In a box of cheese rings from a certain manufacturer, the mass of sodium, $X$ mg, is a continuous random variable with mean $\mu_0$. Following a change in production chain, 15 randomly chosen boxes of cheese rings were analysed. The masses of sodium (in mg) in the boxes are summarised by $\displaystyle \sum (x-1180)=17.5\,,\qquad \displaystyle \sum (x-1180)^2=190.5$.

1. Calculate the sample mean $\bar{x}$ and the unbiased estimate of the population variance $s^2$.

Let $y = x - 1180$. Given $n = 15$, $\sum y = 17.5$ and $\sum y^2 = 190.5$.

Sample mean,
$\bar{x} = 1180 + \frac{\sum y}{n} = 1180 + \frac{17.5}{15} = \frac{7087}{6} \approx 1181.17$ (to 2 d.p.)

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

2. State the null and alternative hypotheses.

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

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

Level of significance: $5\%$

3. State the necessary assumption for the hypothesis test.

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

4. Identify the test statistic and calculate the p-value using a t-test.

Since the population variance is unknown and $n$ is small, we use a $t$-test.
Under $H_0$, the test statistic is $T = \frac{\bar{X} - 1183}{S/\sqrt{15}} \sim t(14)$.

Using a graphing calculator with $\mu_0 = 1183$, $\bar{x} = 1181.1667$, $s_x = \sqrt{\frac{2041}{168}} \approx 3.4855$, and $n = 15$:

$t = -2.037$
$p\text{-value} = 0.0611$ (to 3 s.f.)

5. State 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 from $1183\text{ mg}$.

1. Calculate the sample mean $\bar{x}$ from the given summary data.

$\bar{x} = \frac{\sum (x-1180)}{15} + 1180 = \frac{17.5}{15} + 1180 = 1181\frac{1}{6} \approx 1181.167$

2. Set up the inequality for rejecting $H_0$ in a 1-tailed test.

For the test to conclude that the mass has increased ($H_1: \mu > \mu_0$) at the 5% significance level, the test statistic $Z$ must fall in the critical region $Z > 1.64485$.

$Z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} > 1.64485$

3. Substitute the values and solve for $\mu_0$.

$\frac{1181.1667 - \mu_0}{\frac{6.8}{\sqrt{15}}} > 1.64485$
$\frac{1181.1667 - \mu_0}{1.75573} > 1.64485$
$1181.1667 - \mu_0 > 2.8879$
$\mu_0 < 1178.2788$
$\mu_0 < 1178.3 \text{ (to 1 d.p.)}$

4. Explain whether the normality assumption from part (i) is still necessary.

Yes, the assumption is still necessary. Since the sample size $n = 15$ is small (less than 30), the Central Limit Theorem cannot be used to approximate the distribution of the sample mean $\bar{X}$. We must assume the population is normally distributed for the sample mean to follow a normal distribution.