A manufacturer claims that the mean lifetime of the light bulbs he produces is at least 1200 hours. A random sample of 120 bulbs is taken and the lifetime per bulb, x hours, is measured. The results are summarised by $\sum (x-1200) = -60$ and $\sum (x-1200)^2 = 2014$.

1. Calculate the unbiased estimate of the population mean.

$\bar{x} = 1200 + \frac{\sum (x-1200)}{n}$
$\bar{x} = 1200 + \frac{-60}{120} = 1199.5$

2. Calculate the unbiased estimate of the population variance.

$s^2 = \frac{1}{n-1} \left( \sum (x-1200)^2 - \frac{(\sum(x-1200))^2}{n} \right)$
$s^2 = \frac{1}{119} \left( 2014 - \frac{(-60)^2}{120} \right)$
$s^2 = \frac{1}{119} (2014 - 30) = \frac{1984}{119}$
$s^2 \approx 16.7 \text{ (to 3 s.f.)}$

1. State the null and alternative hypotheses.

Let $\mu$ be the population mean lifetime of the light bulbs, in hours.
The manufacturer claims that $\mu \ge 1200$.
$H_0: \mu = 1200$
$H_1: \mu < 1200$

2. Recall the unbiased estimates from part (i) and state the test statistic.

From part (i), the unbiased estimates are:
$\bar{x} = 1200 + \frac{-60}{120} = 1199.5$
$s^2 = \frac{1}{119} \left( 2014 - \frac{(-60)^2}{120} \right) = \frac{1984}{119} \approx 16.672$

Since the sample size $n = 120$ is large, by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed. The test statistic is:
$Z = \frac{\bar{X} - 1200}{S/\sqrt{120}} \sim N(0, 1) \text{ approximately}$

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

Using a graphing calculator with $\mu_0 = 1200$, $\bar{x} = 1199.5$, $s = \sqrt{\frac{1984}{119}}$, and $n = 120$:
$z = \frac{1199.5 - 1200}{\sqrt{\frac{1984/119}{120}}} \approx -1.3414$
$p\text{-value} = P(Z < -1.3414) \approx 0.089886 \approx 0.0899 \text{ (to 3 s.f.)}$

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

Since the $p$-value $= 0.0899 < 0.10$, we reject $H_0$.
There is sufficient evidence at the $10\%$ level of significance to conclude that the mean lifetime of the light bulbs is less than $1200\text{ hours}$. Therefore, the manufacturer's claim is not valid.

1. State the necessary assumption and the hypotheses for the test.

Necessary assumption: The lifetime of the light bulbs produced by the new process follows a normal distribution.

Let $\mu$ be the population mean lifetime of the bulbs.
The manufacturer's claim is that $\mu \ge 1200$.
$H_0: \mu = 1200$
$H_1: \mu < 1200$

2. Calculate the unbiased estimate of the population variance.

Given the standard deviation of the sample is $9.8$, the sample variance is $9.8^2$.
The unbiased estimate of the population variance is:
$s^2 = \frac{n}{n-1} \times (\text{sample variance}) = \frac{20}{19} \times 9.8^2 = \frac{1920.8}{19} \approx 101.0947$

3. Set up the test statistic and state the condition for not rejecting the null hypothesis.

Under $H_0$, the test statistic is:
$Z = \frac{\bar{X} - \mu}{s / \sqrt{n}} = \frac{k - 1200}{\sqrt{101.0947 / 20}} \sim N(0, 1) \text{ approximately}$
Since the test indicates the manufacturer's claim is valid, we do not reject $H_0$ at the 5% level of significance.
The critical region for this 1-tailed test is $Z \le -1.64485$.
For $H_0$ not to be rejected, the test statistic must lie outside the critical region:
$Z \ge -1.64485$

4. Solve the inequality to find the least possible value of $k$.

$\frac{k - 1200}{\sqrt{101.0947 / 20}} \ge -1.64485$
$k - 1200 \ge -1.64485 \times 2.24827$
$k - 1200 \ge -3.69805$
$k \ge 1196.30195$
Therefore, the least possible value of $k$ is $1196.302$ (correct to 3 decimal places).