Consider our respiratory disturbance index example again.
A reasonable strategy would reject the null hypothesis \(\bar{X}\) was larger than some constant, \(C\).
\(C\) will take into account the variability of \(\bar{X}\).
Typically, \(C\) is chosen so that the probability of a Type I error, \(\alpha\), is \(.05\) (or some other relevant constant), this probability label is a low number. \(5\%\) has emerged as sort of a benchmark in hypothesis testing.
\(\alpha\) = Type I error rate = Probability of rejecting the null hypothesis when, in fact, the null hypothesis is correct.
That’s a bad thing, you don’t want to make these kind of mistakes. But as in our court of law example, you don’t want to set this rate too low, because then we would never reject a null hypothesis. Let’s see if we can choose this constant \(C\), so that probability we would reject is simply tolerably low, say \(5\%\).
Standard error of the mean \(10/\sqrt{100} = 1\), the assumed standard deviation of the population. Here we haven’t drawn a distinction as to whether we estimate, or this is just a number that I’ve given you. Divided by \(\sqrt{100}\), that’s the square root of the sample size, that works out to be 1. Here I just created the settings, so it completely worked out to be 1.
Under \(H_0\) \(\bar{X} \sim N(30, 1)\) (remember the null hypothesis was \(\mu = 30\), and the standard deviation we’ve just calculated).
We want to choose \(C\) so that the \(P(\bar{X} \gt C; H_0)\) is 5%
The \(95th\) percentile of a normal distribution is \(1.645\) standard deviations from the mean
If \(C = \mu + \sigma \times z_{1-\alpha} = 30 + 1 \times 1.645 = 31.645\) we will achieved a cup point, so that the probability that a randomly drawn mean from this population is larger than this is \(5%\).
Then the probability that a \(N(30, 1)\) is larger than it is 5%
So the rule “Reject \(H_0\) when \(\bar{X} \ge 31.645\)” has the probability that the probability of rejection is \(5%\) when \(H_0\) is true (for the \(\mu_0, \sigma\) and \(n\) given)
In general we don’t convert \(C\) back to the original scale
We would just reject because the Z-score; which is how many standard errors the sample mean is above the hypothesized mean
\[ \frac{32 - 32}{10/\sqrt{100}} = 2 \]
is greater than \(1.645\)