To calculate the probability of a Type II error ($\beta$), we need to go through the following steps:

Step 1: Hypotheses

• Null Hypothesis ($H_0$): The mean assembly time with the new method is the same as the current method, i.e., $\mu = 68$ minutes.
• Alternative Hypothesis ($H_1$): The mean assembly time with the new method is less than the current method, i.e., $\mu < 68$ minutes (left-tailed test).

Step 2: Critical value for $\alpha$

Given no significance level ($\alpha$) in the question, we will assume a common $\alpha = 0.05$.

Since this is a left-tailed test, we will find the z-score corresponding to $\alpha = 0.05$: $z_{\alpha} = \text{inverse cumulative probability for 0.05}.$ Using a standard normal table or a calculator, we find: $z_{\alpha} = -1.645.$

Step 3: Test statistic

Now, calculate the test statistic for the sample mean using the following formula: $z = \frac{\bar{X} - \mu_0}{\frac{\sigma}{\sqrt{n}}},$ where:

• $\bar{X} = 62$ (sample mean),
• $\mu_0 = 68$ (population mean),
• $\sigma = 3.1$ (population standard deviation),
• $n = 24$ (sample size).

Plugging in the values: $z = \frac{62 - 68}{\frac{3.1}{\sqrt{24}}} = \frac{-6}{\frac{3.1}{4.899}} = \frac{-6}{0.632653} \approx -9.48.$ Thus, the calculated $z$-statistic is approximately -9.48.

Step 4: Find the critical value of $\mu$ that separates rejection and non-rejection regions

The critical value of $\mu$ can be found by using the formula: $\mu_c = \mu_0 + z_{\alpha} \times \left(\frac{\sigma}{\sqrt{n}}\right).$ Substitute the values: $\mu_c = 68 + (-1.645) \times \left(\frac{3.1}{\sqrt{24}}\right) = 68 + (-1.645) \times 0.632653 \approx 68 - 1.04 \approx 66.96.$ Thus, $\mu_c \approx 66.96$. This is the critical mean assembly time that separates rejection from non-rejection.

Step 5: Calculate the z-score under the alternative hypothesis ($\mu_1 = 62$)

Now, we calculate the z-score for the sample mean under the alternative hypothesis: $z_{\beta} = \frac{66.96 - 62}{\frac{3.1}{\sqrt{24}}} = \frac{66.96 - 62}{0.632653} = \frac{4.96}{0.632653} \approx 7.84.$

Step 6: Find the probability of a Type II error

Finally, the probability of a Type II error ($\beta$) is the area to the left of $z_{\beta}$ in the standard normal distribution. Using standard normal tables or a calculator, the probability corresponding to $z_{\beta} = 7.84$ is practically 1, as the value is far in the tail of the distribution.

Thus, $\beta \approx 1$.

Final Answer:

The probability of a Type II error is approximately $\boxed{1.0000}$.

Do you want further details or have any questions about the steps?

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Here are 5 questions to further understand this concept:

1. How does the choice of significance level ($\alpha$) affect the probability of a Type II error?
2. What is the difference between Type I and Type II errors in hypothesis testing?
3. How can increasing the sample size reduce the probability of a Type II error?
4. Why do we use the z-distribution instead of the t-distribution in this case?
5. What is the role of the critical value in hypothesis testing?

Tip: Increasing the sample size or reducing variability can reduce the chances of a Type II error ($\beta$), improving the test's power.