在某制造商生产的一盒芝士圈中，钠的质量 $X$ 毫克是一个均值为 $\mu_0$ 的连续随机变量。随着生产线的改变，分析了随机抽取的15盒芝士圈。盒子中钠的质量（单位：毫克）汇总如下： $\sum (x-1180)=17.5\,,\qquad \sum (x-1180)^2 =190.5\,.$

1. Calculate the sample mean and unbiased estimate of the population variance.

Let $X$ be the mass of sodium in a box of cheese rings.

Sample mean, $\bar{x} = 1180 + \frac{17.5}{15} = 1181\frac{1}{6} \approx 1181.17$ mg

Unbiased estimate of population variance,
$s^2 = \frac{1}{n-1}\left[ \sum (x-a)^2 - \frac{(\sum (x-a))^2}{n} \right]$
$s^2 = \frac{1}{14}\left( 190.5 - \frac{17.5^2}{15} \right) = \frac{2041}{168} \approx 12.1488$

2. State the hypotheses and the necessary assumption.

Let $\mu$ be the population mean mass of sodium in a box of cheese rings.

$H_0: \mu = 1183$
$H_1: \mu \neq 1183$

Level of significance: $5\%$

Assumption:
Since the sample size ($n=15$) is small, we must assume that the mass of sodium in a box of cheese rings is normally distributed.

3. Identify the test statistic and calculate the p-value.

Since the population variance is unknown and the sample size is small, a $t$-test is appropriate.

Under $H_0$, the test statistic is $T = \frac{\bar{X} - 1183}{S/\sqrt{15}} \sim t(14)$.

Using a graphing calculator:
$p\text{-value} = 0.06105 \approx 0.0611$ (to 3 s.f.)
(Note: The test statistic is $t \approx -2.037$)

4. Draw the conclusion based on the p-value.

Since the $p\text{-value} = 0.0611 > 0.05$, we do not reject $H_0$.

There is insufficient evidence at the $5\%$ level of significance to conclude that the mean mass of sodium in a box of cheese rings has changed.

1. State the hypotheses, identify the sample mean, and establish the test statistic.

Let $\mu$ be the mean mass of sodium in a box of cheese rings.
We are testing:
$H_0: \mu = \mu_0$
$H_1: \mu > \mu_0$ (since the test concludes the mass has increased)

Given the population standard deviation $\sigma = 6.8$ is known, the test statistic is:
$Z = \frac{\bar{X} - \mu_0}{\sigma / \sqrt{n}} \sim \text{N}(0, 1)$

From the data, the sample mean is:
$\bar{x} = 1180 + \frac{17.5}{15} = 1181 \frac{1}{6}$

2. Set up the inequality representing the rejection region for $H_0$.

For the test to conclude that the mass of sodium has increased at the $5\%$ significance level, $H_0$ must be rejected in favour of $H_1$.
The critical value for a $5\%$ upper-tailed test is $z = 1.64485$.
Thus, we require the test statistic to be strictly greater than the critical value:
$\frac{1181 \frac{1}{6} - \mu_0}{6.8 / \sqrt{15}} > 1.64485$

3. Solve the inequality to find the range of values for $\mu_0$.

$1181.1667 - \mu_0 > 1.64485 \times \frac{6.8}{\sqrt{15}}$
$1181.1667 - \mu_0 > 2.8879$
$\mu_0 < 1178.2788$
$\mu_0 < 1178.3 \text{ (to } 1 \text{ d.p.)}$

4. Explain whether the assumption of a normal population distribution is still needed.

Yes, the assumption is still necessary. Even though the population standard deviation is known, the sample size $n = 15$ is small ($n < 30$). The Central Limit Theorem cannot be applied to approximate the distribution of the sample mean $\bar{X}$. Thus, we must assume that the population follows a normal distribution so that $\bar{X}$ is exactly normally distributed and the test is valid.