某巧克力制造公司生产巧克力棒。巧克力棒的包装上注明每根巧克力棒的质量为 200 克。 生产经理收到投诉称巧克力棒的质量被夸大了，他想要进行一次假设检验。

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.