scipy.stats.ttest_ind_from_stats(mean1, std1, nobs1, mean2, std2, nobs2, equal_var=True, alternative='two-sided')[source]#

T-test for means of two independent samples from descriptive statistics.

This is a test for the null hypothesis that two independent samples have identical average (expected) values.


The mean(s) of sample 1.


The corrected sample standard deviation of sample 1 (i.e. ddof=1).


The number(s) of observations of sample 1.


The mean(s) of sample 2.


The corrected sample standard deviation of sample 2 (i.e. ddof=1).


The number(s) of observations of sample 2.

equal_varbool, optional

If True (default), perform a standard independent 2 sample test that assumes equal population variances [1]. If False, perform Welch’s t-test, which does not assume equal population variance [2].

alternative{‘two-sided’, ‘less’, ‘greater’}, optional

Defines the alternative hypothesis. The following options are available (default is ‘two-sided’):

  • ‘two-sided’: the means of the distributions are unequal.

  • ‘less’: the mean of the first distribution is less than the mean of the second distribution.

  • ‘greater’: the mean of the first distribution is greater than the mean of the second distribution.

Added in version 1.6.0.

statisticfloat or array

The calculated t-statistics.

pvaluefloat or array

The two-tailed p-value.


The statistic is calculated as (mean1 - mean2)/se, where se is the standard error. Therefore, the statistic will be positive when mean1 is greater than mean2 and negative when mean1 is less than mean2.

This method does not check whether any of the elements of std1 or std2 are negative. If any elements of the std1 or std2 parameters are negative in a call to this method, this method will return the same result as if it were passed numpy.abs(std1) and numpy.abs(std2), respectively, instead; no exceptions or warnings will be emitted.



Suppose we have the summary data for two samples, as follows (with the Sample Variance being the corrected sample variance):

                 Sample   Sample
           Size   Mean   Variance
Sample 1    13    15.0     87.5
Sample 2    11    12.0     39.0

Apply the t-test to this data (with the assumption that the population variances are equal):

>>> import numpy as np
>>> from scipy.stats import ttest_ind_from_stats
>>> ttest_ind_from_stats(mean1=15.0, std1=np.sqrt(87.5), nobs1=13,
...                      mean2=12.0, std2=np.sqrt(39.0), nobs2=11)
Ttest_indResult(statistic=0.9051358093310269, pvalue=0.3751996797581487)

For comparison, here is the data from which those summary statistics were taken. With this data, we can compute the same result using scipy.stats.ttest_ind:

>>> a = np.array([1, 3, 4, 6, 11, 13, 15, 19, 22, 24, 25, 26, 26])
>>> b = np.array([2, 4, 6, 9, 11, 13, 14, 15, 18, 19, 21])
>>> from scipy.stats import ttest_ind
>>> ttest_ind(a, b)
Ttest_indResult(statistic=0.905135809331027, pvalue=0.3751996797581486)

Suppose we instead have binary data and would like to apply a t-test to compare the proportion of 1s in two independent groups:

                  Number of    Sample     Sample
            Size    ones        Mean     Variance
Sample 1    150      30         0.2        0.161073
Sample 2    200      45         0.225      0.175251

The sample mean \(\hat{p}\) is the proportion of ones in the sample and the variance for a binary observation is estimated by \(\hat{p}(1-\hat{p})\).

>>> ttest_ind_from_stats(mean1=0.2, std1=np.sqrt(0.161073), nobs1=150,
...                      mean2=0.225, std2=np.sqrt(0.175251), nobs2=200)
Ttest_indResult(statistic=-0.5627187905196761, pvalue=0.5739887114209541)

For comparison, we could compute the t statistic and p-value using arrays of 0s and 1s and scipy.stat.ttest_ind, as above.

>>> group1 = np.array([1]*30 + [0]*(150-30))
>>> group2 = np.array([1]*45 + [0]*(200-45))
>>> ttest_ind(group1, group2)
Ttest_indResult(statistic=-0.5627179589855622, pvalue=0.573989277115258)