If we have two samples, $x_1, \ldots , x_n$ and $x’_1, \ldots , x’_n$, and we compute the mean of each one, we might want to ask whether one mean is less than the other. Or more precisely, is their difference significantly less than zero?
- How to compute a confidence interval for a population mean
- How to do a two-sided hypothesis test for a sample mean
- How to do a two-sided hypothesis test for two sample means
- How to do a one-way analysis of variance (ANOVA)
- How to do a hypothesis test for a mean difference (matched pairs)
- How to do a hypothesis test for a population proportion
If we call the mean of the first sample $\bar x_1$ and the mean of the second sample $\bar x_2$, then this is a two-sided test with the null hypothesis $H_0: \bar x_1 - \bar x_2 \ge 0$. We choose a value $0\leq\alpha\leq1$ as the probability of a Type I error (false positive, finding we should reject $H_0$ when it’s actually true). Let’s use $\alpha=0.10$ as an example.
1 2 3 4 5 6 7 8 9 from scipy import stats # Replace these first three lines with the values from your situation. sample1 = [ 6, 9, 7, 10, 10, 9 ] sample2 = [ 12, 14, 10, 17, 9 ] # Run a one-sample t-test and print out alpha, the p value, # and whether the comparison says to reject the null hypothesis. stats.ttest_ind( sample1, sample2, equal_var=False, alternative="less" )
1 Ttest_indResult(statistic=-2.4616581720814326, pvalue=0.02548641870923849)
The output says that the $p$-value is about $0.0255$, which is less than $\alpha=0.10$. Therefore the samples give us enough evidence to reject the null hypothesis at the $\alpha=0.10$ level. That is, the data suggest that $\bar x_1 < \bar x_2$.
equal_var parameter tells SciPy not to assume that the two samples
have equal variances. If in your case they do, you can omit that parameter,
and it will revert to its default value of
Content last modified on 24 July 2023.
Contributed by Nathan Carter (email@example.com)