一家面包店声称每个蛋糕的平均质量为 600 克。对 10 个蛋糕进行了随机抽样检验，并确定了每个蛋糕的质量 $x$ 克。结果发现： $\sum(x-500)=800\quad\text{and}\quad\sum(x-500)^2=65764.$

1. Calculate the unbiased estimate of the population mean.

$\bar{x} = 500 + \frac{\sum(x-500)}{n}$
$\bar{x} = 500 + \frac{800}{10}$
$\bar{x} = 580$

2. Calculate the unbiased estimate of the population variance.

$s^2 = \frac{1}{n-1} \left( \sum(x-500)^2 - \frac{(\sum(x-500))^2}{n} \right)$
$s^2 = \frac{1}{9} \left( 65764 - \frac{800^2}{10} \right)$
$s^2 = \frac{1}{9} (65764 - 64000)$
$s^2 = \frac{1764}{9}$
$s^2 = 196$

1. State the hypotheses and the necessary assumption about the population.

Let $\mu$ be the population mean mass of the cakes.
$H_0: \mu = 600$
$H_1: \mu \neq 600$

Since the sample size $n = 10$ is small ($n < 30$), we must assume that the mass of the cakes follows a normal distribution.

2. Compute the test statistic and find the p-value.

From part (i), the unbiased estimates are $\bar{x} = 580$ and $s^2 = 196$ (so $s = 14$).

Under $H_0$, using the sample variance as an estimate for the population variance, the test statistic is:
$Z = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{580 - 600}{14/\sqrt{10}} \approx -4.5175$

Using a graphing calculator for a two-tailed test, the $p$-value is:
$p\text{-value} = 2 \times P(Z \le -4.5175) \approx 6.26 \times 10^{-6}$

3. Formulate the inequality for the significance level.

Let $\alpha$ be the decimal significance level of the test.
Since there is "significant evidence" that the mass differs from 600 grams, the null hypothesis $H_0$ is rejected.

For $H_0$ to be rejected, the significance level must be greater than or equal to the $p$-value:
$\alpha \ge 6.26 \times 10^{-6}$

1. Set up the test statistic and state the condition for insufficient evidence to reject the null hypothesis.

The hypotheses are:
$H_0: \mu = 600$
$H_1: \mu < 600$

For a 1-tailed test at the 5% significance level, the critical value is $z = -1.64485$.

"Insufficient evidence" means we do not reject $H_0$. Therefore, the test statistic must be greater than or equal to the critical value:
$Z_{calc} \ge -1.64485$

2. Substitute the sample parameters to find the boundary for $m$.

Using the sample standard deviation as an unbiased estimate (or population standard deviation proxy due to large $n$), the test statistic is $Z_{calc} = \frac{m - 600}{\frac{33}{\sqrt{100}}}$.

$\frac{m - 600}{3.3} \ge -1.64485$
$m - 600 \ge 3.3(-1.64485)$
$m - 600 \ge -5.4280$
$m \ge 594.572$

3. State the least value of $m$.

The least value of $m$ is $594.57$ (to 2 d.p.).

4. Determine if any assumptions about the population distribution are necessary, providing a valid reason.

It is not necessary to assume that the masses of cakes are normally distributed.
Reason: Since the sample size $n = 100$ is large ($n \ge 30$), by the Central Limit Theorem, the distribution of the sample mean $\bar{X}$ is approximately normal.