![]() ![]() Unequal variances, the significance level is very nearly 5%. If one uses the Welch test for samples from populations with The parameter var.eq=T leads to use of the pooled test. Notes about R code: (a) The default 2-sample t test in R is the Welch test, which does not assume equal variances. One can quibble whether equal variances are part of the null hypothesis.īut, using the pooled t test, equal variances are essential to a fair test of the null hypothesis. It is clear that unequal variances do change how the test performs. Is unfair to say that the test has "detected" unequal variances, but Have "detected" that means are unequal, because they aren't. Made a difference in how the pooled t test works. Obviously, the change from equal variances to unequal variances has The 'null distribution' (distribution when $H_0$ is true) has changed substantially. Now the test is falsely rejecting about 30% of the time-much more than 5% of the time. Now let's see what happens if we keep everything exactly the same-except that we change the population variances to be unequal, with $\sigma_1^2 = 16$ and $\sigma_2^2 = 1.$ set.seed(818) Just 'as advertised': The pooled 2-sample t test has incorrectly rejected $H_0$ in almost exactly 5% of the tests on one million sets of two samples from Results of a million such pooled 2-sample t tests. We could discuss the theory to show that this However, 5% of the time, a pooled test at the 5% level will makeĪ mistake, rejecting $H_0$ with a P-value $ < 0.05.$ From the simulation, we know that $\mu_1 - \mu_2 = 50.$ (Also that $\sigma_1^2 = \sigma_2^2 = 1.)$Īnd the test has (correctly) failed to reject $H_0.$ True difference in means is not equal to 0Īll is well. We reject $H_0$ atĬomparing two specific such samples, what outputĭo we get from the pooled 2-sample t test? set.seed(1234) Let's consider a sample of size $n_1 = 10$ from In other words, we can assume the sample variances are equal.You're talking about a pooled 2-sample t test, of 05, we fail to reject the null hypothesis. To perform an F-test on these two samples, we can calculate the F test statistic as:Īccording to the F-Distribution Calculator, an F-value of 1.577 with numerator df = n 1-1 = 12 and denominator df = n 2-1 = 12 has a corresponding p-value of 0.22079. Once again suppose we have the following two samples: If the p-value that corresponds to the test statistic is less than some significance level (like 0.05), then we have sufficient evidence to say that the samples do not have equal variances. Where s 1 2 and s 2 2 are the sample variances. The test statistic is calculated as follows: H A: The samples do not have equal variances. Thus, we could proceed to perform Student’s t-test to determine if the two groups have the same mean.Īn F-test is a formal statistical test that uses the following null and alternative hypotheses: Since this ratio is less than 4, we could assume that the variances between the two groups are approximately equal. The ratio of the larger sample variance to the smaller sample variance would be calculated as: ![]() Sample 1 has a variance of 24.86 and sample 2 has a variance of 15.76. So, if the two samples do not have equal variance then it’s best to use the Welch’s t-test.īut how do we determine if the two samples have equal variance?Īs a rule of thumb, if the ratio of the larger variance to the smaller variance is less than 4 then we can assume the variances are approximately equal and use the Student’s t-test.įor example, suppose we have the following two samples: Welch’s t-test:Assumes that both groups of data are sampled from populations that follow a normal distribution, but it does not assume that those two populations have the same variance. Student’s t-test:Assumes that both groups of data are sampled from populations that follow a normal distribution and that both populations have the same variance. When we want to compare the means of two independent groups, we can choose between two different tests: ![]()
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