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: (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 $\sigma^2$ g^2, an appropriate test was carried out which gave a p-value of 0.0529. Another packet of Large-sized packet of flour with mass $h$ g is then added to the sample of 15. The combined sample of Large-sized packets resulted in a rejection of the claim that the mean mass of a Large-sized packet of flour is 2 kg at 4% level of significance.

1. Calculate the unbiased estimate of the population variance using the sample frequency data

From the frequency table, the total number of packets is $n = 50$.
Using a graphing calculator or directly computing the summation values:
$\sum f = 50$
$\sum fx = 49870$
$\sum fx^2 = 49746350$

The unbiased estimate of the population variance, $s^2$, is given by:
$s^2 = \frac{1}{n-1}\left(\sum fx^2 - \frac{(\sum fx)^2}{n}\right)$
$s^2 = \frac{1}{49}\left(49746350 - \frac{49870^2}{50}\right)$
$s^2 = \frac{6012}{49} \approx 123 \text{ (to 3 s.f.)}$

1. State the reasoning based on the context of the problem.

Mr Dough suspects that the Regular-sized packets are underweight. This means he is only interested in whether the mean mass is strictly less than $1\text{ kg}$ (a specific direction). Therefore, he should conduct a one-tail test.

1. Determine the sample mean and unbiased estimate of the population variance (from part (a)).

$n = 50$
$\sum fx = 49870$
$\bar{x} = \frac{49870}{50} = 997.4$
$\sum fx^2 = 49746350$
$s^2 = \frac{1}{49} \left( 49746350 - \frac{49870^2}{50} \right) = \frac{6012}{49} \approx 122.69$

2. State the null and alternative hypotheses.

Let $\mu$ be the population mean mass of Regular-sized packets of flour, in grams.
$H_0: \mu = 1000$
$H_1: \mu < 1000$

3. Calculate the test statistic and $p$-value.

Under $H_0$, since the sample size $n = 50$ is large, by the Central Limit Theorem, the test statistic is $Z = \frac{\bar{X} - 1000}{S / \sqrt{50}} \sim N(0, 1)$ approximately.

Using a graphing calculator with $\mu_0 = 1000$, $\bar{x} = 997.4$, $s = \sqrt{\frac{6012}{49}}$ and $n = 50$:
$z = -1.660 \text{ (to 3 s.f.)}$
$p\text{-value} = 0.0485 \text{ (to 3 s.f.)}$

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

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 mean mass of Regular-sized packets is less than $1\text{ kg}$ (supporting Mr Dough's suspicion that they are underweight).

5. Explain whether the assumption of normality 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 large ($n \ge 30$), by the Central Limit Theorem, the distribution of the sample mean $\bar{X}$ is approximately normal regardless of the population distribution.

1. Determine the sample mean and the probability of the relevant tail for the test statistic.

Let $X$ be the mass of a Large-sized packet in grams.
Null hypothesis, $H_0: \mu = 2000$
Alternative hypothesis, $H_1: \mu \neq 2000$ (2-tailed test since the suspicion is 'not packing correctly')
Sample mean, $\bar{x} = \frac{30075}{15} = 2005$

Since it is a 2-tailed test with a p-value of $0.0529$, the probability in the upper tail is:
$P(\bar{X} \geq 2005) = \frac{0.0529}{2} = 0.02645$

2. Standardise the test statistic, use the inverse normal distribution to find the z-value, and solve for $\sigma$.

Standardising the sample mean:
$P\left(Z \geq \frac{2005 - 2000}{\sigma / \sqrt{15}}\right) = 0.02645$
$P\left(Z < \frac{5\sqrt{15}}{\sigma}\right) = 1 - 0.02645 = 0.97355$

From the standard normal distribution (using inverse normal):
$\frac{5\sqrt{15}}{\sigma} = 1.93567$
$\sigma = \frac{5\sqrt{15}}{1.93567} \approx 10.004$
$\sigma = 10.0 \text{ (correct to 3 s.f.)}$

1. Find the critical $z$-values for a two-tailed test at the 4% level of significance.

Let $X$ be the mass of a Large-sized packet in grams.
$H_0: \mu = 2000$
$H_1: \mu \neq 2000$

Level of significance $\alpha = 0.04$.
For a two-tailed test, the critical area in each tail is $0.02$. The critical $z$-values are $\pm \text{invNorm}(0.98) \approx \pm 2.0537$.

2. Express the test statistic $Z$ for the combined sample in terms of $h$.

With the new packet added, the combined sample size is $n = 16$.
The new sample mean is $\bar{X} = \frac{30075 + h}{16}$.
The test statistic is:
$Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} = \frac{\frac{30075 + h}{16} - 2000}{10 / \sqrt{16}}$
$Z = \frac{30075 + h - 32000}{16 \times 2.5} = \frac{h - 1925}{40}$

3. Formulate inequalities for the rejection region and solve for the range of $h$.

Since the claim is rejected at the 4% level of significance, the test statistic must fall in the critical region:
$Z < -2.0537 \quad \text{or} \quad Z > 2.0537$

For the lower tail:
$\frac{h - 1925}{40} < -2.0537 \implies h < 1925 - 82.148 = 1842.852$

For the upper tail:
$\frac{h - 1925}{40} > 2.0537 \implies h > 1925 + 82.148 = 2007.148$

Rounding the boundaries to the nearest gram, we obtain $h < 1843$ or $h > 2007$.