Under normal continuous use, the average battery life of a PI-99 calculator is claimed to be $k$ hours. A random sample of 13 calculators were obtained, and the battery life, $x$ hours, of each calculator was measured. The results are summarised by $\sum x = 573.39 \quad \text{and} \quad \sum (x - \bar{x})^2 = 42.22$

1. Calculate the unbiased estimate of the population mean

The unbiased estimate of the population mean is given by $\bar{x}$.
$\bar{x} = \frac{\sum x}{n}$
$\bar{x} = \frac{573.39}{13}$
$\bar{x} \approx 44.107 \approx 44.1 \text{ (to 3 s.f.)}$

2. Calculate the unbiased estimate of the population variance

The unbiased estimate of the population variance is given by $s^2$.
$s^2 = \frac{1}{n-1}\sum (x - \bar{x})^2$
$s^2 = \frac{42.22}{13 - 1}$
$s^2 = \frac{42.22}{12}$
$s^2 \approx 3.5183 \approx 3.52 \text{ (to 3 s.f.)}$

1. State the assumption required because of the small sample size

Since the sample size $n = 13$ is small ($n < 30$), the Central Limit Theorem cannot be applied to assume the distribution of the sample mean is approximately normal. Therefore, for the test to be valid, we must assume that the underlying population (battery life of the PI-99 calculators) follows a normal distribution.

1. Define the population parameter.

Let $\mu$ be the population mean battery life of the PI-99 calculator in hours.

2. State the null and alternative hypotheses.

Since the test is to determine if the claim of $\mu = k$ is valid, we use a two-tailed test.
$H_0: \mu = k$
$H_1: \mu \neq k$

1. Recall the sample mean and the unbiased estimate of the population variance from part (i).

From part (i),
$\bar{x} = \frac{573.39}{13} \approx 44.1069$

$s^2 = \frac{1}{13-1} \sum (x - \bar{x})^2 = \frac{42.22}{12} \approx 3.51833$

2. Set up the non-rejection region for the test statistic at the 5% level of significance.

For a two-tailed test at the 5% significance level, the null hypothesis $H_0: \mu = k$ is not rejected if the test statistic $Z$ lies within the non-critical region:
$-1.95996 \le Z \le 1.95996$

Assuming the sample variance $s^2$ is an appropriate estimate for the population variance $\sigma^2$, the test statistic is $Z = \frac{\bar{x} - k}{s / \sqrt{n}}$.

3. Solve the inequality to find the set of values for $k$.

$-1.95996 \le \frac{44.1069 - k}{\sqrt{3.51833 / 13}} \le 1.95996$
$-1.95996 \le \frac{44.1069 - k}{0.52023} \le 1.95996$
$-1.0196 \le 44.1069 - k \le 1.0196$
$44.1069 - 1.0196 \le k \le 44.1069 + 1.0196$
$43.087 \le k \le 45.127$

Rounding to 2 decimal places, we get:
$43.09 \le k \le 45.13$