How to perform two-sample, one-tailed t-test in Python

Question

I want to perform a two-sample, one-tailed t-test to compare two means. For the specific problem I am looking, I want the comparison to only be in one direction. I would like the null hypothesis to be that mu_2 > mu_1 and the alternative hypothesis to be mu_1 <= mu_2. Or should the null hypothesis still be that mu_1 - mu_2 = 0, even for the one-tailed case?

I am working with a large dataset, but if I were to extract and round the parameters, for data_1 it is mu_1 = 4.3, s_1 = 4.8, and n_1 = 40000 and data_2 it is mu_2 = 4.9, s_2 = 4.4, n_2 = 30000. I am using scipy to perform a two-sample t-test:

stats.ttest_ind(data1,
                data2,
                equal_var = False)

Given that scipy only takes into account a two-tail test, I am not sure how to interpret the values. Ttest_indResult(statistic=-19.51646312898464, pvalue=1.3452106729078845e-84). The alpha value is 0.05, and the p-value is much much smaller than that which would mean the null hypothesis is rejected. However, my intuition tells me that the null hypothesis should not be rejected, because mu_2 is clearly larger than mu_1 (at the very minimum I would expect the p-value to be larger). Therefore, I feel like I'm either interpreting the results incorrectly or need to additional calculations to get the correct answer.

I would appreciate any additional help and guidance. Thanks!

Answer 1

You are correct, if you are doing a one sided test, it should have a large p-value. ttest_ind performs a two sided test, which gives the probability that you observe something more extreme than the absolute of your t-statistic.

To do a one sided t test, you can use the cdf, which is the sum of probabilities up to your t statistic.

Modifying this code slightly:

def welch_ttest(x1, x2,alternative):
    n1 = x1.size
    n2 = x2.size
    m1 = np.mean(x1)
    m2 = np.mean(x2)
    v1 = np.var(x1, ddof=1)
    v2 = np.var(x2, ddof=1)
    tstat = (m1 - m2) / np.sqrt(v1 / n1 + v2 / n2)
    df = (v1 / n1 + v2 / n2)**2 / (v1**2 / (n1**2 * (n1 - 1)) + v2**2 / (n2**2 * (n2 - 1)))
    if alternative == "equal":
        p = 2 * t.cdf(-abs(tstat), df)
    if alternative == "lesser":
        p = t.cdf(tstat, df)
    if alternative == "greater":
        p = 1-t.cdf(tstat, df)
    return tstat, df, p

I simulate some data:

import numpy as np
from scipy.stats import ttest_ind
from scipy.stats import t

np.random.seed(seed=123)
data1 = np.random.normal(4.3,4.8,size=40000)
np.random.seed(seed=123)
data2 = np.random.normal(4.9,4.4,size=30000)
ndf = len(data1) +len(data2) - 2
ttest_ind(data1,data2,equal_var = False)

Ttest_indResult(statistic=-16.945279258324227, pvalue=2.8364816571790452e-64)

You get something like your result, we can test the code above for alternative == "equal" which is a two-sided test:

welch_ttest(data1,data2,"equal")

    (<scipy.stats._continuous_distns.t_gen at 0x12472b128>,
     67287.08544468222,
     2.8364816571790452e-64)

You can the same p-value as scipy 2 sided t-test, now we do the one sided test you need:

welch_ttest(data1,data2,"greater")
(<scipy.stats._continuous_distns.t_gen at 0x12472b128>, 67287.08544468222, 1.0)