一位研究人员正在对一种新栽培品种的土豆进行试验。一块田地里有 20 行新栽培品种的土豆植株，每行有 80 株。研究人员打算挖出 8 株并测量每株植株产出的土豆质量。

1. Calculate the sampling interval and specify the random starting point.

Number the plants in the single row from $1$ to $80$.
The sampling interval $k = \frac{N}{n} = \frac{80}{8} = 10$.
Randomly select a starting integer, $r$, between $1$ and $10$ inclusive (using a random number generator).

2. Describe how the remaining plants in the sample are selected.

Select the $r$-th plant, and then pick every $10$-th plant thereafter. The plants chosen will be those numbered $r, r+10, r+20, \dots, r+70$.

3. State an advantage of systematic sampling in this context.

An advantage is that the selected plants are evenly spread out across the entire row, which ensures a more representative sample if there are gradual trends in environmental conditions (such as soil fertility or moisture) along the row. Furthermore, it is easier and quicker to conduct physically in the field than simple random sampling.

1. Calculate the unbiased estimate of the population mean using the coded data summary.

Let the unbiased estimate of the population mean be $\bar{x}$.
$\bar{x} = a + \frac{\sum(x-a)}{n}$
Substituting $a = 150$ and $n = 80$:
$\bar{x} = 150 + \frac{-160}{80}$
$\bar{x} = 150 - 2 = 148$

2. Calculate the unbiased estimate of the population variance using the coded data summary.

Let the unbiased estimate of the population variance be $s^2$.
$s^2 = \frac{1}{n-1} \left( \sum(x-a)^2 - \frac{(\sum(x-a))^2}{n} \right)$
$s^2 = \frac{1}{80-1} \left( 5520 - \frac{(-160)^2}{80} \right)$
$s^2 = \frac{1}{79} \left( 5520 - \frac{25600}{80} \right)$
$s^2 = \frac{1}{79} (5520 - 320)$
$s^2 = \frac{5200}{79} \approx 65.823$
$s^2 \approx 65.8 \text{ (to 3 s.f.)}$

1. State the population parameter and hypotheses

Let $\mu$ be the population mean mass of the new variety of potato, in grams.
$H_0: \mu = 150$
$H_1: \mu < 150$

2. Determine the unbiased estimates of the population mean and variance (from part ii)

$\bar{x} = 150 + \frac{-160}{80} = 148$
$s^2 = \frac{1}{80-1} \left( 5520 - \frac{(-160)^2}{80} \right) = \frac{1}{79}(5520 - 320) = \frac{5200}{79} \approx 65.823$

3. Calculate the test statistic and $p$-value

Under $H_0$, since the sample size $n = 80$ is large ($n \ge 30$), by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed.
Test statistic, $z = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{148 - 150}{\sqrt{\frac{5200}{79 \times 80}}} \approx -2.205$
Using a GC (Z-Test):
$p\text{-value} = P(Z < -2.20489) \approx 0.0137$

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

Since the $p$-value $= 0.0137 > 0.01$, we do not reject $H_0$.
There is insufficient evidence at the 1% significance level to conclude that the population mean mass is less than 150g. Hence, the researcher's claim that the average mass is at least 150g is valid.

1. State the definition of the significance level in the context of the hypotheses.

The significance level is the probability of rejecting the null hypothesis when it is true.

In this context, it means there is a 1% probability of incorrectly concluding that the average mass of the new variety of potato is less than 150g (rejecting $H_0$), when the true average mass is actually 150g ($H_0$ is true).

1. State the hypotheses and identify the appropriate test distribution.

Let $\mu$ be the population mean mass of the new variety of potato.
$H_0: \mu = 150$
$H_1: \mu < 150$
Since the sample size $n=8$ is small and the population variance is unknown, we must assume that the mass of the potatoes is normally distributed to use a $t$-test.

2. Express the test statistic $T$ in terms of $k$.

Given the sample standard deviation is $k$ (i.e. $s_n = k$), the unbiased estimate of the population variance is:
$s^2 = \frac{n}{n-1}s_n^2 = \frac{8}{7}k^2$
The test statistic is:
$T = \frac{\bar{x} - \mu}{s / \sqrt{n}} = \frac{148.5 - 150}{\sqrt{\frac{8}{7}k^2} / \sqrt{8}} = \frac{-1.5}{k/\sqrt{7}} = -\frac{1.5\sqrt{7}}{k}$

3. State the critical region and solve the inequality to find the range of $k$.

For a 1-tailed test at the $1\%$ level of significance with $\nu = 8 - 1 = 7$ degrees of freedom, the critical value is $t_{7, 0.01} = -2.9979$.
For the sample to indicate that the claim is invalid, we must reject $H_0$, which requires the test statistic to fall in the critical region:
$T \le -2.9979$
$-\frac{1.5\sqrt{7}}{k} \le -2.9979$
Since $k$ is a standard deviation, $k > 0$. Multiplying by $k$ and dividing by $-2.9979$ (which reverses the inequality sign) gives:
$k \le \frac{1.5\sqrt{7}}{2.9979}$
$k \le 1.3238$
Hence, the range of values for $k$ is $0 < k \le 1.32$ (to 3 s.f.).

(Note: If $k$ is interpreted strictly as the unbiased estimate of the standard deviation rather than the sample standard deviation, $s=k$, leading to $T = \frac{-1.5}{k/\sqrt{8}} \le -2.9979 \implies k \le 1.42$.)