Over a period of one month, the increase in heights of orchid plants in a nursery has mean 1.34 cm and standard deviation 0.24 cm. Soothing music is introduced in the nursery and one month later, the increase in heights, $x$ cm, is measured for a random sample of 55 orchid plants. A test is carried out, at the 5% level of significance, to determine whether there has been a significant improvement in the increase in heights of the orchid plants.

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.