The graphical and numerical analyses you just performed might
make it pretty clear that the
Exposed group has a higher level of lead, on average, than the
Control group.
But imagine a critic arguing as follows:
"I admit that the average
lead level of the Exposed group is higher than
the average level for the
Control group in your sample. But I believe
this is due solely to chance.
If I were to do my own study, using the
same methodology as you
- but a different selection of subjects, I might,
just by chance, get a set
in which the Exposed group is lower than the
Control group."
Your job is to answer this imaginary critic.
1. State a null hypothesis and alternative hypothesis. You
should do so in terms of the population mean the difference between the
exposed and control children.
2. We have not yet discussed how to go about performing a test of this
hypothesis (but will do so soon and in the next lab). However, one
way to do so is to use these two Stata commands:
ttest x = #
where x should be the difference in lead levels and "#" is whatever value the null hypothesis says the mean should be.
The output will be the observed value of the t-statistic.
3. The t-statistic is random. So we might ask, what's the probability
that, just by chance, if I were to do this experiment again, I'd get an
observed value even more extreme than this one. This
probability is called the p-value and can be calculated by typing:
display tprob(df,t)
where t is the observed value of the t-statistic from your output from
the last command, and df is the degrees of freedom, which you can also
find in the output from the last command.
4. If the p-value is less than 5%, most people would reject the
null hypothesis. Do you reject the null hypothesis?