scipy.stats.ttest |
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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. Parameters: mean1array_likeThe mean(s) of sample 1. std1array_likeThe corrected sample standard deviation of sample 1 (i.e. ddof=1). nobs1array_likeThe number(s) of observations of sample 1. mean2array_likeThe mean(s) of sample 2. std2array_likeThe corrected sample standard deviation of sample 2 (i.e. ddof=1). nobs2array_likeThe number(s) of observations of sample 2. equal_varbool, optionalIf 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’}, optionalDefines 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. New in version 1.6.0. Returns: statisticfloat or arrayThe calculated t-statistics. pvaluefloat or arrayThe two-tailed p-value. See also scipy.stats.ttest_indNotes 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. References [1]https://en.wikipedia.org/wiki/T-test#Independent_two-sample_t-test [2]https://en.wikipedia.org/wiki/Welch%27s_t-test Examples 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 39.0Apply 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 0.175251The 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) |
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