Comparing Two Groups

Previously, we made inferences about a single mean from one group.

Now we extend to comparing two independent groups.

Research question: Is there a difference between the population means of two groups?

Same framework:

• CI: Point estimate ± margin of error
• Hypothesis testing with p-values

Data Structure for Two Means

When comparing two means:

• Response variable: Numerical
• Explanatory variable: Binary categorical (defines the two groups)

This is the same structure as two-proportion inference, but now the response is numerical instead of categorical.

Statistics and Parameters for Two Means

Statistic of interest: \(\bar{x}_1 - \bar{x}_2\) (difference in sample means)

Goal: Make inferences about \(\mu_1 - \mu_2\) (difference in population means)

Pooled Standard Deviation

When comparing two groups, we pool the variability to get a combined estimate of the common population SD.

\[s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}}\]

For birth weights:

\[s_p = \sqrt{\frac{(867-1) \cdot 1.23^2 + (114-1) \cdot 1.6^2}{867+114-2}} = 1.28\]

SE for Difference in Means

The standard error for the difference in sample means:

\[SE = s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\]

For birth weights:

\[SE = 1.28 \times \sqrt{\frac{1}{867} + \frac{1}{114}} = 0.128\]

T-Statistic for Two Means

The T-statistic measures how far the observed difference is from the null value in SE units:

\[T = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{SE}\]

Degrees of freedom: \(df = n_1 + n_2 - 2 = 867 + 114 - 2 = 979\)

For birth weights:

\[T = \frac{0.59 - 0}{0.128} = 4.65\]

Conditions for Two-Sample T-Test

Independence:

• Random sample from each population ✓
• Observations independent within and between groups ✓

Normality:

• Both groups have n ≥ 30, no extreme outliers ✓

Equal variance:

• Similar spread in both groups (check histograms/SDs) ✓

Conditions are met for using the t-distribution.

CI Formula for Difference in Means

When conditions are met:

\[\text{CI} = (\bar{x}_1 - \bar{x}_2) \pm t^*_{df} \times SE\]

Where:

• \(\bar{x}_1 - \bar{x}_2\) = observed difference in sample means
• \(t^*_{df}\) = critical value from t-distribution with df = \(n_1 + n_2 - 2\)
• \(SE = s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\)

Iris Example

Technical Note: Equal Variance

We used Student’s t-test, which assumes equal population variances.

Alternative: Welch’s t-test

• Does NOT assume equal variance
• Uses a different (more complex) df formula
• Jamovi can calculate both versions

For this course, we use the equal variance version. When variances appear very different, Welch’s test is preferred.

What Affects Strength of Evidence?

For hypothesis testing, stronger evidence = smaller p-value = larger |T|

Recall: \(T = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{SE}\) where \(SE = s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\)

Three factors affect |T| (and thus p-value):

1. Effect size: Larger \(|\bar{x}_1 - \bar{x}_2|\) → larger |T| → smaller p-value
2. Sample size: Larger \(n_1\) and \(n_2\) → smaller SE → larger |T| → smaller p-value
3. Variability: Smaller \(s_1\) and \(s_2\) → smaller \(s_p\) → smaller SE → larger |T| → smaller p-value

What Affects CI Width?

For confidence intervals: Width = \(2 \times t^* \times SE\)

Where \(SE = s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\)

Three factors affect width:

1. Confidence level: Higher confidence → larger \(t^*\) → wider CI

2. Sample size: Larger \(n_1\) and \(n_2\) → smaller SE → narrower CI

3. Variability: Smaller \(s_1\) and \(s_2\) → smaller \(s_p\) → smaller SE → narrower CI

Summary

For comparing two independent means:

Conditions: Independence + Normality + Equal variance

References

• Introduction to Modern Statistics (2e) textbook by Mine Çetinkaya-Rundel and Johanna Hardin
• Sections 20.3, 20.4