A chocolate manufacturing company produces chocolate bars. The packaging of the chocolate bar states that the mass of each bar is 200 g. The production manager has been receiving complaints that the mass is overstated and he wants to carry out a hypothesis test.

1. Determine and explain the choice of a 1-tail or 2-tail test.

The manager is specifically testing if the mass is overstated, which means checking if the true mean mass is less than $200\text{ g}$. Since the test is looking for a decrease in one specific direction, a 1-tail test should be carried out.

2. State the hypotheses and define the symbol used.

Let $\mu$ be the population mean mass of the chocolate bars in grams.

$H_0: \mu = 200$
$H_1: \mu < 200$

1. Calculate the unbiased estimate of the population mean.

$\bar{x} = a + \frac{\sum(x-a)}{n}$
$\bar{x} = 200 + \frac{-54}{30} = 200 - 1.8 = 198.2$

2. Calculate the unbiased estimate of the population variance.

$s^2 = \frac{1}{n-1} \left( \sum(x-a)^2 - \frac{(\sum(x-a))^2}{n} \right)$
$s^2 = \frac{1}{30-1} \left( 550 - \frac{(-54)^2}{30} \right)$
$s^2 = \frac{1}{29} \left( 550 - \frac{2916}{30} \right)$
$s^2 = \frac{1}{29} ( 550 - 97.2 )$
$s^2 = \frac{452.8}{29} \approx 15.614 \approx 15.6 \text{ (to 3 s.f.)}$

1. Determine the test statistic and $p$-value using the sample statistics.

From parts (i) and (ii), we are testing $H_0: \mu = 200$ against $H_1: \mu < 200$, with $\bar{x} = 198.2$ and $s^2 = \frac{2264}{145} \approx 15.614$.

Since $n = 30$ is large, we use a $Z$-test. Under $H_0$, the test statistic is:
$z = \frac{\bar{x} - \mu_0}{\sqrt{s^2 / n}} = \frac{198.2 - 200}{\sqrt{\frac{15.6138}{30}}} \approx -2.495$

Using a graphing calculator,
$p\text{-value} = 0.00630 \text{ (to 3 s.f.)}$

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

Level of significance, $\alpha = 0.01$.

Since $p\text{-value} = 0.00630 < 0.01$, we reject $H_0$.

There is sufficient evidence at the 1% level of significance to conclude that the mean mass of the chocolate bars is less than $200\text{ g}$ (i.e. the manager's complaints that the mass is overstated are justified).

1. State the hypotheses and find the critical value for the test.

Let $\mu$ be the population mean mass in grams.
$H_0: \mu = 200$
$H_1: \mu \neq 200$
Since the test is to determine if the mean is indeed 200 g, we use a 2-tail test at the 5% level of significance ($\alpha = 0.05$).
The critical $z$-values are $\pm 1.95996$.

2. Set up the inequality for the non-rejection of the null hypothesis.

Under $H_0$, the test statistic is $Z = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}}$.
For $H_0$ to not be rejected, the test statistic must fall inside the non-rejection region (between the critical values). Given $\bar{x} = 199.2 < 200$, we consider the lower tail limit:
$Z_{\text{stat}} \ge -1.95996$
$\frac{199.2 - 200}{2 / \sqrt{n}} \ge -1.95996$

3. Solve the inequality for $n$.

$\frac{-0.8}{2/\sqrt{n}} \ge -1.95996$
$-0.4\sqrt{n} \ge -1.95996$
$\sqrt{n} \le \frac{-1.95996}{-0.4}$
$\sqrt{n} \le 4.8999$
$n \le 24.009$

4. Determine the maximum integer value for $n$.

Since the sample size $n$ must be an integer, the maximum possible value is 24.