# For a T-test, what would happen if one of my samples was made up of only one observation? [duplicate]

I must be missing something, or under-thinking what goes on with a basic T Test, but I was under the impression that if I do a T-test, and one of my samples was made up of only one observation, the test would fail.

To see if this is indeed the case, I ran some code in Python code. I made up some numbers, where my first sample has 15 values and the second sample has only 1:

import statsmodels.api as sm
print "This is the result of the T-test:"
ttest = sm.stats.ttest_ind(np.array([821,823,814,815,816,817,881,891,234,354,678,765,989,435,657]), np.array([21]), alternative='larger')
print "pvalue:", ttest[1], ", at alpha = 0.05, thus the result is:", ttest[1] <= 0.05


Here's the result:

This is the result of the T-test:
pvalue: 2.72522964545e-08 , at alpha = 0.05, thus the result is: True


I am getting an answer saying that there is a significant difference between the two sample means. Is this simply a bogus answer?

I was under the impression that one observation can't be assumed to belong to a normal distribution, so this assumption for a T-test itself can not be verified. Thus the test should not be appropriate.

• You can assume, sometimes with good reason, that both samples are normal variates with a common variance; but you can't check this assumption with these samples. – Scortchi - Reinstate Monica May 18 '14 at 1:54
• So, even if this assumption is made, is there any purpose to using such a T Test when a sample has only one observation anyway? I feel like it doesn't tell us anything. – tumultous_rooster May 18 '14 at 1:58
• It tells you exactly the same thing as when the sample has many observations: the probability under the null hypothesis of equal means that the t-statistic would exceed that observed. Of course its power will be lower. – Scortchi - Reinstate Monica May 18 '14 at 2:57

The possible issue with the case of one observation, would be "how do we calculate the sample variance?" Obviously, the sample variance is zero (if calculated without the bias-correction factor), and an indeterminate form $0/0$ if we try to calculate it using the bias-correction factor The t-statistic for the equality of means between two samples of unequal sizes and different sample variances ("Welch's t-test"), is
$$t = {\overline{X}_1 - \overline{X}_2 \over s_{\overline{X}_1 - \overline{X}_2}},\qquad s_{\overline{X}_1 - \overline{X}_2} = \sqrt{{s_1^2 \over n_1} + {s_2^2 \over n_2}}$$
$s_1^2, s_2^2$ should be the sample variance expressions with the bias-correction factor, and so there should be a problem. Also the calculation for the appropriate degrees of freedom involves the magnitude "sample size minus one". So normally, the code should not result -but you have to find out exactly which "t-test" the software runs -there are many. In some of them the t-statistics can be calculated even when one of the samples is of size one.
• Aha! I believe I was using the wrong t-test. I have included usevar="unequal" and now have a Welch's t-test on my hands. Thanks! – tumultous_rooster May 18 '14 at 2:42
• One I included the usevar="unequal" option which gave me a Welch's t-test, the result was in line with what we theoretically expected. – tumultous_rooster May 18 '14 at 3:09