某款特定型号汽车的油耗里程 $X$（单位：公里/升汽油）是一个随机变量，其均值为 $\mu$ 公里/升。一家石油公司对他们的汽车汽油进行了改良，旨在提高汽车的油耗里程。该公司使用该特定型号的 13 辆汽车对改良后的汽油进行了初步测试。测试结果汇总如下：
$\sum(x-12) = 6.09, \quad \sum(x-12)^2 = 20.853.$

1. Find the unbiased estimate of the population mean

$\begin{aligned} \bar{x} &= 12 + \frac{\sum(x - 12)}{n} \\ &= 12 + \frac{6.09}{13} \\ &= \frac{16209}{1300} \\ &\approx 12.468 \\ &\approx 12.5 \text{ (to 3 s.f.)} \end{aligned}$

2. Find the unbiased estimate of the population variance

$\begin{aligned} s^2 &= \frac{1}{n-1}\left[\sum(x-12)^2 - \frac{\left(\sum(x-12)\right)^2}{n}\right] \\ &= \frac{1}{13-1}\left[20.853 - \frac{(6.09)^2}{13}\right] \\ &= \frac{1}{12}\left(20.853 - 2.85293\dots\right) \\ &= \frac{18.000069\dots}{12} \\ &= 1.500005\dots \\ &\approx 1.50 \text{ (to 3 s.f.)} \end{aligned}$

1. Calculate the unbiased estimates of the population mean and variance from the given sample data (from part i).

Given $n = 13$, $\sum(x-12) = 6.09$, and $\sum(x-12)^2 = 20.853$.
The unbiased estimate of the population mean, $\bar{x}$, is:
$\bar{x} = 12 + \frac{\sum(x-12)}{n} = 12 + \frac{6.09}{13} \approx 12.4685$
The unbiased estimate of the population variance, $s^2$, is:
$s^2 = \frac{1}{n-1} \left( \sum(x-12)^2 - \frac{(\sum(x-12))^2}{n} \right)$
$s^2 = \frac{1}{12} \left( 20.853 - \frac{6.09^2}{13} \right) = \frac{1}{12} (20.853 - 2.8529) \approx 1.50$

2. State the hypotheses for the test.

Let $\mu$ be the population mean mileage.
The test is whether the modified petrol improves mileage (i.e., its mean is greater than $\mu_0$).
Null hypothesis, $H_0: \mu = \mu_0$
Alternative hypothesis, $H_1: \mu > \mu_0$

3. Formulate the test statistic and the condition for insignificant evidence.

Having "insignificant evidence" means we do not reject $H_0$ at the 5% level of significance.
Using a 1-tailed Z-test, the test statistic is:
$Z = \frac{\bar{X} - \mu_0}{s/\sqrt{n}}$
The critical value for a 5% upper-tail test is $z = 1.6449$.
For $H_0$ not to be rejected, our calculated $Z$ must not fall in the critical region. Thus, we require $Z \le 1.6449$.

4. Solve the inequality to find the range of values for $\mu_0$.

Substitute the calculated values into the inequality:
$\frac{12.4685 - \mu_0}{\sqrt{\frac{1.50}{13}}} \le 1.6449$
$12.4685 - \mu_0 \le 1.6449 \times 0.33968$
$12.4685 - \mu_0 \le 0.5587$
$\mu_0 \ge 11.9098$
Rounding to 3 significant figures, the set of values is $\mu_0 \ge 11.9$.

1. State the null and alternative hypotheses.

Let $\mu$ be the population mean mileage.
$H_0: \mu = 12$
$H_1: \mu > 12$ (1-tailed test for improved mileage)

2. Calculate the test statistic and $p$-value using the estimates from part (i).

From part (i), the sample mean is $\bar{x} = 12 + \frac{6.09}{13} \approx 12.468$ and unbiased variance estimate is $s^2 = 1.5$.

Using a $Z$-test (since the population variance is estimated by $s^2$), the test statistic is:
$z = \frac{\bar{x} - \mu}{s / \sqrt{n}} = \frac{12.468 - 12}{\sqrt{1.5 / 13}} = \frac{6.09 / 13}{\sqrt{1.5 / 13}} \approx 1.3791$

The $p$-value is:
$p\text{-value} = P(Z > 1.3791) \approx 0.0839$

3. Identify the least significance level to reject $H_0$.

For the test to show significant evidence that the modified petrol improves mileage, $H_0$ must be rejected.
This requires the significance level $\alpha$ to be greater than or equal to the $p$-value.
Least significance level $= 0.0839$ (or $8.39\%$).

1. State the necessary assumption for the hypothesis tests to be valid given the small sample size ($n=13$).

Since the sample size $n = 13$ is small ($n < 30$), the Central Limit Theorem cannot be applied to approximate the distribution of the sample mean as normal. Therefore, for the hypothesis tests to be valid, we must assume that the underlying population is normally distributed.