use the t-test to compare the means of two different samples (the formula for the t-test will be provided, as shown in the Mathematical requirements)

Variation

Learning Objective

By the end of this lesson you will be able to use the t‑test to compare the means of two different samples.

Why Study \cdot ariation?

• Variation is the raw material for natural selection.
• Understanding variation helps explain differences in phenotype, disease susceptibility and response to treatment.
• Statistical tools allow us to decide whether observed differences are real or due to random chance.

Statistical Comparison of Two Samples

When we have two independent groups (e.g., two populations of plants grown under different conditions) we often want to know if their average trait values differ. The independent‑samples t‑test provides a method for this.

Formula for the Independent‑Samples t‑Test

The test statistic is calculated as

\$t = \frac{\bar{x}1 - \bar{x}2}{\sqrt{sp^{2}\left(\frac{1}{n1}+\frac{1}{n_2}\right)}}\$

where

\$sp^{2} = \frac{(n1-1)s1^{2} + (n2-1)s2^{2}}{n1+n_2-2}\$

and

• \$\bar{x}1\$, \$\bar{x}2\$ – sample means
• \$s1^{2}\$, \$s2^{2}\$ – sample variances
• \$n1\$, \$n2\$ – sample sizes
• \$s_p^{2}\$ – pooled variance (an estimate of the common population variance)

Step‑by‑Step Procedure

1. State the null hypothesis \$H0\$: the two population means are equal (\$\mu1 = \mu_2\$).
2. Calculate the sample means \$\bar{x}1\$, \$\bar{x}2\$.
3. Calculate the sample variances \$s1^{2}\$, \$s2^{2}\$.
4. Compute the pooled variance \$s_p^{2}\$ using the formula above.
5. Calculate the t‑value.
6. Determine the degrees of freedom \$df = n1 + n2 - 2\$.
7. Compare the calculated t‑value with the critical t‑value from the t‑distribution table at the chosen significance level (e.g., \$\alpha = 0.05\$).
8. Draw a conclusion: reject \$H0\$ if \$|t|\$ exceeds the critical value; otherwise, fail to reject \$H0\$.

Worked Example

Suppose we measure leaf length (cm) of two groups of seedlings.

Calculations:

• \$n1 = n2 = 5\$
• \$\bar{x}_1 = \frac{5.2+5.5+5.1+5.4+5.3}{5}=5.30\$ cm
• \$\bar{x}_2 = \frac{6.1+6.3+5.9+6.2+6.0}{5}=6.10\$ cm
• \$s1^{2}= \frac{\sum (xi-\bar{x}1)^2}{n1-1}=0.025\$ cm\$^{2}\$
• \$s2^{2}= \frac{\sum (xi-\bar{x}2)^2}{n2-1}=0.020\$ cm\$^{2}\$
• \$s_p^{2}= \frac{(5-1)0.025+(5-1)0.020}{5+5-2}=0.0225\$ cm\$^{2}\$
• \$t = \dfrac{5.30-6.10}{\sqrt{0.0225\left(\frac{1}{5}+\frac{1}{5}\right)}} = \dfrac{-0.80}{\sqrt{0.0225\times0.4}} = \dfrac{-0.80}{0.0949}\approx -8.43\$
• \$df = 5+5-2 = 8\$

For a two‑tailed test at \$\alpha = 0.05\$, the critical t‑value for \$df=8\$ is approximately \$2.306\$. Since \$|t| = 8.43 > 2.306\$, we reject the null hypothesis and conclude that the treatment significantly increases leaf length.

Interpretation of Results

• Reject \$H_0\$: The difference between sample means is unlikely to be due to random variation alone.
• Fail to reject \$H_0\$: No evidence that the population means differ; observed difference may be random.

Assumptions and Limitations

• Both samples are independent.
• Data in each group are approximately normally distributed.
• Variances of the two populations are equal (homoscedasticity). If this is not true, a Welch’s t‑test should be used.
• Small sample sizes reduce the power of the test.

Suggested Classroom Activity

Students collect data on a measurable trait (e.g., seed germination time) from two groups of plants grown under different light regimes. They then apply the t‑test steps to determine whether light quality influences germination speed.

Key Take‑aways

• The t‑test provides a quantitative method to compare means of two independent samples.
• Correct calculation of means, variances and pooled variance is essential.
• Always check the assumptions before interpreting the result.