Confidence Interval Length Calculation

To estimate the population mean $\mu$, we construct confidence intervals. The question requires ordering these intervals by their length. For a confidence interval of the mean with a known population standard deviation $\sigma$, the formula is:

$ \bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}} $

The length of the interval is twice the margin of error (ME). The margin of error is calculated as $z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$. Thus, the interval length $L$ is:

$ L = 2 \times ME = 2 \times z_{\alpha/2} \frac{\sigma}{\sqrt{n}} $

For a 95% confidence level, $1 - \alpha = 0.95$, so $\alpha = 0.05$ and $\alpha/2 = 0.025$. The critical z-value $z_{0.025}$ is approximately 1.96. The length calculation depends only on $\sigma$ and $n$.

Interval Length Calculations

Calculate the length $L$ for each case using $L = 2 \times 1.96 \times \frac{\sigma}{\sqrt{n}}$:

• Case A: $\sigma = 3, n = 49$

  Length $L_A = 2 \times 1.96 \times \frac{3}{\sqrt{49}} = 2 \times 1.96 \times \frac{3}{7} \approx 1.6786$

• Case B: $\sigma = 3, n = 81$

  Length $L_B = 2 \times 1.96 \times \frac{3}{\sqrt{81}} = 2 \times 1.96 \times \frac{3}{9} \approx 1.3067$

• Case C: $\sigma = 25, n = 49$

  Length $L_C = 2 \times 1.96 \times \frac{25}{\sqrt{49}} = 2 \times 1.96 \times \frac{25}{7} \approx 14.0000$

• Case D: $\sigma = 25, n = 81$

  Length $L_D = 2 \times 1.96 \times \frac{25}{\sqrt{81}} = 2 \times 1.96 \times \frac{25}{9} \approx 10.8889$

Ordering Intervals by Length

Compare the calculated lengths:

• $L_B \approx 1.3067$
• $L_A \approx 1.6786$
• $L_D \approx 10.8889$
• $L_C \approx 14.0000$

Arranging the intervals from the lowest length to the highest length yields the order: B, A, D, C.