Mr Dough's factory produces packets of flour for sale in local supermarkets packed in 2 sizes -- Regular-sized packets and Large-sized packets. Automated machines weigh and pack the flour. A recent major power outage resulted in all his machines being needed to be recalibrated upon restarting. Regular-sized packets are supposed to have a mean mass of 1 kg. Mr Dough suspects that his machines now pack Regular-sized packets that are underweight. He took a sample of his Regular-sized packets and measured their masses, $x$ g (correct to nearest 5 g). The results are recorded as follows: (see table below) Large-sized packets of flour are known to have a normal distribution with a mean mass of 2 kg. Mr Dough also suspects that his machines are now not packing the Large-sized packets according to the required mass correctly. A sample of 15 Large-sized packets of flour gave a total mass of 30075 g. Assuming Large-sized packets of flour have a population variance of $\sigma^{2}\ \text{g}^{2}$, an appropriate test was carried out which gave a p-value of 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 $Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}$.

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