Question

You intend to estimate a population mean with a confidence interval. You believe the population to have a normal distribution. Your sample size is 6.

While it is an uncommon confidence level, find the critical value that corresponds to a confidence level of 92.5%.
(Report answer accurate to three decimal places with appropriate rounding.)

ta/2 =

Answer 1

To find the critical value \( t_{\alpha/2} \) corresponding to a confidence level of 92.5% with a sample size of 6, you can follow these steps:

1. **Determine \( \alpha \)**: The confidence level is 92.5%, so:
\[
\alpha = 1 - 0.925 = 0.075
\]

2. **Calculate \( \alpha/2 \)**: Since this is a two-tailed test, divide \( \alpha \) by 2:
\[
\frac{\alpha}{2} = \frac{0.075}{2} = 0.0375
\]

3. **Determine the degrees of freedom**: Since you have a sample size of 6, the degrees of freedom \( df \) is:
\[
df = n - 1 = 6 - 1 = 5
\]

4. **Look up the critical value**: You will need to look up the value of \( t_{0.0375} \) with \( df = 5 \). This can be done using a t-table or calculator.

Using the t-distribution table or a calculator, for \( df = 5 \) and \( \alpha/2 = 0.0375 \):

- The critical value \( t_{0.0375, 5} \) is found. Typically, you would look for the closest probability (1 - 0.0375 = 0.9625) in the t-table.

By checking a source like a t-distribution table, or if you're using software or a calculator, you would find:

\[
t_{0.0375, 5} \approx 2.571
\]

Thus, the critical value \( t_{\alpha/2} \) for a confidence level of 92.5% with 5 degrees of freedom is:

\[
\boxed{2.571}
\]