A study$^1$ found that students at the National University of Singapore and the Nanyang Technological University sleep an average of 6.2 hours a day. Andy surveys 50 students from university $S$ to estimate each person's daily sleeping hours. The sample mean is $\bar{t}$ hours and the sample standard deviation is 2.2 hours. A test is to be carried out at the 5% level of significance to draw conclusions on the average daily sleeping hours of students from university $S$. $^1$[Source: https://www.straitstimes.com/singapore/lack-of-sleep-is-nothing-to-yawn-about]

1. Define the population mean and state the null and alternative hypotheses.

Let $\mu$ be the population mean daily sleeping hours of students from university $S$.
$H_0: \mu = 6.2$
$H_1: \mu > 6.2$

2. Formulate the test statistic and identify the critical region for rejecting the null hypothesis.

Under $H_0$, since the sample size $n = 50$ is large, by the Central Limit Theorem, the sample mean $\bar{T}$ is approximately normally distributed.
The test statistic is:
$Z = \frac{\bar{T} - \mu}{s/\sqrt{n}} = \frac{\bar{T} - 6.2}{2.2/\sqrt{50}} \sim N(0,1) \text{ approximately.}$
For Andy to conclude that the students sleep more than $6.2$ hours on average, $H_0$ must be rejected.
At the $5\%$ level of significance for a 1-tailed test, the critical region is $z \ge 1.64485$.

3. Solve the inequality to find the set of values of $\bar{t}$.

$\frac{\bar{t} - 6.2}{2.2/\sqrt{50}} \ge 1.64485$
$\bar{t} \ge 6.2 + 1.64485 \left(\frac{2.2}{\sqrt{50}}\right)$
$\bar{t} \ge 6.7117$
$\bar{t} \ge 6.71 \text{ (to 3 s.f.)}$
The set of values of $\bar{t}$ is $\{ \bar{t} : \bar{t} \ge 6.71 \}$.

1. Identify the critical value for a two-tailed test at the 5% significance level.

For a two-tailed test at the $5\%$ level of significance ($H_1: \mu \neq 6.2$), the critical $z$-values are $\pm 1.95996$.

2. Set up and solve the inequalities to find the rejection region for $\bar{t}$.

We reject $H_0$ if $z < -1.95996$ or $z > 1.95996$.
$\frac{\bar{t} - 6.2}{\frac{2.2}{\sqrt{50}}} < -1.95996 \quad \text{or} \quad \frac{\bar{t} - 6.2}{\frac{2.2}{\sqrt{50}}} > 1.95996$
$\bar{t} < 6.2 - 1.95996 \left(\frac{2.2}{\sqrt{50}}\right) \quad \text{or} \quad \bar{t} > 6.2 + 1.95996 \left(\frac{2.2}{\sqrt{50}}\right)$
$\bar{t} < 6.2 - 0.6098 \quad \text{or} \quad \bar{t} > 6.2 + 0.6098$
$\bar{t} < 5.5902 \quad \text{or} \quad \bar{t} > 6.8098$
To 3 significant figures, the set of values is $\{ \bar{t} : \bar{t} < 5.59 \text{ or } \bar{t} > 6.81 \}$.