在一个月的时间里，花圃中兰花植物的高度增长量的均值为 1.34 cm，标准差为 0.24 cm。在花圃中播放舒缓音乐，一个月后，测量了 55 株兰花植物的随机样本，高度增长量为 $x$ cm。 在 5% 的显著性水平下进行检验，以确定兰花植物高度的增长是否有了显著改善。

1. State the appropriate null and alternative hypotheses.

Let $\mu$ be the population mean increase in heights of the orchid plants after soothing music is introduced.

The test is to determine if there has been an improvement (which corresponds to an increase in heights), so we use a one-tailed test:
$H_0: \mu = 1.34$
$H_1: \mu > 1.34$

2. Determine if any assumption is needed and provide a reason based on the sample size.

No assumption about the population distribution (e.g. assuming it is normally distributed) is needed.

Reason: Since the sample size $n = 55$ is large ($n \ge 30$), by the Central Limit Theorem, the distribution of the sample mean $\bar{X}$ is approximately normal regardless of the population's distribution.

1. Identify the condition for not rejecting the null hypothesis based on the critical $z$-value.

From part (i), this is a right-tailed test ($H_1: \mu > 1.34$).
At the $5\%$ level of significance, the critical value is $z = 1.6449$.
To not reject $H_0$, the test statistic $Z$ must not fall in the critical region, so we require:
$Z \le 1.6449$

2. Set up the inequality using the standard test statistic formula for the sample mean.

Substitute $Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ into the inequality:
$\frac{\bar{x} - 1.34}{\frac{0.24}{\sqrt{55}}} \le 1.6449$

3. Solve for $\bar{x}$.

$\bar{x} \le 1.34 + 1.6449 \left(\frac{0.24}{\sqrt{55}}\right)$
$\bar{x} \le 1.34 + 0.0532$
$\bar{x} \le 1.3932$
Therefore, the set of values is $\bar{x} \le 1.39$ (to 3 s.f.).

1. State the meaning of the level of significance in the specific context of the question.

The level of significance is the probability of rejecting the null hypothesis $H_0$ when $H_0$ is actually true.
In this context, $H_0$ represents the scenario where the mean increase in height is 1.34 cm (no improvement), and rejecting it means concluding there is an improvement.
Therefore, a 5% level of significance means there is a 5% probability of concluding that there is a significant improvement in the increase in heights of the orchid plants, when in fact there is no improvement.