I am studying the problem to predict popularity of a tweet, and want to test null hypothesis: there is no relationships between favorite_counts and another set of variables, like number of friends of users.

I am not sure if normalise or standardise the variables, because I am thinking how to model popularity and don't know how the distributions of likes and friends among users are (please advise).

So I tried the two, and tried an independent t_test.

I get very different results:

from sklearn.preprocessing import StandardScaler, MinMaxScaler
do_scaled = pd.DataFrame(StandardScaler().fit_transform(do[columns].values), columns=columns)

ttest_ind(do_scaled.favorite_count, do_scaled.user_favourites_count)
#Ttest_indResult(statistic=-1.682257624164912e-16, pvalue=0.9999999999999999)

#pvalue is about 1 : the association is likely due to pure chance 

here a boxplot to show the distribution of outliers (StandardScaler)

enter image description here

from sklearn.preprocessing import StandardScaler, MinMaxScaler
do_scaled = pd.DataFrame(MinMaxScaler().fit_transform(do[columns].values), columns=columns)

ttest_ind(do_scaled.favorite_count, do_scaled.user_favourites_count)
#Ttest_indResult(statistic=-5.999028611045575, pvalue=2.3988962933916377e-09)

#pvalue is almost 0 (less than 1%) : there is an association between predictor and response.

here a boxplot to show the distribution of outliers (MinMaxScaler)

enter image description here

I don't understand why I get opposite results and don't know how to interpret them. Can you please advice ? Can you please help to approach the problem ?

  • $\begingroup$ hypothesis testing is quite different from scaling. The question is poorly designed . Could you rectify for a correct answer ? $\endgroup$ Jan 22, 2022 at 12:59
  • 1
    $\begingroup$ The question is completely fine. To use Student's test itself in such a way is unusual, something like Spearman's rank correlation would probably be more appropriate here, but o/w the question is fine. And as @user305883 mentions in a comment below, it does not make sense to t-test against standartized populations. $\endgroup$ Jan 23, 2022 at 16:31

1 Answer 1


First, T-test null hypothesis is that there are no differences between means of two samples. And p-value is the probability to observe the data, given that the null hypothesis is correct, so if p-value is small - you are likely to reject the null hypothesis. So in your case it is actually vice-versa to what you wrote:

  • In case of StandardScaler your test says that "two samples are taken from the distributions with the same mean"^[1].
  • And in case of MinMaxScaler it says that "two samples are unlikely to be taken from the distributions with the same mean".

Now to the second part, why you get this result. The answer is actually quite straightforward. To compute Student's statistics one use 3 parameters (6 in case of comparing means of two samples): Mean of the sample, Variance (or standard deviation) of the sample and the size of the sample^[2]. StandardScaler applies z-scoring:

$$ X_{\text{standartized}} = \frac{X - \text{mean}(X)}{\text{std}(X)} $$

thus, after standartization both of the columns have zero mean and unit variance, therefore Student t-test says that means of two samples are the same (because they are the same and equal to 0).

Conversely, MinMaxScaler:

$$ X_{\text{minmax}} = \frac{X - \text{min}(X)}{\text{max(X)} - \text{min(X)}} $$

does not make niether mean, nor variance of two samples to be equal (it makes minimal value of the sample to be equal to 0 and maximal to 1), therefore Student t-test says that they are different.

[1] To be more precise your results says that you can not reject the null hypothesis (you could never accept a null hypothesis in statistical testing).

[2] You could check the technicalities of the t-test on wiki page for Welch's t-test https://en.wikipedia.org/wiki/Welch%27s_t-test (Unpaired independent T-test for two samples of different sizes with different variance, which is an appropriate version of the test in your case)

  • 1
    $\begingroup$ Thank you! So, it does not make sense to t-test against standardised populations, correct ? because t-test = (mean1 - mean2) / (std(differences between paired data) / sqrt(sample size) ) and mean1 equals mean2 that equal to zero, in the case above. $\endgroup$
    – user305883
    Jan 15, 2022 at 17:38
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    $\begingroup$ Exactly right. Since the goal of t-test is to compare the means of two populations if their means are equal you will always end up with p-value equal to 1 for any degrees of freedom (=independent of populations sizes) $\endgroup$ Jan 15, 2022 at 17:44
  • $\begingroup$ @Anvar The hypothesis does not depend upon type of scaling as you claim. $\endgroup$ Jan 22, 2022 at 13:03

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