A t-test for two independent samples allows you to determine if there's a statistically significant difference between the means of two independent groups. The calculation involves several steps, depending on whether you assume equal variances between the two groups.
1. Formulate Hypotheses
- Null Hypothesis (H0): There is no significant difference between the means of the two populations. (μ1 = μ2)
- Alternative Hypothesis (Ha): There is a significant difference between the means of the two populations. This can be two-tailed (μ1 ≠ μ2), right-tailed (μ1 > μ2), or left-tailed (μ1 < μ2).
2. Check Assumptions
Before performing the t-test, ensure these assumptions are reasonably met:
- Independence: The data points within each group are independent of each other.
- Normality: The data in each group are approximately normally distributed. You can assess this using histograms, Q-Q plots, or statistical tests like the Shapiro-Wilk test.
- Equality of Variances (Homogeneity of Variance): This assumption is only required if you're using the pooled variance t-test. You can check this using Levene's test. If Levene's test is significant (p < α, where α is your significance level), you'll use the Welch's t-test (which does not assume equal variances).
3. Calculate the Test Statistic (t)
There are two main formulas for the t-statistic, depending on whether you assume equal variances:
a) Equal Variances Assumed (Pooled Variance t-test)
This test assumes that the variances of the two populations are equal. Use this if Levene's test is not significant.
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Pooled Variance (sp2):
sp2 = [(n1 - 1)s12 + (n2 - 1)s22] / (n1 + n2 - 2)Where:
- n1 = sample size of group 1
- n2 = sample size of group 2
- s12 = sample variance of group 1
- s22 = sample variance of group 2
-
t-statistic:
t = (x̄1 - x̄2) / [sp √(1/n1 + 1/n2)]Where:
- x̄1 = sample mean of group 1
- x̄2 = sample mean of group 2
- sp = square root of the pooled variance (sp2)
-
Degrees of Freedom (df): df = n1 + n2 - 2
b) Equal Variances Not Assumed (Welch's t-test)
This test does not assume that the variances of the two populations are equal. Use this if Levene's test is significant.
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t-statistic:
t = (x̄1 - x̄2) / √(s12/n1 + s22/n2)Where:
- x̄1 = sample mean of group 1
- x̄2 = sample mean of group 2
- s12 = sample variance of group 1
- s22 = sample variance of group 2
- n1 = sample size of group 1
- n2 = sample size of group 2
-
Degrees of Freedom (df): This is calculated using a more complex formula (Welch-Satterthwaite equation). Many statistical software packages calculate this automatically. A simplified approximation is:
df ≈ ((s12/n1 + s22/n2)^2) / (((s12/n1)^2/(n1-1)) + ((s22/n2)^2/(n2-1)))4. Determine the p-value
Using the calculated t-statistic and degrees of freedom, find the p-value associated with your test from a t-distribution table or using statistical software. The p-value represents the probability of observing a test statistic as extreme as, or more extreme than, the one calculated, assuming the null hypothesis is true. The specific calculation depends on whether you are doing a one-tailed or two-tailed test.
5. Make a Decision
Compare the p-value to your chosen significance level (α), typically 0.05.
- If p ≤ α: Reject the null hypothesis. There is a statistically significant difference between the means of the two groups.
- If p > α: Fail to reject the null hypothesis. There is not enough evidence to conclude that there is a statistically significant difference between the means of the two groups.
Example
Let's say you want to compare the test scores of two different teaching methods.
- Group 1 (Method A): n1 = 25, x̄1 = 80, s12 = 100
- Group 2 (Method B): n2 = 30, x̄2 = 75, s22 = 90
Assume Levene's test indicates that the variances are approximately equal. Therefore, you would use the pooled variance t-test.
- Pooled Variance: sp2 = [(25-1)100 + (30-1)90] / (25 + 30 - 2) = 94.74
- t-statistic: t = (80 - 75) / [√94.74 √(1/25 + 1/30)] = 2.09
- Degrees of Freedom: df = 25 + 30 - 2 = 53
- P-value: Assuming a two-tailed test and α = 0.05, using a t-distribution table or statistical software, you'd find a p-value of approximately 0.041.
- Decision: Since p (0.041) ≤ α (0.05), you would reject the null hypothesis and conclude that there is a statistically significant difference in test scores between the two teaching methods.
