A teacher, Mr. Ku, suspects that the average time a student spends on his or her mobile phone per day is $\mu_0$ minutes. He selected a random sample of 97 students in the school who own mobile phones and recorded the amount of time each student spent on his or her phone in a randomly selected day. The results are displayed in the table below.

Time spent per day (to nearest minute)	60	65	72	90	110	180
Number of people	11	20	32	18	10	6

1. Calculate the unbiased estimate of the population mean.

From the frequency table, we can compute the total number of students ($n$), $\sum fx$, and $\sum fx^2$:
$n = 11 + 20 + 32 + 18 + 10 + 6 = 97$
$\sum fx = 11(60) + 20(65) + 32(72) + 18(90) + 10(110) + 6(180) = 8064$

$\bar{x} = \frac{\sum fx}{n} = \frac{8064}{97} \approx 83.1 \text{ (to 3 s.f.)}$

2. Calculate the unbiased estimate of the population variance.

Calculate the sum of squares, $\sum fx^2$:
$\sum fx^2 = 11(60^2) + 20(65^2) + 32(72^2) + 18(90^2) + 10(110^2) + 6(180^2) = 751188$

$s^2 = \frac{1}{n-1} \left( \sum fx^2 - \frac{(\sum fx)^2}{n} \right)$
$s^2 = \frac{1}{96} \left( 751188 - \frac{8064^2}{97} \right)$
$s^2 = \frac{653095}{776} \approx 842 \text{ (to 3 s.f.)}$

1. Calculate the sample mean ($\bar{x}$) and unbiased estimate of the population variance ($s^2$) as required from part (i).

$\sum x = 11(60) + 20(65) + 32(72) + 18(90) + 10(110) + 6(180) = 8064$

$\sum x^2 = 11(60^2) + 20(65^2) + 32(72^2) + 18(90^2) + 10(110^2) + 6(180^2) = 751188$

$\bar{x} = \frac{8064}{97} \approx 83.1340$

$s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right) = \frac{1}{96} \left( 751188 - \frac{8064^2}{97} \right) \approx 841.6035$

2. Formulate the test statistic and state the rejection criteria for a two-tailed test at the 5% significance level.

$H_0: \mu = \mu_0$
$H_1: \mu \neq \mu_0$

Since $n = 97 \ge 30$, by the Central Limit Theorem, $\bar{X}$ is approximately normally distributed.
The standardized test statistic is:
$Z = \frac{\bar{X} - \mu_0}{s/\sqrt{n}} \sim N(0, 1) \text{ approximately}$

For a two-tailed test at the 5% level of significance, the critical value is $z_{0.025} \approx 1.960$.
We reject $H_0$ if $Z \le -1.960$ or $Z \ge 1.960$.

3. Substitute the calculated values and solve the inequalities to find the range of $\mu_0$.

$\frac{83.1340 - \mu_0}{\sqrt{841.6035 / 97}} \le -1.960 \quad \text{or} \quad \frac{83.1340 - \mu_0}{\sqrt{841.6035 / 97}} \ge 1.960$

$\frac{83.1340 - \mu_0}{2.9456} \le -1.960 \quad \text{or} \quad \frac{83.1340 - \mu_0}{2.9456} \ge 1.960$

$83.1340 - \mu_0 \le -5.773 \quad \text{or} \quad 83.1340 - \mu_0 \ge 5.773$

$\mu_0 \ge 88.907 \quad \text{or} \quad \mu_0 \le 77.361$

Therefore, to 3 significant figures, $H_0$ is rejected if $\mu_0 \le 77.4$ or $\mu_0 \ge 88.9$.

1. State the meaning of the level of significance (probability of making a Type I error) in the context of the problem.

The level of significance is the probability of rejecting the null hypothesis when it is actually true. In this context, it means there is a $5\%$ probability of incorrectly concluding that the average time a student spends on his or her mobile phone per day differs from $\mu_0$ minutes, when the average time is actually $\mu_0$ minutes.

1. Relate the rejection of the null hypothesis at the 5% level to the p-value.

Rejecting $H_0$ at the 5% significance level means that the $p$-value is less than $0.05$.

2. Compare with the condition required for the 1% level of significance and state the conclusion.

To reject $H_0$ at the 1% significance level, we would need the $p$-value to be strictly less than $0.01$. Because the $p$-value could be in the range $0.01 \le p\text{-value} < 0.05$, we cannot conclude for certain that $H_0$ will be rejected at the 1% level.