scipy.stats.ttest_ind_from_stats¶
- 
scipy.stats.ttest_ind_from_stats(mean1, std1, nobs1, mean2, std2, nobs2, equal_var=True)[source]¶
- T-test for means of two independent samples from descriptive statistics. - This is a two-sided test for the null hypothesis that two independent samples have identical average (expected) values. - Parameters
- mean1array_like
- The mean(s) of sample 1. 
- std1array_like
- The standard deviation(s) of sample 1. 
- nobs1array_like
- The number(s) of observations of sample 1. 
- mean2array_like
- The mean(s) of sample 2. 
- std2array_like
- The standard deviations(s) of sample 2. 
- nobs2array_like
- 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]. 
 
- Returns
- statisticfloat or array
- The calculated t-statistics. 
- pvaluefloat or array
- The two-tailed p-value. 
 
 - See also - Notes - New in version 0.16.0. - References - Examples - Suppose we have the summary data for two samples, as follows: - 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): - >>> 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 0.16 Sample 2 200 45 0.225 0.174375 - 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(0.16), nobs1=150, ... mean2=0.225, std2=np.sqrt(0.17437), nobs2=200) Ttest_indResult(statistic=-0.564327545549774, pvalue=0.5728947691244874) - 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) 
