Please consider the Table below
| Item | Sample Mean | Sample count, n | Sample Standard Deviation | Standard Error SD/sqrt(n) |
|---|---|---|---|---|
| Unit A | 95.9461 PSI | n=430 | 3.8397 | 0.185166776 |
| Unit B | 94.488 PSI | n=25 | 2.5344 | 0.50688 |
Now when I want to show if these two tests are significantly different from each other.
$$ \frac{94.488 - 95.9461}{ \sqrt{0.50688^2+0.185166776^2}} \approx -2.7 $$
Now according to this Website these two units are significantly different!? Even if I have the true standard deviation (in this case its probably similar). I would still get similar results. Dividing by the sample count completely destroys any kind of SE, as you can see: as n approaches infinity, SE approaches 0. You could make any two averages significantly different from each other with a large enough sample size, simply because you are dividing by a smaller and smaller number. Intuitively, I don' think there is any difference.
Suggestion 1: I don't think this addresses the problem I am getting at. When doing a Z-test, we would be using standard deviation, but with a t-test, or in this case, we divide by n. This doesn't make sense to me, we should inherently have a larger SE with a sample than compared to a population because we don't have as many data points.
Edit: for those asking for the formula from the website.
If the Z-statistic is less than 2, the two samples are the same.
If the Z-statistic is between 2.0 and 2.5, the two samples are marginally different
If the Z-statistic is between 2.5 and 3.0, the two samples are significantly different
If the Z-statistic is more then 3.0, the two samples are highly signficantly different
$$ Z = \frac{\bar{X_1}-\bar{X_2}}{\sqrt{\sigma^{2}_{x_1}+\sigma^{2}_{x_2}}} $$
Edit: To further clarify, below is a histogram of my tests. How can we possibly look at these two distributions and say that they are significantly different? Especially when, I only have 25 samples for one!
