# 10.1 - Z-Test: When Population Variance is Known

10.1 - Z-Test: When Population Variance is KnownLet's start by acknowledging that it is completely unrealistic to think that we'd find ourselves in the situation of knowing the population variance, but not the population mean. Therefore, the hypothesis testing method that we learn on this page has limited practical use. We study it only because we'll use it later to learn about the "power" of a hypothesis test (by learning how to calculate Type II error rates). As usual, let's start with an example.

## Example 10-1

Boys of a certain age are known to have a mean weight of \(\mu=85\) pounds. A complaint is made that the boys living in a municipal children's home are underfed. As one bit of evidence, \(n=25\) boys (of the same age) are weighed and found to have a mean weight of \(\bar{x}\) = 80.94 pounds. It is known that the population standard deviation \(\sigma\) is 11.6 pounds (the unrealistic part of this example!). Based on the available data, what should be concluded concerning the complaint?

### Answer

The null hypothesis is \(H_0:\mu=85\), and the alternative hypothesis is \(H_A:\mu<85\). In general, we know that if the weights are normally distributed, then:

\(Z=\dfrac{\bar{X}-\mu}{\sigma/\sqrt{n}}\)

follows the standard normal \(N(0,1)\) distribution. It is actually a bit irrelevant here whether or not the weights are normally distributed, because the same size \(n=25\) is large enough for the Central Limit Theorem to apply. In that case, we know that \(Z\), as defined above, follows at least approximately the standard normal distribution. At any rate, it seems reasonable to use the test statistic:

\(Z=\dfrac{\bar{X}-\mu_0}{\sigma/\sqrt{n}}\)

for testing the null hypothesis

\(H_0:\mu=\mu_0\)

against any of the possible alternative hypotheses \(H_A:\mu \neq \mu_0\), \(H_A:\mu<\mu_0\), and \(H_A:\mu>\mu_0\).

For the example in hand, the value of the test statistic is:

\(Z=\dfrac{80.94-85}{11.6/\sqrt{25}}=-1.75\)

The critical region approach tells us to reject the null hypothesis at the \(\alpha=0.05\) level if \(Z<-1.645\). Therefore, we reject the null hypothesis because \(Z=-1.75<-1.645\), and therefore falls in the rejection region:

As always, we draw the same conclusion by using the \(p\)-value approach. Recall that the \(p\)-value approach tells us to reject the null hypothesis at the \(\alpha=0.05\) level if the \(p\)-value \(\le \alpha=0.05\). In this case, the \(p\)-value is \(P(Z<-1.75)=0.0401\):

As expected, we reject the null hypothesis because the \(p\)-value \(=0.0401<\alpha=0.05\).

By the way, we'll learn how to ask Minitab to conduct the \(Z\)-test for a mean \(\mu\) in a bit, but this is what the Minitab output for this example looks like this:

N | Mean | SE Mean | 95% Upper Bound | Z | P |
---|---|---|---|---|---|

25 | 80.9400 | 2.3200 | 84.7561 | -1.75 | 0.040 |