一家工厂生产长度为 10 cm 的红肠卷。为了进行质量控制，生产主管希望从一台特定机器生产的红肠卷中抽取一个随机样本。

1. State the definition of a simple random sample using the specific context of the question.

A sample is random if every sausage roll produced by the particular machine has an equal probability (or chance) of being selected for the sample.

1. State a valid practical or contextual reason for coding the data by subtracting 10.

The factory aims to produce sausage rolls of exactly 10 cm. Summarising the data as $(x-10)$ measures the deviation from this target length.
Consequently, since all values of $x$ are expected to be close to 10, $(x-10)$ will be close to 0. Working with these smaller numbers simplifies the manual calculation of the mean and variance and prevents the loss of significance that can occur when squaring and summing large values.

1. Calculate the unbiased estimate of the population mean

$\bar{x} = 10 + \frac{\sum (x-10)}{n}$
$\bar{x} = 10 + \frac{-2.5}{60}$
$\bar{x} = \frac{239}{24} \approx 9.9583 \approx 9.96 \text{ cm (to 3 s.f.)}$

2. Calculate the unbiased estimate of the population variance

$s^2 = \frac{1}{n-1} \left( \sum (x-10)^2 - \frac{(\sum (x-10))^2}{n} \right)$
$s^2 = \frac{1}{60-1} \left( 4.3 - \frac{(-2.5)^2}{60} \right)$
$s^2 = \frac{1}{59} \left( 4.3 - \frac{6.25}{60} \right)$
$s^2 = \frac{1007}{14160} \approx 0.071116 \approx 0.0711 \text{ cm}^2 \text{ (to 3 s.f.)}$

1. State the null and alternative hypotheses, and define the symbol used.

Let $\mu$ be the population mean length, in cm, of the sausage rolls produced by the machine.
$H_0: \mu = 10$
$H_1: \mu < 10$

2. Identify the unbiased estimates from part (c) and define the test statistic.

From part (c), the unbiased estimates are $\bar{x} = 10 - \frac{2.5}{60} = \frac{239}{24} \approx 9.9583$ and $s^2 = \frac{1}{59}\left(4.3 - \frac{(-2.5)^2}{60}\right) = \frac{1007}{14160} \approx 0.071116$.

Since $n = 60$ is large, by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed. Under $H_0$, the test statistic is:
$Z = \frac{\bar{X} - 10}{s/\sqrt{n}} \approx \frac{9.9583 - 10}{\sqrt{0.071116 / 60}} \approx -1.2103$

3. Calculate the $p$-value.

Using a graphing calculator, the $p$-value for this 1-tail test is:
$p\text{-value} = P(Z < -1.2103) \approx 0.11308$

4. Compare the $p$-value with the level of significance and state the conclusion in context.

Since the $p$-value of $0.113 > 0.10$, we do not reject $H_0$.
There is insufficient evidence at the $10\%$ level of significance to conclude that the sausage rolls produced by the machine are shorter than 10 cm on average. Hence, the manager's claim is not valid.

1. Formulate the null and alternative hypotheses and determine the critical values for a 2-tail test.

Let $\mu$ be the population mean length of the sausage rolls.
$H_0: \mu = 10$
$H_1: \mu \neq 10$
Level of significance, $\alpha = 0.10$
Since this is a two-tailed test, the rejection region is split into two $5\%$ tails. The critical values from the standard normal distribution are $z = \pm 1.64485$.

2. Set up the standardisation formula for the test statistic.

Since the population variance $\sigma^2 = 5$ is known and $n = 80$ is large, by the Central Limit Theorem, the sample mean follows a normal distribution under $H_0$:
$\bar{X} \sim \text{N}\left(10, \frac{5}{80}\right)$
The test statistic is given by:
$Z = \frac{\bar{x} - 10}{\sqrt{\frac{5}{80}}} = \frac{\bar{x} - 10}{0.25}$

3. Formulate the inequality for the non-rejection region and solve for the sample mean.

For $H_0$ not to be rejected, the test statistic must fall within the non-rejection (acceptance) region:
$-1.64485 \le Z \le 1.64485$
$-1.64485 \le \frac{\bar{x} - 10}{0.25} \le 1.64485$
$-0.41121 \le \bar{x} - 10 \le 0.41121$
$9.5888 \le \bar{x} \le 10.4112$
$9.59 \le \bar{x} \le 10.4 \text{ (to 3 s.f.)}$