在一个实验室中，检测员测试一块电路板所需的时间（以分钟为单位）是一个连续随机变量 $X$。$X$ 的标准差为 0.63，且在普通条件下，$X$ 的期望值为 5.82。由于引入了激励计划，实验室主管声称检测员进行测试的速度更快了。结果发现，在一个含有 150 次测试的随机样本中，测试电路板所需的平均时间为 5.74。 两年后，一名质量控制官员抽取了 20 次测试的随机样本，发现测试电路板所需的平均时间为 5.88。他声称测试电路板所需的平均时间与 $\mu_0$ 分钟不同。

1. State the null and alternative hypotheses based on the supervisor's claim that the test is carried out more quickly (i.e. the mean time taken is reduced).

$H_0: \mu = 5.82$
$H_1: \mu < 5.82$

2. Define the parameter $\mu$ clearly in the context of the question.

where $\mu$ is the population mean time taken, in minutes, for an inspector to test an electrical circuit board.

1. State the test parameters and calculate the value of the test statistic.

From part (a), the hypotheses are:
$H_0: \mu = 5.82$
$H_1: \mu < 5.82$

Given $n = 150$, sample mean $\bar{x} = 5.74$, and population standard deviation $\sigma = 0.63$.
Under $H_0$, the test statistic is:
$z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} = \frac{5.74 - 5.82}{0.63 / \sqrt{150}}$
$z \approx -1.5552$

2. Determine the $p$-value for the one-tailed test.

Using the standard normal distribution:
$p\text{-value} = P(Z \le -1.5552) \approx 0.05994$

3. Compare the $p$-value to the level of significance and state the final conclusion in context.

Since $p\text{-value} = 0.0599 > 0.05$, we do not reject $H_0$.

There is insufficient evidence at the $5\%$ level of significance to conclude that the mean time taken to test an electrical circuit board is less than $5.82$ minutes. Thus, the laboratory supervisor's claim is not justified.

1. Explain the relevance of the sample size and the Central Limit Theorem to the need for knowing the population distribution.

Since the sample size used by the supervisor is $n = 150$, which is considered large (i.e., $n \ge 30$), the Central Limit Theorem applies. This means that the distribution of the sample mean $\bar{X}$ will be approximately normally distributed, regardless of the underlying population distribution of $X$. Therefore, the supervisor does not need to know anything about the population distribution of $X$.

1. State the assumption regarding the population distribution required due to the small sample size

Since the sample size $n = 20$ is small (less than 30), the Central Limit Theorem cannot be applied. Therefore, we must assume that the underlying population of the time taken to test an electrical circuit board, $X$, follows a normal distribution.

2. State the assumption regarding the population variance required to conduct a Z-test

A hypothesis test for the population mean using a Normal distribution requires the population variance to be known. Since the standard deviation of $0.63$ was established two years ago, we must assume the population standard deviation remains unchanged at $0.63$ minutes.

1. Identify the critical values and set up the inequality for the non-rejection of the null hypothesis.

Since the alternative hypothesis is $H_1: \mu \neq \mu_0$, this is a two-tailed test. At the $1\%$ level of significance, the critical value for the standard normal distribution is $z \approx 2.5758$.

For $H_0$ not to be rejected, the test statistic $Z$ must fall within the non-rejection region:
$-2.5758 \le Z \le 2.5758$

2. Substitute the test statistic formula using the given sample data.

The test statistic is given by $Z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}$.

Given the sample mean $\bar{x} = 5.88$, population standard deviation $\sigma = 0.63$ (from the problem stem), and sample size $n = 20$:
$-2.5758 \le \frac{5.88 - \mu_0}{0.63 / \sqrt{20}} \le 2.5758$

3. Solve the inequality to find the range of $\mu_0$.

Multiply all parts by the standard error $\frac{0.63}{\sqrt{20}} \approx 0.14087$:
$-2.5758(0.14087) \le 5.88 - \mu_0 \le 2.5758(0.14087)$
$-0.36286 \le 5.88 - \mu_0 \le 0.36286$

Subtract $5.88$ from all parts:
$-6.24286 \le -\mu_0 \le -5.51714$

Multiply by $-1$ and reverse the inequality signs:
$5.5171 \le \mu_0 \le 6.2429$

Rounding to 3 significant figures, we get:
$5.52 \le \mu_0 \le 6.24$