To solve this, let’s determine the correct order of the statements to derive the confidence interval for a $t$-distribution. Here’s the approach:

1. Starting Point: The pivotal quantity $T = \frac{\bar{X} - \mu}{s / \sqrt{n}}$ has a $t$-distribution with $n - 1$ degrees of freedom. This is in statement A.

2. Probability Statement: We use the $t$-distribution to set up a probability statement involving $T$: $P\left( -t_{\alpha/2, n-1} < T < t_{\alpha/2, n-1} \right) = 1 - \alpha$ This is represented by E.

3. Rewriting in Terms of $T$: Substitute $T = \frac{\bar{X} - \mu}{s / \sqrt{n}}$ into the inequality: $P\left( -t_{\alpha/2, n-1} < \frac{\bar{X} - \mu}{s / \sqrt{n}} < t_{\alpha/2, n-1} \right) = 1 - \alpha$ This matches F.

4. Multiplying by $s / \sqrt{n}$: Solve for $\mu$ by rearranging the inequality: $P\left( -t_{\alpha/2, n-1} \frac{s}{\sqrt{n}} < \bar{X} - \mu < t_{\alpha/2, n-1} \frac{s}{\sqrt{n}} \right) = 1 - \alpha$ This step is in D.

5. Isolating $\mu$: Add $\bar{X}$ to all parts of the inequality to solve for $\mu$: $P\left( \bar{X} - t_{\alpha/2, n-1} \frac{s}{\sqrt{n}} < \mu < \bar{X} + t_{\alpha/2, n-1} \frac{s}{\sqrt{n}} \right) = 1 - \alpha$ This is expressed in B.

6. Final Confidence Interval: This gives us the $1 - \alpha$ confidence interval for $\mu$: $\bar{X} - t_{\alpha/2, n-1} \frac{s}{\sqrt{n}} < \mu < \bar{X} + t_{\alpha/2, n-1} \frac{s}{\sqrt{n}}$ This final form is given by G.

So, the correct order of the statements to derive the confidence interval is:

A, E, F, D, B, G

Would you like further details or have any questions?

Related Questions:

1. What is a $t$-distribution and how does it differ from a normal distribution?
2. Why do we use $t$-distributions for confidence intervals when the population standard deviation is unknown?
3. How does the confidence level $1 - \alpha$ impact the width of the confidence interval?
4. How does sample size $n$ affect the confidence interval for the mean?
5. What are the assumptions required to use a $t$-distribution for confidence intervals?

Tip:

The width of the confidence interval for the mean decreases as the sample size $n$ increases, due to the $\sqrt{n}$ in the denominator. This reflects increased precision in estimating the mean with larger samples.