S大学的研究人员声称，通常存在于土壤和植物中的一种真菌菌株——真菌A，能够分解难以回收的塑料聚丙烯。对于经过紫外光或热预处理的聚丙烯，在90天内，每100克聚丙烯中至少能被分解27克。 M大学的研究人员认为S大学的研究人员夸大了他们的结论，并决定对一个由120批聚丙烯（每批质量为100克）组成的随机样本进行假设检验。研究人员记录了在90天内，每一批中真菌A能够分解的聚丙烯质量 $x$ 克。结果汇总如下：
$\sum (x-27) = -81, \sum (x-27)^2 = 2070.8$

1. Calculate the unbiased estimate of the population mean using the shifted sum.

$\bar{x} = a + \frac{\sum (x-a)}{n}$
$\bar{x} = 27 + \frac{-81}{120}$
$\bar{x} = 27 - 0.675 = 26.325$

2. Calculate the unbiased estimate of the population variance using the shifted sums.

$s^2 = \frac{1}{n-1} \left( \sum (x-a)^2 - \frac{(\sum (x-a))^2}{n} \right)$
$s^2 = \frac{1}{120-1} \left( 2070.8 - \frac{(-81)^2}{120} \right)$
$s^2 = \frac{1}{119} \left( 2070.8 - \frac{6561}{120} \right)$
$s^2 = \frac{1}{119} \left( 2070.8 - 54.675 \right)$
$s^2 = \frac{2016.125}{119} \approx 16.942$

1. Define the population mean $\mu$ and formulate the null and alternative hypotheses.

Let $\mu$ be the population mean mass of polypropylene (in grams) per 100 grams that Fungus A was able to break down over a 90-day period.

Since University M believes the claim of 'at least 27 grams' is overstated, we test for a mean of less than 27 grams (a 1-tail test).
$H_0: \mu = 27$
$H_1: \mu < 27$

2. State the test statistic and its distribution under the null hypothesis.

Level of significance: $\alpha = 0.05$.
Under $H_0$, since the sample size $n = 120$ is large ($n \ge 30$), by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed. The test statistic is:
$Z = \frac{\bar{X} - 27}{S/\sqrt{120}} \sim \text{N}(0, 1) \text{ approximately.}$

3. Calculate the sample mean, unbiased estimate of variance, and the $p$-value.

Using the summary statistics from part (a):
Sample mean, $\bar{x} = 27 + \frac{-81}{120} = 26.325$
Unbiased estimate of population variance, $s^2 = \frac{1}{119}\left(2070.8 - \frac{(-81)^2}{120}\right) = \frac{2016.125}{119} \approx 16.942$

Using a graphing calculator for a 1-tail $Z$-test:
$z \approx -1.796$
$p\text{-value} \approx 0.0362 \text{ (to 3 s.f.)}$

4. State the conclusion in the context of the question.

Since the $p$-value $= 0.0362 < 0.05$, we reject $H_0$.

There is sufficient evidence at the $5\%$ level of significance to conclude that the population mean mass of polypropylene broken down is less than 27 grams. Hence, the researchers at University S have overstated their claim.

1. State the definition of the significance level in the context of the hypothesis test

The 5% level of significance is the probability of rejecting the null hypothesis when it is actually true.
In this context, it means there is a probability of 0.05 (or 5%) of incorrectly concluding that the population mean mass of polypropylene broken down by Fungus A per 100 grams is less than 27 grams ($H_1: \mu < 27$), when in fact the population mean is 27 grams ($H_0: \mu = 27$).

1. State whether the assumption is necessary and justify using the sample size and Central Limit Theorem

Since the sample size $n = 120$ is large ($n \ge 30$), by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed regardless of the population distribution.
Therefore, it is not necessary to assume that the population follows a normal distribution.

1. State the hypotheses and the test statistic formula.

$H_0: \mu = 27$
$H_1: \mu < 27$
Under $H_0$, since the population variance is known, the test statistic is $Z = \frac{\bar{Y} - \mu}{\sigma/\sqrt{n}} \sim \text{N}(0,1)$

2. Set up the condition for not rejecting the null hypothesis.

For $H_0$ to not be rejected at the 5% level of significance in a lower-tailed test, the test statistic must be greater than or equal to the critical value:
$Z \ge -1.64485$

3. Substitute the given values and solve for $\bar{y}$.

$\frac{\bar{y} - 27}{4/\sqrt{50}} \ge -1.64485$
$\bar{y} \ge 27 - 1.64485 \left(\frac{4}{\sqrt{50}}\right)$
$\bar{y} \ge 26.0695$
$\bar{y} \ge 26.1 \text{ (to 3 s.f.)}$