How to perform two-sample one-tailed t-test with numpy/scipy
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because the one-sided tests can be backed out from the two-sided tests. (With symmetric distributions one-sided p-value is just half of the two-sided pvalue)
It goes on to say that scipy always gives the test statistic as signed. This means that given p and t values from a two-tailed test, you would reject the null hypothesis of a greater-than test when p/2 < alpha and t > 0, and of a less-than test when p/2 < alpha and t < 0.
After trying to add some insights as comments to the accepted answer but not being able to properly write them down due to general restrictions upon comments, I decided to put my two cents in as a full answer.
First let's formulate our investigative question properly. The data we are investigating is
A = np.array([0.19826790, 1.36836629, 1.37950911, 1.46951540, 1.48197798, 0.07532846])
B = np.array([0.6383447, 0.5271385, 1.7721380, 1.7817880])
with the sample means
A.mean() = 0.99549419
B.mean() = 1.1798523
I assume that since the mean of B is obviously greater than the mean of A, you would like to check if this result is statistically significant.
So we have the Null Hypothesis
H0: A >= B
that we would like to reject in favor of the Alternative Hypothesis
H1: B > A
Now when you call scipy.stats.ttest_ind(x, y), this makes a Hypothesis Test on the value of x.mean()-y.mean(), which means that in order to get positive values throughout the calculation (which simplifies all considerations) we have to call
stats.ttest_ind(B,A)
instead of stats.ttest_ind(B,A). We get as an answer
t-value = 0.42210654140239207p-value = 0.68406235191764142
and since according to the documentation this is the output for a two-tailed t-test we must divide the p by 2 for our one-tailed test. So depending on the Significance Level alpha you have chosen you need
p/2 < alpha
in order to reject the Null Hypothesis H0. For alpha=0.05 this is clearly not the case so you cannot reject H0.
An alternative way to decide if you reject H0 without having to do any algebra on t or p is by looking at the t-value and comparing it with the critical t-value t_crit at the desired level of confidence (e.g. 95%) for the number of degrees of freedom df that applies to your problem. Since we have
df = sample_size_1 + sample_size_2 - 2 = 8
we get from a statistical table like this one that
t_crit(df=8, confidence_level=95%) = 1.860
We clearly have
t < t_crit
so we obtain again the same result, namely that we cannot reject H0.