一位教师顾老师怀疑，学生每天使用手机的平均时间为 $\mu_0$ 分钟。他随机抽取了该校 97 名拥有手机的学生，并记录了每位学生在随机选择的一天中玩手机的时间。结果如下表所示。

每天所花时间（精确到分钟）	60	65	72	90	110	180
人数	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.