A hospital introduces a new medication regimen to help patients recover from a viral infection more quickly. Previously, the average recovery time was 7.2 days. After adopting the new regimen, a random sample of 40 patients is observed, and their recovery times, $x$, in days, are recorded. The results are summarised below.
$\sum (x - 7.2) = -42 \quad \text{and} \quad \sum (x - 7.2)^2 = 648$

1. Determine whether the test is directional or non-directional based on the context.

The hospital wants to test if the new medication helps patients recover 'more quickly'. This implies they are only interested in seeing if the average recovery time is less than 7.2 days, making it a directional hypothesis, which requires a one-tailed test.

1. Calculate the unbiased estimate of the population mean.

Let the unbiased estimate of the population mean be $\bar{x}$.
$\bar{x} = a + \frac{\sum (x-a)}{n}$
$\bar{x} = 7.2 + \frac{-42}{40}$
$\bar{x} = 6.15$

2. Calculate the unbiased estimate of the population variance.

Let the unbiased estimate of the population variance be $s^2$.
$s^2 = \frac{1}{n-1} \left( \sum (x-a)^2 - \frac{(\sum (x-a))^2}{n} \right)$
$s^2 = \frac{1}{39} \left( 648 - \frac{(-42)^2}{40} \right)$
$s^2 = \frac{1}{39} (648 - 44.1)$
$s^2 = \frac{603.9}{39} \approx 15.485 \approx 15.5 \text{ (to 3 s.f.)}$

1. Retrieve the sample mean and unbiased estimate of the population variance (from part b).

From part (b), the unbiased estimates are:
Sample mean, $\bar{x} = \frac{\sum (x - 7.2)}{40} + 7.2 = \frac{-42}{40} + 7.2 = 6.15$
Unbiased estimate of population variance, $s^2 = \frac{1}{39} \left( 648 - \frac{(-42)^2}{40} \right) = \frac{2013}{130} \approx 15.485$

2. Define the parameter and formulate the hypotheses.

Let $\mu$ be the population mean recovery time (in days) under the new regimen.
Null hypothesis, $H_0 : \mu = 7.2$
Alternative hypothesis, $H_1 : \mu < 7.2$

3. State the test statistic, compute its value, and find the $p$-value.

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. We use a 1-tail $Z$-test.
Under $H_0$, the test statistic is $Z = \frac{\bar{X} - 7.2}{S / \sqrt{40}} \sim N(0,1)$ approximately.
Using a graphing calculator:
$z = \frac{6.15 - 7.2}{\sqrt{\frac{2013}{130} / 40}} \approx -1.6876$
$p\text{-value} \approx 0.0457$

4. Make a conclusion based on the $p$-value.

Since the $p\text{-value} = 0.0457 < 0.05$, we reject $H_0$.
There is sufficient evidence at the 5% level of significance to conclude that the average recovery time of patients under the new regimen has reduced (i.e., is less than 7.2 days).

1. State the meaning of the significance level in the context of the hypothesis test

The significance level is the probability of rejecting the null hypothesis ($H_0$) when it is true.

In this context, $H_0$ is that the true average recovery time is 7.2 days, and $H_1$ is that the average recovery time is less than 7.2 days.

Therefore, at the 5% significance level, there is a 5% (or 0.05) probability of concluding that the average recovery time under the new regimen has decreased, when in fact it has not changed from 7.2 days.

1. Explain the limitation of a small sample size when the population distribution is unknown.

The population distribution of recovery times is unknown (it is not stated to be normal). Since the sample size of 10 is small ($n < 30$), the Central Limit Theorem cannot be applied to approximate the distribution of the sample mean as a normal distribution. Therefore, the test cannot be carried out, and taking a sample of 10 is not sufficient.

1. State the hypotheses and the condition on the test statistic for not rejecting $H_0$.

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

Since $n = 30$ is sufficiently large, by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed. The test statistic is:
$Z = \frac{\bar{X} - \mu}{s/\sqrt{n}} = \frac{k - 7.2}{1.7/\sqrt{30}}$

For a lower-tailed test at the $1\%$ significance level, the critical value is $z = -2.3263$.
Because the test concludes that the average recovery time did not improve, there is insufficient evidence to reject $H_0$. Thus, the test statistic must be greater than or equal to the critical value:
$z \ge -2.3263$

2. Substitute the given values into the inequality and solve for $k$.

$\frac{k - 7.2}{1.7 / \sqrt{30}} \ge -2.3263$
$k - 7.2 \ge -2.3263 \left( \frac{1.7}{\sqrt{30}} \right)$
$k - 7.2 \ge -0.7220$
$k \ge 6.4780$

The set of values is $k \ge 6.48$ (to 3 s.f.).