2.10 Example

Now let’s go through the men and women height example to see what happens.

I’ll use the (fake) data from section 2.10.1. The heights of 25 men, and 25 women. We first compute the averages for each group

1. Calculate the difference in the means

The means are ${\mu }_m=70.228$ and ${\mu }_w=64.702$. So $\mathrm{\Delta }\mu ={\mu }_m-{\mu }_w=5.527$.

2. Calculate the estimated variance of this difference

We use eqn. 2.11 to do get $Var(\mathrm{\Delta }\mu )$. First we calculate ${\sigma }_m^2=7.7115$ and ${\sigma }_2^2=5.7873$. So

	$Var(\mathrm{\Delta }\mu )=({\displaystyle \frac{(n_m-1){\sigma }_m^2+(n_w-1){\sigma }_w^2)}{n_m+n_w-2}})({\displaystyle \frac{1}{n_m}}+{\displaystyle \frac{1}{n_w}})$		(2.15)
	$=({\displaystyle \frac{(25-1)\times 7.7115+(25-1)\times 5.7873}{25+25-2}})({\displaystyle \frac{1}{25}}+{\displaystyle \frac{1}{25}})=0.54$

3. Form t

$t=\mathrm{\Delta }\mu /\sqrt{Var(\mathrm{\Delta }\mu )}=5.527/.54=10.235$.

3. Calculate likelihood

We have $25+25-1=49$ degrees of freedom. So we look up the area under the t-distribution with $49$ degrees of freedom, starting at We can use handy online tools like a calculator that’ll tell you the area under the curve In this particular one, it tells you the area from the left so you enter in $-10.235$ instead. In this case, the answer is $0.0000$.

4. Verdict

This is very significant statistically. The probability that you’d get this from the null hypothesis is $0$ to at least for decimal places.