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

See all solutions.

If we have two samples, $x_1, \ldots , x_n$ and $x’_1, \ldots , x’_n$, and we compute the mean of each one, we might want to ask whether one mean is less than the other. Or more precisely, is their difference significantly less than zero?

## Solution

If we call the mean of the first sample $\bar x_1$ and the mean of the second sample $\bar x_2$, then this is a left-tailed test with the null hypothesis $H_0: \bar x_1 - \bar x_2 \ge 0$. 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).

1
2
3
4
5
6
7
8
# 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, alternative = "less" )

1
2
3
4
5
6
7
8
9
10
Welch Two Sample t-test

data:  sample1 and sample2
t = -2.4617, df = 5.7201, p-value = 0.02549
alternative hypothesis: true difference in means is less than 0
90 percent confidence interval:
-Inf -1.605229
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.

1
2
3
# Is there enough evidence to reject the null hypothesis?
result <- t.test( sample1, sample2, conf.level=1-alpha, alternative = "less" )
result$p.value < alpha  1 [1] TRUE  In this case, the samples give us enough evidence to reject the null hypothesis at the$\alpha=0.10$level. The data suggest that$\bar x_1 < \bar 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.