在 AquaV 游泳培训学校中，学员游完一圈泳池所需的平均时间为 35 秒。该学校采用了一项新的国际培训计划 Breakthru，为期 3 个月，并希望分析 Breakthru 在缩短学员用时方面是否有效。 抽取了 30 名学员作为样本，他们游完一圈泳池所需的时间 $x$ 秒汇总如下： $\sum (x-30)=94$, $\sum (x-30)^2=758$ 。

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

Let $\mu$ be the population mean time taken, in seconds, by a trainee to swim a lap of the pool after 3 months of the Breakthru training programme.

Since an improvement in swimming timings means the time taken has decreased, the hypotheses are:
$H_0: \mu = 35$
$H_1: \mu < 35$

2. Calculate the sample mean $\bar{x}$.

$\bar{x} = \frac{\sum (x-30)}{n} + 30$
$\bar{x} = \frac{94}{30} + 30 = \frac{497}{15} \approx 33.133 \text{ s}$

3. Calculate the unbiased estimate of the population variance $s^2$.

$s^2 = \frac{1}{n-1}\left( \sum (x-30)^2 - \frac{(\sum (x-30))^2}{n} \right)$
$s^2 = \frac{1}{29}\left( 758 - \frac{94^2}{30} \right)$
$s^2 = \frac{6952}{435} \approx 15.982$

4. Determine the $p$-value using a 1-tail Z-test.

Since the sample size $n=30$ is large, by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed.
Under $H_0$, the test statistic is $Z = \frac{\bar{X} - 35}{s/\sqrt{30}} \sim N(0,1)$ approximately.

Using a graphing calculator for a 1-tail $Z$-test:
$\text{Test statistic } z \approx -2.5575$
$p\text{-value} \approx 0.00527 \text{ (to 3 s.f.)}$

5. Compare the $p$-value with the significance level and state the conclusion in context.

Level of significance, $\alpha = 0.05$.
Since $p\text{-value} = 0.00527 < 0.05$, we reject $H_0$.
There is sufficient evidence at the $5\%$ significance level to conclude that the mean time taken to swim a lap of the pool has decreased, meaning that the timings of the trainees have improved.

1. State a necessary condition for the sample data to be representative and for the Central Limit Theorem to apply.

Assume that the sample of 30 trainees is a random sample, which implies that the times taken by each trainee are independent of one another.

1. State the null and alternative hypotheses.

Let $\mu$ be the population mean time taken to swim a lap in AquaZ.
$H_0: \mu = 38$
$H_1: \mu < 38$

2. Identify the test statistic and the critical value for the rejection region.

Since the population follows a normal distribution and the sample size $n=30$ is sufficiently large, we can substitute the unbiased estimate of the population standard deviation ($s=4$) and use the standard normal distribution $Z$ (a $t$-test is also fully acceptable and exact here).

The test statistic is:
$Z = \frac{\bar{Y} - \mu}{s/\sqrt{n}} = \frac{\bar{Y} - 38}{4/\sqrt{30}}$

At the 5% level of significance for a lower-tailed test, the critical region is $Z \le -1.6449$.

3. Set up and solve the inequality to find the set of values for $\bar{y}$.

To reject $H_0$, the test statistic must fall within the critical region:
$\frac{\bar{y} - 38}{4/\sqrt{30}} \le -1.64485$
$\bar{y} - 38 \le -1.64485 \times \frac{4}{\sqrt{30}}$
$\bar{y} - 38 \le -1.2012$
$\bar{y} \le 36.798$

Rounding to 3 significant figures, we get $\bar{y} \le 36.8$.

(Note: Using the exact $t$-distribution with $n-1=29$ degrees of freedom gives a critical value of $t = -1.6991$, leading to $\bar{y} \le 36.759$, which also correctly rounds to $36.8$.)

1. Calculate the sample mean $\bar{y}$.

$\bar{y} = \frac{\sum (y-30)}{n} + 30$

$\bar{y} = \frac{234}{30} + 30 = 7.8 + 30 = 37.8$

2. Evaluate the test statistic and state the conclusion.

Under $H_0: \mu = 38$, the test statistic is:
$z = \frac{\bar{y} - \mu}{s / \sqrt{n}} = \frac{37.8 - 38}{4 / \sqrt{30}} \approx -0.2739$

The critical value for a 5% significance level (1-tailed) is $-1.645$.

Since $z = -0.2739 > -1.645$ (equivalently, $\bar{y} = 37.8$ lies outside the rejection region $\bar{y} < 36.8$ derived in part (c)), we do not reject $H_0$.

Conclusion: There is insufficient evidence at the 5% level of significance to conclude that there is an improvement in the swimming times for trainees in AquaZ.