一家洗手液公司 Cleanser 声称，其每瓶洗手液的平均容量为 15 ml。消费者协会随机抽取了 80 瓶样本，并测量了每瓶的容量。结果发现样本均值和样本方差分别为 $\bar{x}$ 和 $2\ \text{ml}^2$。在 5% 的显著性水平上进行了假设检验，以检验 Cleanser 是否夸大了其洗手液的平均容量。 结果发现，没有足够的证据表明 Cleanser 夸大了其洗手液的平均容量。

1. State the definition of a random sample incorporating the context (size 80, Cleanser hand sanitiser bottles)

A random sample in this context means that every possible sample of 80 bottles of Cleanser hand sanitiser has an equal chance of being selected.

1. State the meaning of the level of significance in terms of the null and alternative hypotheses within the given context.

The level of significance is the probability of rejecting the null hypothesis when it is true. In this context, $H_0$ is that the mean volume is 15 ml, and rejecting it implies concluding that Cleanser has overstated the mean volume (mean volume < 15 ml).

Therefore, at the 5% level of significance, there is a 5% probability of concluding that Cleanser has overstated the mean volume of its hand sanitiser, when in fact the true mean volume is exactly 15 ml.

1. Calculate the unbiased estimate of the population variance using the sample size and sample variance

$s^2 = \frac{n}{n-1} \times \text{sample variance}$
$s^2 = \frac{80}{80-1} \times 2$
$s^2 = \frac{160}{79} \approx 2.03 \text{ (to 3 s.f.)}$

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

Let $\mu$ be the population mean volume of a bottle of Cleanser hand sanitiser, in ml.
$H_0: \mu = 15$
$H_1: \mu < 15$

2. State the test statistic using the unbiased estimate of the population variance.

From part (iii), the unbiased estimate of the population variance is $s^2 = \frac{80}{79}(2) = \frac{160}{79}$.
Since $n = 80 \ge 30$, by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed under $H_0$. The test statistic is:
$Z = \frac{\bar{X} - 15}{\sqrt{\frac{160/79}{80}}} = \frac{\bar{X} - 15}{\sqrt{2/79}} \sim N(0, 1)$

3. Determine the critical value and the inequality for not rejecting the null hypothesis.

For a $1$-tailed test at the $5\%$ level of significance, the critical value is $z = -1.64485$.
Since there is insufficient evidence to conclude that the company overstated the mean volume (i.e., that $\mu < 15$), we do not reject $H_0$. Thus, the test statistic must lie outside the critical region:
$Z \ge -1.64485$

4. Solve the inequality to find the minimum required value for the sample mean.

$\frac{\bar{x} - 15}{\sqrt{2/79}} \ge -1.64485$
$\bar{x} - 15 \ge -1.64485 \sqrt{\frac{2}{79}}$
$\bar{x} \ge 15 - 0.26171$
$\bar{x} \ge 14.738$
Correct to 2 decimal places, the range of values is $\bar{x} \ge 14.74$.

1. Find the unbiased estimate of the population mean

Entering the frequency table into a graphing calculator, we obtain the sample mean, which is the unbiased estimate for the population mean:
$\hat{\mu} = \bar{x} = 14.9125$

2. Find the unbiased estimate of the population variance

From the graphing calculator, the unbiased estimate of the population variance $s^2$ is:
$s^2 = 1.72642\dots \approx 1.73 \text{ (to 3 s.f.)}$
(Alternatively, using the formula $s^2 = \frac{1}{n-1}\left(\sum fx^2 - \frac{(\sum fx)^2}{n}\right)$:
$\sum fx = 1193$, $\sum fx^2 = 17927$
$s^2 = \frac{1}{79} \left( 17927 - \frac{1193^2}{80} \right) \approx 1.73$)

1. Identify the relationship between the variances and the test statistics

From earlier parts, the identical sample mean is $\bar{x} = 14.9125 < 15$. The unbiased estimate of the population variance for Germsfree ($s_2^2 \approx 1.73$) is smaller than that for Cleanser ($s_1^2 \approx 2.03$).

Since $n=80$ for both, the standard error for Germsfree is smaller. With the numerator of the test statistic being negative ($14.9125 - 15 < 0$), a smaller denominator means the test statistic for Germsfree is more negative than that for Cleanser ($z_2 < z_1 < 0$).

2. Conclude the inequality relating the $p$-values

The test is a lower-tail test ($H_1: \mu < 15$) since we are testing if the company has overstated the volume (i.e. if the true mean is less than claimed).
The $p$-value is the lower-tail probability, $P(Z < z)$.
Since $z_2 < z_1$, it follows that $P(Z < z_2) < P(Z < z_1)$.
Thus, $p_2 < p_1$.