The waiting time for a patient to see a doctor in a clinic is $T$ minutes. The waiting times of a randomly chosen sample of 120 patients are summarised by $\sum (t-15) = 123$, and $\sum (t-15)^2 = 2504$.

1. Calculate the unbiased estimate of the population mean using the coded sum.

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

2. Calculate the unbiased estimate of the population variance using the coded sums and show it rounds to 19.983.

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

1. State the null and alternative hypotheses

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

2. Calculate the sample mean and the test statistic/p-value

From part (i),
$\bar{t} = 15 + \frac{123}{120} = 16.025$
$s^2 \approx 19.98256$

Under $H_0$, since the sample size $n = 120$ is large, by the Central Limit Theorem the test statistic is approximately:
$Z = \frac{\bar{T} - \mu}{s/\sqrt{n}} \sim \text{N}(0, 1)$

$z = \frac{16.025 - 15}{\sqrt{19.98256 / 120}} \approx 2.5118$

Using a graphing calculator, the $p$-value for a two-tailed test is:
$p\text{-value} = 2 \times \text{P}(Z > 2.5118) \approx 0.0120$

3. State the decision and conclusion in context

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 Central Limit Theorem.

Let $\mu$ be the population mean waiting time.
$H_0: \mu = 15$
Since the sample mean $\bar{t} = 15.5 > 15$, the appropriate one-tail test is $H_1: \mu > 15$.
Using the unbiased estimate of the population variance from part (i), $s^2 = 19.983$.
Under $H_0$, since $n \ge 50$, the Central Limit Theorem applies and $\bar{T} \sim \text{N}\left(15, \frac{19.983}{n}\right)$ approximately.
The test statistic is $Z = \frac{\bar{T} - 15}{\sqrt{19.983 / n}}$.

2. Set up the inequality for the null hypothesis not being rejected.

For a $6\%$ level of significance in a right-tailed test, the critical $z$-value is such that $P(Z < z) = 0.94$. Using a graphing calculator, $z_{0.06} \approx 1.5548$.
Since $H_0$ is not rejected, the test statistic must be less than or equal to the critical value:
$\frac{15.5 - 15}{\sqrt{\frac{19.983}{n}}} \le 1.5548$

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 representing the number of patients, and the question specifies $n \ge 50$, the required range is:
$50 \le n \le 193$