在一家工厂中，两台独立的机器 A 和 B 将面粉装入包装袋中，包装袋上标明每袋 1.5 kg。在一次质量控制检查中，生产经理随机抽取了由机器 A 填装的 40 袋面粉进行检查。这 40 袋面粉的质量（$x$ kg）记录如下。 $\sum x = 60.32 \quad \sum x^2 = 90.993$

1. State the definition of a random sample tailored to the specific context.

A random sample in this context means that every bag of flour filled by machine A has an equal chance of being chosen for the inspection.

1. Calculate the unbiased estimate of the population mean.

$\bar{x} = \frac{\sum x}{n}$
$\bar{x} = \frac{60.32}{40} = 1.508$

2. Calculate the unbiased estimate of the population variance.

$s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right)$
$s^2 = \frac{1}{39} \left( 90.993 - \frac{(60.32)^2}{40} \right)$
$s^2 = \frac{1}{39} (90.993 - 90.96256)$
$s^2 = \frac{0.03044}{39} \approx 0.000781 \text{ (to 3 s.f.)}$

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

Sample mean, $\bar{x} = \frac{\sum x}{n} = \frac{60.32}{40} = 1.508$

Unbiased estimate of variance, $s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right)$
$s^2 = \frac{1}{39} \left( 90.993 - \frac{60.32^2}{40} \right) = \frac{0.03044}{39} \approx 0.00078051$

2. State the hypotheses and define the population mean parameter.

Let $\mu$ be the population mean mass (in kg) of the bags filled by machine A.

$H_0: \mu = 1.5$
$H_1: \mu > 1.5$

3. State the test statistic, invoking the Central Limit Theorem, and evaluate the $p$-value using the graphing calculator.

Since the sample size $n = 40 \ge 30$ is large, by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed.
We carry out a 1-tail $Z$-test at the $5\%$ level of significance.

Test statistic: $Z = \frac{\bar{X} - 1.5}{s/\sqrt{n}} \sim N(0, 1)$ approximately.

Using a graphing calculator with $\mu_0 = 1.5$, $\bar{x} = 1.508$, $s = \sqrt{0.00078051} \approx 0.027938$, and $n = 40$:
$z = 1.8110 \approx 1.81$
$p\text{-value} = 0.03506 \approx 0.0351$

4. Conclude the hypothesis test in the context of the question.

Since $p\text{-value} = 0.0351 < 0.05$, we reject $H_0$.
There is sufficient evidence at the $5\%$ level of significance to conclude that machine A produces overweight bags.

1. State the contextual meaning of the $p$-value.

The $p$-value is the probability of obtaining a sample mean mass as extreme as or more extreme than the observed value ($\bar{x} = \frac{60.32}{40} = 1.508\text{ kg}$), assuming the null hypothesis ($\mu = 1.5\text{ kg}$) is true.

In context: It is the probability of obtaining a sample mean mass of $1.508\text{ kg}$ or more, assuming that the true population mean mass of the bags filled by machine A is $1.5\text{ kg}$.

1. State the hypotheses and the critical region for a two-tailed test at the $5\%$ significance level.

Let $\mu$ be the population mean mass of the bags in kg.
$H_0: \mu = 1.5$
$H_1: \mu \neq 1.5$

For a two-tailed test at the $5\%$ level of significance, we reject $H_0$ if the test statistic $z \le -1.960$ or $z \ge 1.960$.

2. Express the test statistic in terms of $n$ and set up the rejection inequality.

Under $H_0$, $Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \sim \mathrm{N}(0, 1)$.

Given $\bar{x} = 1.489$, $\mu = 1.5$, and $\sigma^2 = 0.00198$:
$z = \frac{1.489 - 1.5}{\sqrt{0.00198/n}} = \frac{-0.011\sqrt{n}}{\sqrt{0.00198}}$

Since $\bar{x} < 1.5$, we require $z \le -1.960$ for there to be sufficient evidence to reject $H_0$:
$\frac{-0.011\sqrt{n}}{\sqrt{0.00198}} \le -1.960$

3. Solve the inequality for $n$.

$\sqrt{n} \ge \frac{1.960 \sqrt{0.00198}}{0.011}$
$\sqrt{n} \ge 7.9286$
$n \ge 62.86$

Since the sample size $n$ must be an integer, $n \ge 63$.