Working through a basic Statistics problem and I cannot figure out how to get the proper value of the difference in sample means for $\bar y - \bar x$.
Here is the relevant info:
$X \sim N(1200, 90)$ distribution $Y \sim N(1215, 110)$ distribution Sample Size: $100$
I know that the answer is $N(-15, 14.2)$.
I can figure out the $-15$ easy enough, however, I cannot figure out how to get $14.2$.
I thought the correct way to approach this would be to do $(110 / \sqrt{100}) - (90 / \sqrt{100})$ however this is not the case.
Can you please point me in the right direction and explain what I need to do differently? Thank you very much
If what you say you though was correct were correct then you'd get a negative number for the standard deviation if you subtracted in the other direction: $\displaystyle \frac{90}{\sqrt{100}} - \frac{110}{\sqrt{100}}.$ But a standard deviation cannot be negative.
You need $$ \sqrt{\left( \frac{90}{\sqrt{100}} \right)^2 + \left( \frac{110}{\sqrt{100}} \right)^2} = \sqrt{9^2 + 11^2} = \sqrt{81+121} = \sqrt{202} = \cdots $$
The fact that variances can be added in that way is the reason why standard deviations rather than mean absolute deviations are used as a measure of dispersion.