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"null.value" "alternative" "method" "data.name" "statistic" "parameter" "p.value" "conf.int" "estimate" For t.test it's easy to figure out what we want: > ttest = t.test(x,y) In addition, for some hypothesis tests, you may need to pass the object from the hypothesis test to the summary function and examine its contents. A general method for a situation like this is to use the class and names functions to find where the quantity of interest is. For this function, the R help page has a detailed list of what the object returned by the function contains.
#WHY DOES MINITAB 18 NOT ASSUME EQUAL VARIANCES HOW TO#
T = 1.4896, df = 15.481, p-value = 0.1564Īlternative hypothesis: true difference in means is not equal to 0īefore we can use this function in a simulation, we need to find out how to extract the t-statistic (or some other quantity of interest) from the output of the t.test function. Let's test it out on a simple example, using data simulated from a normal distribution. The function t.test is available in R for performing t-tests. Before we can explore the test much further, we need to find an easy way to calculate the t-statistic. There is also a widely used modification of the t-test, known as Welch's t-test that adjusts the number of degrees of freedom when the variances are thought not to be equal to each other. It is known that under the null hypothesis, we can calculate a t-statistic that will follow a t-distribution with n1 + n2 - 2 degrees of freedom. The null hypothesis is that the two means are equal, and the alternative is that they are not. The assumption for the test is that both groups are sampled from normal distributions with equal variances. One of the most common tests in statistics is the t-test, used to determine whether the means of two groups are equal to each other. Originally for Statistics 133, by Phil Spector t-tests
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Thenįind the critical value for a 5% level test, and perform the intermediateĬomputations necessary to get $T^\prime.
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Notice that $\nu = 22 $ you should use the formula to verify this result. T table to find the critical value for a test at the 5% level, verify the value of $S_p.$ and toĭo the other intermediate computations required to find the $T$ statistic.įor the Welch test the Minitab output is as follows: Two-Sample T-Test Is the alternative hypothesis is $H_a: \mu_1 \ne \mu_2.)$ I will leave it to you to consult a printed (This P-value is for a two-sided test that Results from Minitab for the pooled test are as follows: Two-Sample T-Testīecause the P-value exceeds 5%, you would not reject the null hypothesisĪt the 5% level of significance. Sizes are unequal and the smaller sample has the larger variance. The bad behavior of the pooled t test can be especially serious if sample In your textbook, you can read about it and find the formula for the degrees of freedom on Wikipedia. $\nu$ nearer the larger value if $S_1^2 \approx S_2^2$ and nearer the smaller value if the sample variances differ greatly. The result of theįormula is that $\min(n_1 - 1, n_2 - 1) \le \nu \le n_1 + n_2 - 2,$ with $$S_p^2 = \frac,$$ which has approximately Student's t distribution withĭegrees of freedom $\nu$ found according to a formula. (1) The pooled test makes the assumption that the two population variances are equal, computes the 'pooled' variance estimate as There are two versions of the two-sample t test.