Hypothesis testing, chance

Question

To find out whether the inhabitants of two South Pacific islands may be regarded as having the same racial ancestry, an anthropologist determines the cephalie indices of six adult males from each island, getting $x_1$ = 77.4 and $x_2$ = 72.2 and the corresponding standard deviation $s_1 = 3.3$ and $s_2 = 2.1$. Test at $0.01$ level of significance whether the difference between the two sample means can reasonably be attributed to chance. Assume that the populations sampled are normal and have equal variances.

I am confused on what the hypotheses should be: should it be $H_0: \mu_1 - \mu_2 = 0$ and $H_1: \mu_1-\mu_2 \neq 0$?

Answer 1

Pooled 2-sample t test. Minitab is one of the relatively few statistical packages that will perform a pooled 2-sample t test on summarized data (that is, without seeing the 12 actual data values). You have correctly used 6 + 6 -2 = 10 degrees of freedom to get your critical value 3.169. Also, Minitab gets $T = 3.26$ for the test statistic, which agrees with your result. Here is Minitab output.

Two-Sample T-Test and CI 

Sample  N   Mean  StDev  SE Mean
1       6  77.40   3.30      1.3
2       6  72.20   2.10     0.86

Difference = μ (1) - μ (2)
Estimate for difference:  5.20
95% CI for difference:  (1.64, 8.76)
T-Test of difference = 0 (vs ≠): 
  T-Value = 3.26  P-Value = 0.009  DF = 10
Both use Pooled StDev = 2.7659

Welch 2-sample t-test. However, as a matter of statistical practice, I would not be willing to assume the two population variances are equal. (If the population means might be different, why not also the population variances?) The Welch version of the 2-sample t test does not assume equal population variances. Standard statistical practice is not to assume equal variances unless you have strong prior evidence of equality. (In this situation, that might come from measuring cephalic indexes for many other Pacific islands and having found that means vary, but that variances don't.)

Typically the Welch 2-sample t test uses a reduced degrees of freedom to compensate for possibly unequal variances. A formula involving sample sizes and sample standard deviations is used to find the adjusted degrees of freedom. Here is Minitab output for the Welch 2-sample t test.

Two-Sample T-Test and CI 

Sample  N   Mean  StDev  SE Mean
1       6  77.40   3.30      1.3
2       6  72.20   2.10     0.86

Difference = μ (1) - μ (2)
Estimate for difference:  5.20
95% CI for difference:  (1.52, 8.88)
T-Test of difference = 0 (vs ≠): 
  T-Value = 3.26  P-Value = 0.012  DF = 8

Notice that df = 8 here instead of 10 for the pooled test. Also, notice that there is no note at the end about using a pooled standard deviation. The P-value is a little larger for the Welch test, but is still below 0.05, so $H_0$ is rejected here as well.