多夫先生（Mr Dough）的工厂生产供当地超市销售的面粉，包装分为两种规格——普通装和大号装。自动机器负责称重并包装面粉。最近发生了一次重大停电，导致他所有的机器在重新启动时都需要重新校准。 普通装面粉的平均质量应为 1 kg。多夫先生怀疑他的机器现在包装的普通装面粉分量不足。他抽取了普通装面粉的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： （见下表） 已知大号装面粉的质量服从正态分布，平均质量为 2 kg。多夫先生还怀疑他的机器现在未能按照要求的质量正确包装大号装面粉。抽取 15 袋大号装面粉的样本，测得总质量为 30075 g。 假设大号装面粉的总体方差为 $\sigma^{2}\ \text{g}^{2}$，进行适当的检验后，得到的 p 值为 0.0529。

1. Determine the unbiased estimate of the population variance directly from the grouped data using a graphing calculator.

Entering the data into a graphing calculator (with masses as $x$ and number of packets as frequency):
$s^2 = \frac{6012}{49} = 122.6938...$
$s^2 \approx 123 \text{ (to 3 s.f.)}$

1. Identify the direction of the test from the context and conclude.

Since Mr Dough suspects that the Regular-sized packets are underweight, he is testing for a decrease in the mean mass (i.e. the alternative hypothesis is $H_1: \mu < 1000\text{ g}$). Therefore, a one-tail test should be conducted.

1. Formulate the null and alternative hypotheses.

Let $X$ be the mass of a Regular-sized packet in grams, and $\mu$ be the population mean mass.

$H_0: \mu = 1000$
$H_1: \mu < 1000$
(Note: A one-tailed test is used because the suspicion is that the packets are 'underweight'.)

2. Calculate the test statistic and $p$-value from the sample data.

From the data provided (and calculations typical for part a):
$n = 50$
$\bar{x} = \frac{\sum f x}{n} = \frac{49870}{50} = 997.4$ g
$s^2 = \frac{1}{n-1}\left(\sum f x^2 - \frac{(\sum fx)^2}{n}\right) = \frac{1}{49}\left(49746350 - \frac{49870^2}{50}\right) = \frac{6012}{49} \approx 122.69$

Under $H_0$, the test statistic is $Z = \frac{\bar{X} - \mu}{s/\sqrt{n}} \approx \frac{997.4 - 1000}{\sqrt{122.69/50}} \approx -1.660$

Using a graphing calculator for a $Z$-test:
$p\text{-value} = P(Z < -1.660) \approx 0.0485$ (to 3 s.f.)

3. Make a decision and interpret the result in the context of the problem.

Since the $p$-value $= 0.0485 < 0.05$, we reject $H_0$.
There is sufficient evidence at the 5% level of significance to conclude that the population mean mass of Regular-sized packets is less than 1 kg, hence Mr Dough's suspicion that his machines are packing underweight packets is justified.

4. Explain whether the assumption of a normal distribution is necessary.

It is not necessary to assume that the mass of the packets follows a normal distribution.
Since the sample size $n = 50$ is sufficiently large ($n \ge 30$), by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed.

1. Calculate the sample mean and formulate the hypotheses to determine the type of test.

Let $X$ be the mass of a Large-sized packet.
Sample mean, $\bar{x} = \frac{30075}{15} = 2005 \text{ g}$

Population mean, $\mu = 2000 \text{ g}$ (2 kg)
$H_0: \mu = 2000$
$H_1: \mu \neq 2000$
The test is two-tailed since Mr Dough suspects the mass is not packed 'correctly' (rather than purely 'underweight' or 'overweight').

2. Use the given p-value for a two-tailed test to find the corresponding z-value.

For a two-tailed test, the $p$-value is $2 \times \text{P}(Z > z) = 0.0529$.
$\text{P}(Z > z) = 0.02645$
$\text{P}(Z < z) = 1 - 0.02645 = 0.97355$

Using the standard normal distribution:
$z \approx 1.93568$

3. Equate the z-value to the standard normal test statistic formula to solve for the population standard deviation, $\sigma$.

The test statistic for the sample mean is given by $Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}$.

$1.93568 = \frac{2005 - 2000}{\sigma / \sqrt{15}}$
$1.93568 = \frac{5\sqrt{15}}{\sigma}$
$\sigma = \frac{5\sqrt{15}}{1.93568}$
$\sigma \approx 10.004 \text{ g}$
$\sigma = 10.0 \text{ (correct to 3 s.f.)}$