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I was performing paired t-tests and came across the example from the R function. They show that addition of strong outlier can make the test more compliant about differences:

t.test(1:10, y = c(7:20))      # P = .00001855

t.test(1:10, y = c(7:20, 200)) # P = .1245    -- NOT significant anymore

How is this possible ? Does that mean that a t-test shouldn't be used in the case of outliers presence in the data ?

Renaud Bied-charreton
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1 Answers1

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The answer that every statistician says: it depends.
The t-tests you have here are Two-Sample T-Tests. If you were doing paired T-Tests, your $x$ and $y$ inputs would have to be the same length. Here, we are testing whether the true difference in means for these independent datasets is $0$ or not. In other words, is there a statistically significant difference between the two means?
Note: The code you have assumes that both $x$ and $y$ are statistically independent samples, so this would not be paired.
 As for your outlier question, t-tests are looking at the sample means and comparing them. Outliers with a large enough magnitude will skew the mean, which can then result in a significant difference in the sample means. When you notice outliers in the data, you need to see if it's a legitimate statistic -- some outliers can be outright removed because they were input errors, etc; however, you should not remove outliers just because they "ruin" your data. Without more information, it's difficult to say whether this outlier can be removed or imputed, but understand that t-tests are looking at the sample means, which can be influenced by outliers in the data.
To use a parametric t-test, we need to have normality for both samples. In your first line, we can check normality for samples $x$ and $y$:

shapiro.test(1:10) -- p-value = 0.8924
shapiro.test(7:20) -- p-value = 0.7964
shapiro.test(c(7:20, 200)) -- p-value = 3.206e-07

Assuming 1:10 and 7:20 are independent datasets (not paired), then we can use a parametric t-test just like we have in the first line of your code. We can see that we get a significant result for the third normality test, which means that sample is not normal. Thus, you would be required to use non-parametric methods to do a test if you were using the dataset with an outlier (assuming that outlier is valid).

cwbenton
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