The four steps and the critical-value route

The skeleton

Every test on the paper is marked in four steps, in this order:

1. Hypotheses. $H_0$ and $H_a$, with $\mu$ defined.
2. Statistic. How many standard errors the estimate sits from the null value.
3. Table. A critical value (this lesson) and/or a p-value (L10.4), at the stated $\alpha$.
4. Decision and sentence. Reject or do not reject $H_0$ at $\alpha$, then a sentence in the business context.

With $\sigma$ known, the statistic is the z-score of $\bar{x}$ under $H_0$:

$$z=\frac{\bar{x}-{\mu }_0}{\sigma /\sqrt{n}}$$

It is Module 8's standardisation with ${\mu }_0$ — the null value — in place of $\mu$. If $H_0$ is true, $z$ is a draw from $N(0,1)$; a $z$ far out in the tail named by $H_a$ is evidence against $H_0$.

Worked example — did the new process cut days-to-pay?

Kitchener Ridge Logistics introduced a new collections process. Historically $\mu =45$ days with $\sigma =8$. A sample of $n=64$ invoices under the new process has $\bar{x}=43.1$. Test at $\alpha =0.05$ whether the mean has fallen.

Step 1. $H_0:\mu =45$, $H_a:\mu <45$, where $\mu$ is the mean days-to-pay under the new process. Lower-tailed, from the word fallen.

Step 2. $\mathrm{SE}=8/\sqrt{64}=1$;

$$z=\frac{43.1-45}{1}=-1.90$$

The sample mean is 1.9 standard errors below the null value.

Step 3. The critical value is the $z$ that cuts off area $\alpha$ in the tail named by $H_a$. Lower tail, $\alpha =0.05$: $-z_{0.05}=-1.645$. The rejection region is $z<-1.645$.

Step 4. $-1.90<-1.645$: the statistic is in the rejection region. Reject $H_0$ at $\alpha =0.05$. There is evidence, at the 5 % level, that the new process has reduced mean days-to-pay below 45 days.

Condition, on the side: $n=64>30$, so $\bar{X}$ is normal by the CLT and $\sigma$ is known — a z-test is the right tool.

The sign follows the tail

The critical value is not always 1.645 with a plus sign:

Had the question been has the mean changed? — two-sided — the same $z=-1.90$ would be compared with $\pm 1.96$: $|-1.90|=1.90<1.96$, so do not reject. The same data give a different decision under a different question, which is why the tail is settled in step 1 and never revisited. A two-sided test splits $\alpha$ between the tails and is harder to reject.

What the statistic means

$z=-1.90$ says the sample mean lies 1.9 SEs below the value $H_0$ asserts. It does not say how far $\bar{x}$ is from the true $\mu$ — unknown — and it is not a probability of anything. The probability arrives in L10.4, when the same $z$ is turned into a p-value.