A farmer claims that the mean weight of the melons grown in his farm is at least 1.2 kg. A random sample of 10 melons is chosen and the weight, $x$ kg, of each melon is recorded. The results are as follows:

1.037	0.914	1.019	1.234	1.110
1.417	1.008	0.846	1.105	1.331

Assuming the weight of a randomly chosen melon follows a normal distribution, carry out an appropriate test of the farmer's claim at the 10% significance level. A buyer concludes that the sample could not provide sufficient evidence, at $\alpha\%$ significance level, that the mean weight differs from 1.2 kg. State the range of values of $\alpha$ for him to draw such a conclusion. Another random sample of 40 melons is weighed and the results are as follows: $\sum x = 45.738$ and $\sum x^{2} = 72.576$

1. Calculate $\sum x$ and $\sum x^2$ for the first sample

For the first sample of $n_1 = 10$ melons:
$\sum_{1} x = 1.037 + 0.914 + \dots + 1.331 = 11.021$
$\sum_{1} x^2 = 1.037^2 + 0.914^2 + \dots + 1.331^2 = 12.436237$

2. Find the combined sums for the overall sample

For the combined sample ($n = 10 + 40 = 50$):
$\sum x = 11.021 + 45.738 = 56.759$
$\sum x^2 = 12.436237 + 72.576 = 85.012237$

3. Calculate the unbiased estimates of the population mean and variance

Unbiased estimate of population mean, $\bar{x}$:
$\bar{x} = \frac{\sum x}{n} = \frac{56.759}{50} = 1.13518 \approx 1.14 \text{ (3 s.f.)}$

Unbiased estimate of population variance, $s^2$:
$s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right)$
$s^2 = \frac{1}{49} \left( 85.012237 - \frac{(56.759)^2}{50} \right)$
$s^2 = \frac{1}{49} \left( 85.012237 - 64.43168162 \right)$
$s^2 = \frac{20.58055538}{49} = 0.420011\dots \approx 0.420 \text{ (3 s.f.)}$

1. Calculate the sample mean and unbiased variance of the combined sample (required from part (i)).

For the first sample ($n_1=10$):
$\sum x = 1.037 + 0.914 + \dots + 1.331 = 11.021$
$\sum x^2 = 1.037^2 + 0.914^2 + \dots + 1.331^2 = 12.436237$

For the second sample ($n_2=40$):
$\sum x = 45.738$, $\sum x^2 = 72.576$

Combined sample ($n=50$):
$\sum x = 11.021 + 45.738 = 56.759$
$\sum x^2 = 12.436237 + 72.576 = 85.012237$

Unbiased estimate of population mean:
$\bar{x} = \frac{56.759}{50} = 1.13518$

Unbiased estimate of population variance:
$s^2 = \frac{1}{49}\left(85.012237 - \frac{56.759^2}{50}\right) \approx 0.420011$

2. Set up the hypotheses and state the non-rejection condition for the test.

Let $\mu$ be the population mean weight of the melons.
The farmer claims the mean is at least $m$ kg, so we test to see if there is significant evidence that it is less.
$H_0: \mu = m$
$H_1: \mu < m$

Since $n = 50$ is large, by the Central Limit Theorem, the test statistic is:
$Z = \frac{\bar{X} - m}{s / \sqrt{50}} \sim N(0, 1)$ approximately.

At the $5\%$ significance level for a 1-tailed test, the critical value is $z = -1.64485$.
For the farmer's claim not to be rejected, the test statistic must fall outside the critical region (i.e. to the right of the critical value):
$Z \ge -1.64485$

3. Solve the inequality to find the set of values of m.

Substitute the calculated values into the inequality:
$\frac{1.13518 - m}{\sqrt{0.420011 / 50}} \ge -1.64485$

$\frac{1.13518 - m}{0.091653} \ge -1.64485$

$1.13518 - m \ge -1.64485 \times 0.091653$
$1.13518 - m \ge -0.150755$
$m \le 1.13518 + 0.150755$
$m \le 1.2859$

Therefore, to 3 significant figures, $m \le 1.29$.