Dough先生的工厂生产在当地超市销售的袋装面粉，包装分为两种规格：普通装和大装。自动机器对称重和包装进行处理。最近一次严重停电导致所有机器在重启时都需要重新校准。 普通装面粉的平均质量应该为 1 kg。Dough先生怀疑他的机器现在包装的普通装面粉分量不足。他抽取了一批普通装面粉作为样本，并测量了它们的质量 $x$ g（精确到最接近的 5 g）。结果记录如下：

质量, $x$ (g)	普通装面粉袋数
980	5
985	6
990	7
995	7
1000	9
1005	6
1010	5
1015	3
1020	2

1. Compute the unbiased estimate of the population variance from the frequency table

By entering the given mass and frequency data into a graphing calculator:
$n = 50$
$s^2 = 122.693...$
$s^2 \approx 123 \text{ (to 3 s.f.)}$

(Alternatively, using the formula $s^2 = \frac{1}{n-1}\left[\sum fx^2 - \frac{(\sum fx)^2}{n}\right]$ with $\sum fx = 49870$ and $\sum fx^2 = 49746350$, we get $s^2 = \frac{1}{49}\left[ 49746350 - \frac{49870^2}{50} \right] = \frac{6012}{49}$)

1. State the test type with the reason based on the context

Mr Dough should conduct a one-tail test because he specifically suspects that the packets are underweight, meaning he is testing if the population mean mass is less than 1 kg (i.e., testing for a directional decrease in mean mass).

1. Set up the null and alternative hypotheses.

Let $\mu$ be the population mean mass of Regular-sized packets in grams.
$H_0: \mu = 1000$
$H_1: \mu < 1000$
Level of significance $\alpha = 0.05$

2. Use the sample statistics from earlier to compute the test statistic and $p$-value.

From part (a), the sample size $n = 50$, sample mean $\bar{x} = 997.4$, and the unbiased estimate of population variance $s^2 = \frac{6012}{49} \approx 122.694$.

Since $n$ is large, the test statistic is:
$Z = \frac{\bar{X} - 1000}{s/\sqrt{n}} \sim N(0, 1) \text{ approximately.}$
Using a graphing calculator's Z-Test:
$Z_{\text{calc}} \approx -1.66$
$p\text{-value} = 0.0485 \text{ (to 3 s.f.)}$

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

Since the $p\text{-value} = 0.0485 < 0.05$, we reject $H_0$.

There is sufficient evidence at the 5% level of significance to conclude that the mean mass of the Regular-sized packets of flour is less than 1000 g (i.e. they are underweight).

4. Explain whether the assumption of a normal population is required.

It is not necessary to assume that the mass of Regular-sized packets follows a normal distribution.

Since the sample size $n = 50$ is large ($n \ge 30$), by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed regardless of the underlying population's distribution.

1. Calculate the sample mean and define the null and alternative hypotheses.

Let $\mu$ be the population mean mass of Large-sized packets in grams.
The required mass is $2\text{ kg} = 2000\text{ g}$.
Sample mean, $\bar{x} = \frac{30075}{15} = 2005\text{ g}$

$H_0: \mu = 2000$
$H_1: \mu \neq 2000$ (since he suspects the mass is not packed correctly)

2. Relate the $p$-value to the test statistic and set up the equation for a two-tailed test.

Since the population follows a normal distribution and the population variance $\sigma^2$ is considered as a parameter, the test statistic is $Z = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}} \sim \mathrm{N}(0, 1)$

Since it is a two-tailed test and $\bar{x} > 2000$, the $p$-value is given by:
$p\text{-value} = 2 \times \mathrm{P}\left(Z > \frac{2005 - 2000}{\sigma/\sqrt{15}}\right) = 0.0529$
$\mathrm{P}\left(Z > \frac{5\sqrt{15}}{\sigma}\right) = \frac{0.0529}{2} = 0.02645$

3. Solve for $\sigma$ using the inverse normal distribution.

Using a graphing calculator, the $z$-value for an upper tail probability of $0.02645$ is approximately $1.93566$.

$\frac{5\sqrt{15}}{\sigma} = 1.93566$
$\sigma = \frac{5\sqrt{15}}{1.93566}$
$\sigma \approx 10.003$
$\sigma = 10.0 \text{ (shown correct to 3 s.f.)}$

1. Formulate the test statistic for the combined sample in terms of $h$.

The combined sample has size $n = 16$ and total mass $\sum x = 30075 + h$.
The new sample mean is $\bar{x} = \frac{30075 + h}{16}$.
Since we are testing the claim that the mean mass is 2 kg ($2000$ g) against the alternative that it is not (as established by the two-tailed context in the previous part), the hypotheses are $H_0: \mu = 2000$ and $H_1: \mu \neq 2000$.
The test statistic is:
$Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{\frac{30075 + h}{16} - 2000}{\frac{10.0}{\sqrt{16}}} = \frac{\frac{30075 + h}{16} - 2000}{2.5}$

2. Determine the critical $z$-values and set up the inequalities for rejection of $H_0$.

At the $4\%$ level of significance for a two-tailed test, the critical region is split symmetrically, leaving $2\%$ in each tail.
Using the standard normal distribution, the critical $z$-values are $\pm 2.0537$.
For $H_0$ to be rejected, the test statistic must fall in the critical region:
$Z \le -2.0537 \quad \text{or} \quad Z \ge 2.0537$
$\frac{\frac{30075 + h}{16} - 2000}{2.5} \le -2.0537 \quad \text{or} \quad \frac{\frac{30075 + h}{16} - 2000}{2.5} \ge 2.0537$

3. Solve the inequalities for $h$ and round the boundaries to the nearest gram.

$\frac{30075 + h}{16} - 2000 \le -5.134 \quad \text{or} \quad \frac{30075 + h}{16} - 2000 \ge 5.134$
$\frac{30075 + h}{16} \le 1994.866 \quad \text{or} \quad \frac{30075 + h}{16} \ge 2005.134$
$30075 + h \le 31917.85 \quad \text{or} \quad 30075 + h \ge 32082.15$
$h \le 1842.85 \quad \text{or} \quad h \ge 2007.15$
Rounding the boundaries to the nearest gram, the possible range of values is:
$h \le 1843 \quad \text{or} \quad h \ge 2007$