Two-Sample T-Test for Comparing Means
Conduct an independent two-sample t-test to compare the mean test scores of students taught using two different methods.
Problem Scenario
A school district wants to compare two teaching methods for an algebra course. Method A (traditional lecture) is used with 15 students, yielding a mean score of 78.4 and standard deviation of 8.2. Method B (active learning) is used with 17 students, yielding a mean score of 84.6 and standard deviation of 7.5. At the 0.05 significance level, is there a significant difference in mean scores between the two methods? Assume equal variances.
Given Data
Requirements
- State the null and alternative hypotheses
- Calculate the pooled standard deviation and the t-test statistic
- Determine degrees of freedom, find the p-value, and state the conclusion
Solution
Step 1:
State the hypotheses. H_0: mu_1 = mu_2 (no difference in mean scores). H_a: mu_1 != mu_2 (there is a difference in mean scores). This is a two-tailed test.
Step 2:
Calculate the pooled variance: s_p^2 = [(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2] / (n_1 + n_2 - 2) = [(14)(67.24) + (16)(56.25)] / 30 = [941.36 + 900.00] / 30 = 1841.36 / 30 = 61.379. So s_p = sqrt(61.379) = 7.834.
Step 3:
Calculate the standard error of the difference: SE = s_p * sqrt(1/n_1 + 1/n_2) = 7.834 * sqrt(1/15 + 1/17) = 7.834 * sqrt(0.06667 + 0.05882) = 7.834 * sqrt(0.12549) = 7.834 * 0.3542 = 2.774.
Step 4:
Calculate the t-statistic: t = (x-bar_1 - x-bar_2) / SE = (78.4 - 84.6) / 2.774 = -6.2 / 2.774 = -2.235. Degrees of freedom: df = n_1 + n_2 - 2 = 15 + 17 - 2 = 30.
Step 5:
Find the p-value. For a two-tailed test with t = -2.235 and df = 30, the p-value = 2 * P(T < -2.235) approximately equals 2 * 0.0165 = 0.033.
Step 6:
Decision: Since p-value (0.033) < alpha (0.05), we reject H_0. There is statistically significant evidence that the two teaching methods produce different mean algebra scores.
Final Answer
t = -2.235, df = 30, p-value = 0.033. At the 0.05 significance level, we reject H_0. There is a statistically significant difference in mean algebra scores between the two teaching methods, with active learning (Method B) yielding higher scores on average (84.6 vs. 78.4).
Key Takeaways
- ✓The pooled two-sample t-test assumes equal population variances. If this assumption is questionable, use the Welch t-test which does not assume equal variances.
- ✓Statistical significance does not necessarily imply practical significance. A 6.2-point difference should be evaluated in the context of the grading scale and educational goals.
- ✓With small sample sizes, the normality assumption for each group becomes more important. Check with histograms or normality tests if possible.
Common Errors to Avoid
- ✗Using the wrong degrees of freedom. For the pooled t-test, df = n_1 + n_2 - 2, not n_1 + n_2 or (n_1 - 1) + (n_2 - 1) separately.
- ✗Assuming equal variances without checking. If the ratio of the larger to smaller sample standard deviation exceeds about 2, consider using the Welch t-test instead.
- ✗Forgetting that the two-tailed p-value is twice the one-tailed p-value. If your test is two-tailed, always multiply the single-tail probability by 2.
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Common questions about this problem type
The pooled t-test assumes both populations have equal variances and uses a pooled estimate of the common variance. The Welch t-test does not assume equal variances and adjusts the degrees of freedom accordingly (using the Welch-Satterthwaite approximation). When in doubt, the Welch t-test is the safer choice.
Yes, but the normality assumption becomes critical with very small samples. If the data are clearly non-normal (heavily skewed or with outliers), consider a nonparametric test like the Mann-Whitney U test instead.