How to do a two-sided hypothesis test for two sample means (in R)

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Task

If we have two samples, $x_1,\dots ,x_n$ and $x_1^{\prime },\dots ,x_m^{\prime }$, and we compute the mean of each one, we might want to ask whether the two means seem approximately equal. Or more precisely, is their difference statistically significant at a given level?

Related tasks:

• How to compute a confidence interval for a population mean
• How to do a two-sided hypothesis test for a sample mean
• How to do a one-way analysis of variance (ANOVA)
• How to do a one-sided hypothesis test for two sample means
• How to do a hypothesis test for a mean difference (matched pairs)
• How to do a hypothesis test for a population proportion

Solution

If we call the mean of the first sample ${\overline{x}}_1$ and the mean of the second sample ${\overline{x}}_2$, then this is a two-sided test with the null hypothesis $H_0:{\overline{x}}_1={\overline{x}}_2$. We choose a value $0\le \alpha \le 1$ as the probability of a Type I error (false positive, finding we should reject $H_0$ when it’s actually true).

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# Replace these first three lines with the values from your situation.
alpha <- 0.10
sample1 <- c( 6, 9, 7, 10, 10, 9 )
sample2 <- c( 12, 14, 10, 17, 9 )

# Run a one-sample t-test and print out alpha, the p value,
# and whether the comparison says to reject the null hypothesis.
t.test( sample1, sample2, conf.level=1-alpha )

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	Welch Two Sample t-test

data:  sample1 and sample2
t = -2.4617, df = 5.7201, p-value = 0.05097
alternative hypothesis: true difference in means is not equal to 0
90 percent confidence interval:
 -7.0057683 -0.7942317
sample estimates:
mean of x mean of y 
      8.5      12.4 

Although we can deduce the answer to our question from the above output, by comparing the $p$ value with $\alpha$ manually, we can also ask R to do it.

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# Is there enough evidence to reject the null hypothesis?
result <- t.test( sample1, sample2, conf.level=1-alpha )
result$p.value < alpha

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[1] TRUE

In this case, the samples give us enough evidence to reject the null hypothesis at the $\alpha =0.10$ level. The data suggest that ${\overline{x}}_1\ne {\overline{x}}_2$.

Here we did not assume that the two samples had equal variance. If in your case they do, you can pass the parameter var.equal=TRUE to t.test.

Content last modified on 24 July 2023.

See a problem? Tell us or edit the source.

Contributed by Nathan Carter (ncarter@bentley.edu)