在某诊所中，患者看医生的等待时间为 $T$ 分钟。随机抽取的 120 名患者的等待时间数据汇总如下：
$\sum (t-15) = 123, \quad \text{and} \quad \sum (t-15)^2 = 2504.$

1. Find the unbiased estimate of the population mean.

$\bar{t} = \frac{\sum (t-15)}{n} + 15$
$\bar{t} = \frac{123}{120} + 15 = 16.025$

2. Calculate the unbiased estimate of the population variance and round to show the given value.

$s^2 = \frac{1}{n-1} \left( \sum (t-15)^2 - \frac{(\sum (t-15))^2}{n} \right)$
$s^2 = \frac{1}{120-1} \left( 2504 - \frac{123^2}{120} \right)$
$s^2 = \frac{1}{119} \left( 2504 - 126.075 \right)$
$s^2 = \frac{2377.925}{119} = 19.98256\dots$
$s^2 \approx 19.983 \text{ (shown)}$

1. Calculate the sample mean from the data provided in part (i)

From part (i), the sample mean is:
$\bar{t} = \frac{\sum (t-15)}{n} + 15 = \frac{123}{120} + 15 = 16.025$

2. State the null and alternative hypotheses

Let $\mu$ be the population mean waiting time in minutes.
$H_0: \mu = 15$
$H_1: \mu \neq 15$

3. Calculate the test statistic and $p$-value

Since $n = 120$ is large, by the Central Limit Theorem, the sample mean $\bar{T}$ is approximately normally distributed.
Under $H_0$, the test statistic is $Z = \frac{\bar{T} - 15}{S/\sqrt{n}} \sim N(0, 1)$ approximately.
Using the unbiased estimate of population variance $S^2 \approx 19.983$ from part (i):
$z = \frac{16.025 - 15}{\sqrt{19.983 / 120}} \approx 2.5118$
Using a graphing calculator, the $p$-value for a 2-tailed test is:
$p\text{-value} = 2 \times P(Z > 2.5118) \approx 0.0120$

4. Make a conclusion based on the $p$-value

Compare the $p$-value with the level of significance $\alpha = 0.05$.
Since $p\text{-value} = 0.0120 < 0.05$, we reject $H_0$.
There is sufficient evidence at the $5\%$ level of significance to conclude that the mean waiting time differs from $15$ minutes.

1. State the hypotheses and the test statistic, using the variance estimate from part (i).

Let $\mu$ be the population mean waiting time.
$H_0: \mu = 15$
$H_1: \mu > 15$ (since the sample mean $\bar{t} = 15.5 > 15$ and a one-tail test was conducted)

Since $n \ge 50$ is large, by the Central Limit Theorem, the sample mean $\bar{T}$ is approximately normally distributed. We use the unbiased estimate of the population variance from part (i), $s^2 = 19.983$.

Under $H_0$, the test statistic is:
$Z = \frac{\bar{T} - \mu}{s / \sqrt{n}} \sim \mathrm{N}(0, 1) \text{ approximately}$

2. Formulate an inequality for $n$ based on the non-rejection condition at the 6% significance level.

For a one-tail test at the $6\%$ level of significance, the critical value is given by $z = 1.5548$.

Since the null hypothesis was not rejected, the calculated test statistic must be less than or equal to the critical value:
$z_{\text{calc}} \le 1.5548$
$\frac{15.5 - 15}{\sqrt{19.983 / n}} \le 1.5548$

3. Solve the inequality to find the range of values for $n$.

$\frac{0.5 \sqrt{n}}{\sqrt{19.983}} \le 1.5548$
$\sqrt{n} \le \frac{1.5548 \times \sqrt{19.983}}{0.5}$
$\sqrt{n} \le 13.900$
$n \le 193.22$

Since $n$ must be an integer and the question states $n \ge 50$, the range of values for $n$ is:
$50 \le n \le 193$