A company produces reinforcing bars for use in construction. The production manager claims that the mean tensile strength of the reinforcing bars is at least 620 MPa. The company director wishes to investigate this claim and selects a random sample of 80 reinforcing bars. The tensile strengths, $x$, in MPa, of the random sample are summarised by $\sum (x-620) = -300,\qquad \sum (x-620)^2 = 36680.$ The quality control manager wishes to test whether the mean tensile strength differs from 620 MPa. The quality control manager finds that the mean tensile strength of 10 randomly chosen reinforcing bars is 628.9 MPa. He carries out a hypothesis test at the 5% level of significance.

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:
$0 < \sigma^2 < 206 \text{ (to 3 s.f.)}$