Illustrating comment, using R:
set.seed(2020)
x1 = rnorm(500, 100, 15)
x2 = rnorm(100, 105, 17)
summary(x1); length(x1); sd(x1)
Min. 1st Qu. Median Mean 3rd Qu. Max.
53.32 89.29 98.93 99.18 109.45 148.02
[1] 500 # size sample 1
[1] 15.96929 # sample SD sample 1
summary(x2); length(x2); sd(x2)
Min. 1st Qu. Median Mean 3rd Qu. Max.
59.74 94.62 104.05 104.77 114.88 146.67
[1] 100
[1] 17.11946
The two sample means $\bar X_1 = 99.18$ and $\bar X_2 = 104.77$ differ.
The question is whether, in view of the variability of the data, this
difference is large enough to be 'statistically significant' at the 5% level.
In the boxplots below, boxes are of different widths, as a reminder that
sample sizes are quite different. The fact that the 'notches' in the sides of
the boxplots do not overlap, is a preliminary clue that sample means may be
significantly different.
boxplot(x1, x2, varwidth=T, col="skyblue2", pch=20, notch=T)
A Welch t test (used because population variances are unequal), the small
P-value $0.003 < 0.05$ indicates significant difference at the 5% level.
This is not "proof" that the population means differ. However, we are unlikely
to get such different sample means if the population means are the same.
t.test(x1, x2)
Welch Two Sample t-test
data: x1 and x2
t = -3.0129, df = 135.64, p-value = 0.003089
alternative hypothesis:
true difference in means is not equal to 0
95 percent confidence interval:
-9.257022 -1.920342
sample estimates:
mean of x mean of y
99.18129 104.76998
Addendum per comment. Here is a one-sided test. If $\bar X_1 > \bar X_2,$
then the test of $H_0: \mu_1 = \mu_2$ against $H_0: \mu_1 < \mu_2$ will have
a P-value half the size of the two-sided test.
t.test(x1, x2, alt="less")
Welch Two Sample t-test
data: x1 and x2
t = -3.0129, df = 135.64, p-value = 0.001544
alternative hypothesis:
true difference in means is less than 0
95 percent confidence interval:
-Inf -2.516599
sample estimates:
mean of x mean of y
99.18129 104.76998
