1. ## Two sample t-test

We have an experiment on plant growth. There are 10 different locations used and at each location there are two plants grown, one with fertilizer (y) and one without (x) such that:

$\Sigma x=224,\ \Sigma x^2=5270,\ \Sigma y=244,\ \Sigma y^2=6222$

If we assume that the two samples are independent, then test the null hypothesis (there is no difference) against the alternative (there is a difference).

For x:

$\bar{x}=22.4\ \ s_x^2=252.4/9=28.04....$

For y:

$\bar{y}=24.4\ \ s_y^2=268.4/9=29.82....$

The estimate of $s^2$ for the paired variance is

$s^2=\frac{(9\times 28.04)+((9\times 29.82)}{18}=28.93$

thus $s=5.379$

The t statistic is then: $t=\frac{\bar{x}-\bar{y}}{s\sqrt(\frac{1}{9}+\frac{1}{9}})}=0.7887. ..$

Is what I have done so far correct? Do i now do on to check the degree's of freedom and the levels of significance?

I am then asked to do the same if we assumed they were paired samples? what do I do here?

Thanks for any help

2. ## Re: Two sample t-test

for the paired difference experiment you need the original x's and y's
$\bar d=\bar X-\bar Y$ but the estimate of the variance won't come from
the sample variances of the X's ad Y's. You need all of the di's.