Proving equality of mean between ANOVA with 2 levels and t-test

Question

I want to prove that there is an equality of the ANOVA with 2 levels and t-test. So:

From the t-test we know that:

$$t=\frac{\overline{x}_1-\overline{x}_2}{\sqrt{\frac{S_1^2}{n_1}-\frac{S_1^2}{n_1}}} \sim t(n_1+n_2-2)$$ And we know for a fact that if a statistic is t-distributed => $\mu=0$.

Now for ANOVA with 2 levels we should have a null hypothesis: $$H_0:\mu_0=\mu_1$$

And I guess from this hypothesis we should get something out but I have no idea.

Can someone please show me a prove that this statement is true ?

Answer 1

I'm guessing that you have never looked carefully at the formulas for a pooled two-sample t test or for a one-way ANOVA. (And that one purpose of assigning this problem is to get you to do that.) Below both procedures are computed in Minitab statistical software. The data are ten normal observations in group A and ten in group B. I suggest you look at the formulas and use them to compute what Minitab has computed. Then I'm guessing you will understand what you need to do.

Data. Fake data were simulated as follows: A ~ NORM(100, 10), B ~ NORM(105, 10). So $\mu_1 \ne \mu_2$ and $H_0$ is false. However, with population SD $\sigma = 10$ there is not enough information with only $n = 10$ replications in each group to detect that $H_0$ is false.

A
    99   103    92   105   103    97    98   113    90   112

B
   109    91   110   103   106   100   101   112   114   110

Pooled (equal pop variances) two-sided two-sample t test. You should compute pooled SD $S_p$ and T-value.

Two-Sample T-Test and CI: A, B 

Two-sample T for A vs B

    N    Mean  StDev  SE Mean
A  10  101.20   7.60      2.4
B  10  105.60   6.95      2.2

Difference = μ (A) - μ (B)
Estimate for difference:  -4.40
T-Test of difference = 0 (vs ≠): 
    T-Value = -1.35  P-Value = 0.193  DF = 18
Both use Pooled StDev = 7.2801

One-factor ANOVA, two levels of factor. You should compute MS(Factor) and MS(Error). In the ANOVA table below these values are slightly rounded. Note the relationships of MS(Factor) to the numerator of the T statistic (above) and MS(Error) to the denominator of the T statistic. The F statistic is the ratio of MS(Factor) to MS(Error). Verify that the pooled SD is the square root of MS(Error). Notice that the P-value is exactly the same here as for the t test above.

One-way ANOVA: A, B 

Method

Null hypothesis         All means are equal
Alternative hypothesis  At least one mean is different

Equal variances were assumed for the analysis.

Factor Information

Factor  Levels  Values
Factor       2  A, B

Analysis of Variance

Source  DF       SS      MS  F-Value  P-Value
Factor   1    96.80   96.80     1.83    0.193
Error   18   954.00   53.00
Total   19  1050.80

Pooled StDev = 7.28011