The mileage, $X$ km per litre of petrol, of a particular model of car is a random variable with mean $\mu$ km per litre. A petroleum company had modifications done to their car petrol with the intention of improving mileage in cars. The company conducted a pilot test of the modified petrol with 13 cars of that particular model. The results are summarised by
$\sum(x-12) = 6.09, \quad \sum(x-12)^2 = 20.853.$

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.