一家医院引入了一种新的药物疗法，以帮助患者更快地从病毒感染中康复。此前，平均康复时间为 7.2 天。 采用新疗法后，观察了 40 名患者的随机样本，并记录了他们的康复时间 $x$（天）。结果总结如下。 $\sum (x-7.2) = -42 \quad\text{and}\quad \sum (x-7.2)^2 = 648$

1. Determine the direction of the test from the context

The phrase 'recover from a viral infection more quickly' indicates that the hospital is looking for a decrease in the average recovery time (i.e., whether the new mean is strictly less than $7.2$ days). Therefore, a one-tailed test is appropriate.

1. Calculate the unbiased estimate of the population mean.

$\bar{x} = a + \frac{\sum (x-a)}{n}$
$\bar{x} = 7.2 + \frac{-42}{40}$
$\bar{x} = 7.2 - 1.05 = 6.15$

2. Calculate the unbiased estimate of the population variance.

$s^2 = \frac{1}{n-1} \left( \sum (x-a)^2 - \frac{(\sum (x-a))^2}{n} \right)$
$s^2 = \frac{1}{40-1} \left( 648 - \frac{(-42)^2}{40} \right)$
$s^2 = \frac{1}{39} (648 - 44.1)$
$s^2 = \frac{603.9}{39} \approx 15.4846...$
$s^2 \approx 15.5 \text{ (to 3 s.f.)}$

1. Define the population parameter and state the null and alternative hypotheses.

Let $\mu$ be the population mean recovery time, in days, of patients on the new regimen.

Since the new medication is intended to help patients recover more quickly (a decrease in average time), we carry out a left-tailed test:
$H_0: \mu = 7.2$
$H_1: \mu < 7.2$

2. State the test statistic and compute its value along with the $p$-value.

From part (b), the sample mean is $\bar{x} = 7.2 + \frac{-42}{40} = 6.15$ and the unbiased estimate of the population variance is $s^2 = \frac{1}{39}\left(648 - \frac{(-42)^2}{40}\right) = \frac{201.3}{13} \approx 15.485$.

Under $H_0$, since the sample size $n = 40$ is large ($n \ge 30$), by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed. The test statistic is $Z = \frac{\bar{X} - 7.2}{S/\sqrt{40}}$.

Using a $Z$-test on a graphing calculator:
Test statistic $z = -1.688$ (to 3 d.p.)
$p\text{-value} = 0.0457$ (to 3 s.f.)

3. Compare the $p$-value to the significance level and state the conclusion in the context of the question.

Level of significance, $\alpha = 0.05$.

Since $p\text{-value} = 0.0457 < 0.05$, we reject the null hypothesis $H_0$.

There is sufficient evidence at the 5% level of significance to conclude that the new medication regimen helps patients recover more quickly (i.e., the average recovery time is less than 7.2 days).

1. State the meaning of the 5% significance level using the specific context of the null and alternative hypotheses.

The significance level is the probability of rejecting $H_0$ when it is true. In this context, it means there is a 5% (or 0.05) chance of incorrectly concluding that the average recovery time is less than 7.2 days (i.e. the new medication regimen is effective), when in reality the average recovery time is still 7.2 days.

1. Explain why a sample size of 10 is insufficient based on the Central Limit Theorem and the unknown population distribution.

No, it would not be sufficient. The underlying population distribution of the recovery times is unknown. Since the sample size of 10 is small ($n < 30$), the Central Limit Theorem cannot be applied to assume that the sample mean is approximately normally distributed, which is a necessary condition for carrying out the hypothesis test.

1. State the hypotheses and the distribution of the sample mean.

Let $\mu$ be the population mean recovery time under the new regimen.
$H_0: \mu = 7.2$
$H_1: \mu < 7.2$

Under $H_0$, since the sample size $n = 30$ is sufficiently large, by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed:
$\bar{X} \sim \text{N}\left(7.2, \frac{1.7^2}{30}\right)$

2. Formulate the inequality for the non-rejection of the null hypothesis.

The test concludes that the average recovery time did not improve, which means there is insufficient evidence to reject $H_0$.
For a one-tailed test at the 1% significance level, $H_0$ is not rejected if the $p$-value is greater than or equal to $0.01$.
$\text{P}(\bar{X} \le k) \ge 0.01$

3. Solve the inequality for $k$ using standardisation or inverse normal.

Using standardisation:
$\frac{k - 7.2}{1.7/\sqrt{30}} \ge -2.3263$
$k - 7.2 \ge -2.3263 \left( \frac{1.7}{\sqrt{30}} \right)$
$k \ge 7.2 - 0.7220$
$k \ge 6.4779...$
$k \ge 6.48 \text{ (to 3 s.f.)}$

(Alternatively, using a graphing calculator: $k \ge \text{invNorm}\left(0.01, 7.2, \frac{1.7}{\sqrt{30}}\right) \approx 6.48$)