For all two-sample tests, the sample sizes must be equal in the two groups

We are testing the hypothesis that the population means are equal for the two samples. We assume that the variances for the two samples are equal.

H0:  μ1 = μ2
Ha:  μ1 ≠ μ2

Test statistic:  T = -12.62059
Pooled standard deviation:  sp = 6.34260
Degrees of freedom:  ν = 326
Significance level:  α = 0.05
Critical value (upper tail):  t1-α/2,ν = 1.9673
Critical region: Reject H0 if |T| > 1.9673

The absolute value of the test statistic for our example, 12.62059, is greater than the critical value of 1.9673, so we reject the null hypothesis and conclude that the two population means are different at the 0.05 significance level.

In general, there are three possible alternative hypotheses and rejection regions for the one-sample t-test:

Alternative HypothesisRejection RegionHa: μ1 ≠ μ2|T| > t1-α/2,νHa: μ1 > μ2T > t1-α,νHa: μ1 < μ2T < tα,ν

For our two-tailed t-test, the critical value is t1-α/2,ν = 1.9673, where α = 0.05 and ν = 326. If we were to perform an upper, one-tailed test, the critical value would be t1-α,ν = 1.6495. The rejection regions for three posssible alternative hypotheses using our example data are shown below.

Without doing any testing, we can see that the averages for men and women in our samples are not the same. But how different are they? Are the averages “close enough” for us to conclude that mean body fat is the same for the larger population of men and women at the gym? Or are the averages too different for us to make this conclusion?

We'll further explain the principles underlying the two sample t-test in the statistical details section below, but let's first proceed through the steps from beginning to end. We start by calculating our test statistic. This calculation begins with finding the difference between the two averages:

$ 22.29 - 14.95 = 7.34 $

This difference in our samples estimates the difference between the population means for the two groups.

Next, we calculate the pooled standard deviation. This builds a combined estimate of the overall standard deviation. The estimate adjusts for different group sizes. First, we calculate the pooled variance:

$ s_p^2 = \frac{((n_1 - 1)s_1^2) + ((n_2 - 1)s_2^2)} {n_1 + n_2 - 2} $

$ s_p^2 = \frac{((10 - 1)5.32^2) + ((13 - 1)6.84^2)}{(10 + 13 - 2)} $

$ = \frac{(9\times28.30) + (12\times46.82)}{21} $

$ = \frac{(254.7 + 561.85)}{21} $

$ =\frac{816.55}{21} = 38.88 $

Next, we take the square root of the pooled variance to get the pooled standard deviation. This is:

$ \sqrt{38.88} = 6.24 $

We now have all the pieces for our test statistic. We have the difference of the averages, the pooled standard deviation and the sample sizes.  We calculate our test statistic as follows:

$ t = \frac{\text{difference of group averages}}{\text{standard error of difference}} = \frac{7.34}{(6.24\times \sqrt{(1/10 + 1/13)})} = \frac{7.34}{2.62} = 2.80 $

Does the sample size of each group must be the same for a two sample t interval?

What are the two assumptions for a two

Is a two

When the sample sizes are equal the pooled variance of the two groups is the average of the 2 sample variances True False?