Solution.

1.

(a)

Set up the hypothesis. “Correctly set” means the mean is neither too high nor too low, so the alternative is two-sided: \[H_0:\ \text {the machine is correctly set},\qquad H_1:\ \text {the machine is not correctly set}\] or, in symbols, \[H_0:\ \mu =1000\,\text {g}\qquad \text {against}\qquad H_1:\ \mu \neq 1000\,\text {g}.\]

(b)

Set up the level of significance \(\alpha =5\%\)

(c)

Test statistic. The standard deviation of \(5\) g is a property of the machine, so \(\sigma \) is known and the statistic is \(Z\) — not \(t\), despite \(n=9\) being small. What a small sample does require is that the population be normal, which is given. \begin {align*} Z &=\frac {\overline {X}-\mu }{\sigma /\sqrt {n}}=\frac {1003-1000}{5/\sqrt {9}}=\frac {3}{5/3}=1.80\\ Z_{0.025}&=1.96 \end {align*}

(d)

Critical region

Interpretation: our statistic \(Z=1.80\) lies between \(-1.96\) and \(1.96\), in the acceptance region, so we fail to reject \(H_0\). At the \(5\%\) level there is no evidence that the machine is incorrectly set. The \(P\)-value is \(2P(Z>1.80)=0.072\), just above \(0.05\).

2.

Now ask a different question: is the machine set too high? Naming a direction makes the test one-tailed. \[H_0:\ \text {the machine is correctly set},\qquad H_1:\ \text {the machine over-fills}\] \[H_0:\ \mu =1000\,\text {g}\qquad \text {against}\qquad H_1:\ \mu >1000\,\text {g}\] with \(\alpha =5\%\). The statistic is unchanged, \(Z=1.80\), but the critical value is not: all \(5\%\) now sits in one tail, so it is \(Z_{0.05}=1.645\) rather than \(1.96\).

\(Z=1.80\) exceeds \(Z_{0.05}=1.645\), so it falls in the critical region and we reject \(H_0\). The \(P\)-value is \(P(Z>1.80)=0.036\), below \(0.05\).

At the \(5\%\) level, then, the machine is over-filling.

Remark 3.3. The same data, the same statistic \(Z=1.80\), and two different answers — because the two questions are different. Asking ”is it correctly set?” leaves \(5\%\) to be split between two tails, so the bar is \(1.96\) and \(1.80\) does not clear it. Asking ”is it set too high?” puts the whole \(5\%\) in one tail, the bar drops to \(1.645\), and the same evidence now clears it.

Nothing is contradictory here, but the direction must be chosen before the data are seen. Running the two-tailed test, seeing it fail, and then switching to the one-tailed version to obtain a rejection is not a legitimate procedure — it doubles the true error rate.

Solution.

1.

Set up the hypothesis. The association suspects the wait is longer, which names a direction, so the test is one-tailed: \[H_0:\ \text {the average wait is as claimed},\qquad H_1:\ \text {the average wait is longer}\] \[H_0:\ \mu =60\ \text {minutes}\qquad \text {against}\qquad H_1:\ \mu >60\ \text {minutes}.\]

2.

Set up the level of significance at \(\alpha =1\%\).

3.

Test statistic. Here \(\sigma \) is unknown and estimated by \(S\), but \(n=50>30\) is large, so \(Z\) serves: \begin {align*} Z &=\frac {\overline {X}-\mu }{S/\sqrt {n}}=\frac {64-60}{6.3/\sqrt {50}}=\frac {4}{0.891}=4.49 \end {align*}

(If \(S\) had been computed with divisor \(n\) rather than \(n-1\), the equivalent form is \(S/\sqrt {n-1}=6.3/7=0.9\), giving \(Z=4.44\). The two conventions differ trivially here and not at all in the conclusion.)

4.

Critical region

5.

Conclusion. \(Z=4.49\) lies far beyond the critical value \(Z_{0.01}=2.326\), so we reject \(H_0\). There is very strong evidence, at the \(1\%\) level, that the average waiting time exceeds the \(60\) minutes the clinic states. The \(P\)-value is about \(0.0000036\) — if the true average really were \(60\) minutes, a sample mean this high would occur roughly once in \(280\,000\) samples.

Two things the test does not establish, worth stating because they are easy to slide into. It does not say the wait is much longer: the estimated excess is \(4\) minutes, and with \(n=50\) even a small excess is detectable. Significance measures the strength of the evidence, not the size of the effect. And it says nothing about why the wait is longer than claimed — these are observed times, not an experiment.