某公司生产用于建筑的钢筋。生产经理声称钢筋的平均抗拉强度至少为 620 MPa。公司董事希望调查这一声称，并随机抽取了 80 根钢筋作为样本。该随机样本的抗拉强度 $x$（单位：MPa）汇总如下： $\sum (x-620) = -300,\qquad \sum (x-620)^2 = 36680.$ 质量控制经理希望检验平均抗拉强度是否不同于 620 MPa。质量控制经理发现，随机抽取的 10 根钢筋的平均抗拉强度为 628.9 MPa。他在 5% 的显著性水平下进行假设检验。

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

$\hat{\mu} = \bar{x} = 620 + \frac{\sum (x-620)}{n}$
$\hat{\mu} = 620 + \frac{-300}{80}$
$\hat{\mu} = 616.25$

2. Calculate the unbiased estimate of the population variance using the given sums of squares.

$s^2 = \frac{1}{n-1} \left( \sum (x-620)^2 - \frac{(\sum (x-620))^2}{n} \right)$
$s^2 = \frac{1}{80-1} \left( 36680 - \frac{(-300)^2}{80} \right)$
$s^2 = \frac{1}{79} \left( 36680 - 1125 \right)$
$s^2 = \frac{35555}{79} \approx 450 \text{ (to 3 s.f.)}$

1. Define the population parameter and state the null and alternative hypotheses.

Let $\mu$ be the population mean tensile strength of the reinforcing bars, in MPa.

$H_0: \mu = 620$
$H_1: \mu < 620$

2. State the sample statistics derived from part (i) and the distribution of the sample mean.

From part (i),
The unbiased estimate of the population mean is $\bar{x} = 620 + \frac{-300}{80} = 616.25$.
The unbiased estimate of the population variance is $s^2 = \frac{1}{79} \left( 36680 - \frac{(-300)^2}{80} \right) = \frac{35555}{79} \approx 450.063$.

Since $n = 80$ is large ($n \ge 30$), by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed. Under $H_0$, $\bar{X} \sim N\left(620, \frac{s^2}{80}\right)$ approximately.

3. Calculate the value of the test statistic.

The test statistic is $Z = \frac{\bar{X} - \mu_0}{s/\sqrt{n}}$.

$z = \frac{616.25 - 620}{\sqrt{\frac{35555/79}{80}}} \approx -1.5810$

4. Determine the p-value and compare it to the level of significance.

For a lower-tail test,
$p\text{-value} = P(Z < -1.5810) = 0.0569$

Level of significance, $\alpha = 0.05$.
Since $p\text{-value} = 0.0569 > 0.05$, we do not reject $H_0$.

5. State the conclusion in the context of the problem.

There is insufficient evidence at the 5% level of significance to conclude that the mean tensile strength is less than 620 MPa. The production manager's claim that the mean tensile strength is at least 620 MPa is not rejected.

1. State the meaning of the 5% level of significance in the context of the hypothesis test

The 5% level of significance means that there is a probability of $0.05$ (or $5\%$) of incorrectly concluding that the mean tensile strength of the reinforcing bars is less than $620\text{ MPa}$ (rejecting $H_0$), when the true mean tensile strength is in fact $620\text{ MPa}$ ($H_0$ is true).

1. State the assumption required for a small sample hypothesis test to be valid.

The quality control manager uses a sample size of $n = 10$. Since $n$ is small ($n < 30$), the Central Limit Theorem cannot be applied to approximate the distribution of the sample mean.

Therefore, we must assume that the underlying population (the tensile strength of the reinforcing bars) follows a normal distribution for the test to be valid.

1. Express the test statistic $z$ in terms of the unknown population variance $\sigma^2$.

Let the population variance be $\sigma^2$.
The test statistic for a sample from a normal population with known variance is given by:
$z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{628.9 - 620}{\frac{\sigma}{\sqrt{10}}} = \frac{8.9}{\frac{\sigma}{\sqrt{10}}}$

2. State the critical value for a two-tailed test at the 5% level of significance.

Since the alternative hypothesis is $\mu \neq 620$ ("differs from"), this is a 2-tail test.
At the 5% level of significance, the null hypothesis $H_0$ is rejected if $z \le -1.95996$ or $z \ge 1.95996$.

3. Formulate and solve the inequality to find the possible values of $\sigma^2$.

Since $\bar{x} = 628.9 > 620$, we have $z > 0$, so the rejection condition is $z \ge 1.95996$.
$\frac{8.9}{\frac{\sigma}{\sqrt{10}}} \ge 1.95996$
$\frac{\sigma}{\sqrt{10}} \le \frac{8.9}{1.95996}$
$\frac{\sigma^2}{10} \le \left( \frac{8.9}{1.95996} \right)^2$
$\frac{\sigma^2}{10} \le 20.6198$
$\sigma^2 \le 206.198$
Since a population variance must be strictly positive, the set of possible values is:
$0 < \sigma^2 < 206 \text{ (to 3 s.f.)}$