Tim is a student in a school with a large student population. As part of his project, Tim would like to interview some of his schoolmates to obtain some feedback about the school.

1. Identify the relevant strata in the population based on the demographics provided.

The population can be divided into $2 \times 2 = 4$ equal strata: Year 1 males, Year 1 females, Year 2 males, and Year 2 females.

2. Determine the sample size for each stratum and state the sampling method within each.

Since the 4 strata are of equal size and the total sample size is 80, Tim should obtain a simple random sample of $\frac{80}{4} = 20$ students from each stratum.

1. Explain the choice of a one-tailed test based on the context.

The problem states that "Tim believes that students are spending more time on their homework." This corresponds to a directional alternative hypothesis ($H_1: \mu > 3$), meaning he is only interested in whether the mean has increased, which justifies a one-tailed test.

1. Calculate the unbiased estimate of the population mean.

$\bar{x} = c + \frac{\sum (x-c)}{n}$
$\bar{x} = 3 + \frac{16}{80} = 3.2$

2. Calculate the unbiased estimate of the population variance.

$s^2 = \frac{1}{n-1} \left( \sum (x-c)^2 - \frac{(\sum (x-c))^2}{n} \right)$
$s^2 = \frac{1}{80-1} \left( 164 - \frac{16^2}{80} \right)$
$s^2 = \frac{1}{79} (164 - 3.2)$
$s^2 = \frac{160.8}{79} \approx 2.0354 \approx 2.04 \text{ (to 3 s.f.)}$

1. Formulate the null and alternative hypotheses, and set a suitable level of significance.

Let $\mu$ be the population mean time (in hours) students spend on homework each day.

$H_0: \mu = 3$
$H_1: \mu > 3$

Let the level of significance be $\alpha = 0.05$ (or $5\%$).

2. State the test distribution and test statistic.

Since the sample size $n = 80$ is large ($n \ge 30$), by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed.

Under $H_0$, the test statistic is $Z = \frac{\bar{X} - 3}{S/\sqrt{80}} \sim N(0, 1)$ approximately.

3. Calculate the $p$-value using the unbiased estimates from part (c).

From part (c), $\bar{x} = 3.2$ and $s^2 = \frac{160.8}{79} \approx 2.0354$.

Using the 1-tail Z-test on a graphing calculator with $\mu_0 = 3$, $\bar{x} = 3.2$, $s = \sqrt{\frac{160.8}{79}}$ and $n = 80$:

Test statistic $z \approx 1.2538$
$p\text{-value} = 0.10495 \dots \approx 0.105$ (to 3 s.f.)

4. Conclude the test in the context of the question.

Since the $p\text{-value} = 0.105 > 0.05$, we do not reject $H_0$.

There is insufficient evidence at the $5\%$ level of significance to conclude that students spend more than $3$ hours each day on their homework.
Therefore, Tim's belief is not accurate.

1. State the necessary assumption for a small sample hypothesis test

Assumption: The time spent each day on homework by the students follows a normal distribution.

2. Determine the critical value and state the test statistic for the hypothesis test

For a two-tail test at the $2\%$ level of significance, the critical value is $z = 2.3263$.
The test statistic under $H_0 : \mu = 3$ is given by $Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}$.
Substituting the given values:
$Z = \frac{3.2 - 3}{\sigma / \sqrt{10}} = \frac{0.2\sqrt{10}}{\sigma}$

3. Set up and solve the inequality for the rejection region to find the range of $\sigma$

Since the null hypothesis is rejected and $\bar{x} > \mu$, the test statistic must lie in the upper critical region:
$\frac{0.2\sqrt{10}}{\sigma} > 2.3263$
$\sigma < \frac{0.2\sqrt{10}}{2.3263}$
$\sigma < 0.27187$
Since the standard deviation must be strictly positive:
$0 < \sigma < 0.272 \text{ (to 3 s.f.)}$