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M2 Lecture 3: Independent T Test

t Test for Independent Groups

  • Purpose: Compare the means of two independent groups of subjects (unrelated groups)

  • Application example: Does drinking orange juice (OJ) daily affect memory scores?

  • Independent groups: OJ drinkers vs non-OJ drinkers

  • Dependent variable (DV): memory score (level of measurement: interval/ratio)

  • Assumptions

    • Independence of observations between groups

    • Normality of the DV within each group

    • Homogeneity of variances (equal variances across groups)

    • DV is interval/ratio

  • Level of significance used in the example: \alpha = 0.05

Hypothesis Testing Framework EXAMPLE

  • Null hypothesis (H0): There will be no difference in memory score between the OJ drinkers and the non-OJ drinkers.

  • Research/alternative hypothesis (H1): There will be a difference in memory score between the OJ drinkers and the non-OJ drinkers.

  • Test Statistic: The t-test will be utilized to determine whether the means of the two groups are significantly different from each other. The significance level (alpha) will be set at 0.05, indicating that we will reject the null hypothesis if the p-value is less than this threshold.

  • Six-step testing procedure (recalled from the review)

    1. Develop null and research hypotheses

    2. Choose a level of significance (\alpha)

    3. Determine which statistical test is appropriate

    4. Run analysis to obtain test statistic and p-value

    5. Make a decision about rejecting or failing to reject the null hypothesis

    6. Make a conclusion

Levene’s Test for Equality of Variances

  • Purpose: Assess whether the variances in the two groups are equal

  • Null hypothesis (Levene): There is no difference in variances between the two groups

  • If Levene’s test is non-significant (p > 0.05), proceed with the standard t-test assuming equal variances

  • If Levene’s test is significant (p ≤ 0.05), variances are heterogeneous, and a version of the t-test that does not assume equal variances (Welch’s t-test) is appropriate

Example Data: Memory and OJ Drinking

  • Groups: OJ drinkers vs non-OJ drinkers (two independent groups)

  • Reported results (summary statistics from the example):

    • OJ drinkers: mean memory score \overline{X}{\text{OJ}} = \text{11.42}$, standard deviation s{\text{OJ}} = \text{2.07}

    • Non-OJ drinkers: mean memory score \overline{X}{\text{nonOJ}} = \text{9.50}, standard deviation s{\text{nonOJ}} = \text{2.15}

    • Sample sizes: likely n1 = n2 = 12 (total df reported as t(22), consistent with two groups of 12)

  • Test statistic reported: t(22) = 2.225, p = 0.037

  • Difference in means: \Delta\overline{X} = \overline{X}{\text{OJ}} - \overline{X}{\text{nonOJ}} = 11.42 - 9.50 = 1.92

  • 95% confidence interval for the mean difference (not explicitly given in slides, but typically reported with t-testing): not provided in the transcript

  • Conclusion in the transcript: p = 0.037 < \alpha = 0.05, therefore reject the null hypothesis; there is a difference in memory scores between the two groups

  • Final stated interpretation (from slide): On average, memory scores of OJ drinkers were significantly higher than those of non-OJ drinkers (OJ: M=11.42, SD=2.07 vs non-OJ: M=9.50, SD=2.15), with t(22) = 2.225, p = 0.037

Formulas and Calculations

  • Pooled-variance independent samples t-test (assuming equal variances)

    • Pooled variance: Sp^2 = \frac{(n1 - 1)s1^2 + (n2 - 1)s2^2}{n1 + n2 - 2}

    • Test statistic: t = \frac{\overline{X}1 - \overline{X}2}{\sqrt{ Sp^2\left(\frac{1}{n1} + \frac{1}{n2}\right) }}

    • Degrees of freedom (df): df = n1 + n2 - 2

  • Welch’s t-test (no assumption of equal variances)

    • Test statistic: t = \frac{\overline{X}1 - \overline{X}2}{\sqrt{ \frac{s1^2}{n1} + \frac{s2^2}{n2} }}

    • Degrees of freedom (approximate, Welch-Satterthwaite):
      df \approx \frac{\left(\frac{s1^2}{n1} + \frac{s2^2}{n2}\right)^2}{\frac{\left(\frac{s1^2}{n1}\right)^2}{n1 - 1} + \frac{\left(\frac{s2^2}{n2}\right)^2}{n2 - 1}}

  • Interpretation of p-value

    • If p < \alpha, reject the null hypothesis and conclude a difference between group means

    • If p \ge \alpha, fail to reject the null hypothesis

Connections to Practice and Implications

  • Practical implication: If daily orange juice consumption is associated with higher memory scores, consider potential confounders (e.g., overall diet, caffeine intake, education, baseline memory) before causal claims are made

  • Ethical/philosophical note: Correlation does not imply causation; randomized controlled trials are needed to establish causality

  • Relevance to evidence-based practice: Use appropriate statistical tests based on data characteristics (independence, normality, equal variances) and report test statistics, degrees of freedom, and exact p-values

Quick Reference from the Transcript

  • Study design: t Test for Independent Groups

  • Assumptions: independence, normality, homogeneity of variances, DV=interval/ratio

  • Null hypothesis: no difference in memory scores between OJ drinkers and non-OJ drinkers

  • Alternative hypothesis: there is a difference

  • Significance level: \alpha = 0.05

  • Levene’s test: Null is equality of variances; significance guides whether to assume equal variances

  • Reported result: t(22) = 2.225, p = 0.037; p < 0.05; reject H0

  • Means and standard deviations (example conclusion): OJ \overline{X}{\text{OJ}} = 11.42, s{\text{OJ}} = 2.07; non-OJ \overline{X}{\text{nonOJ}} = 9.50, s{\text{nonOJ}} = 2.15

  • Interpretation statement: On average, OJ drinkers scored higher on memory tests than non-OJ drinkers by about 1.92$$ points, difference being statistically significant

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