicit method. Another
group of 17 students was given the same course by means of another method. At the
end of the semester, the same test was administered to each group. The 15 students
under method A made an average of 85 with a standard deviation of 4, while the 17
students under method B made an average of 81 with a standard deviation of 5. Test
the null hypothesis of no significant difference in the performance of the two groups of
students at 5% level of significance.
Solution:
1. H0:
H1:
2. Level of Significance:
3.Test:
4. Critical Region:
5. Solution:
6. Decision:
7. Conclusion:
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! 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.
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