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: