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 18.

While it is an uncommon confidence level, find the critical value that corresponds to a confidence level of 86.6%.
(Report answer accurate to three decimal places with appropriate rounding.)

ta/2 =

Answer 1

To find the critical value \( t_{\alpha/2} \) for a confidence level of 86.6% with a sample size of 18, we first need to determine how much area is in each tail of the t-distribution.

1. **Calculate \( \alpha \)**: The confidence level is 86.6%, so:
\[
\alpha = 1 - 0.866 = 0.134
\]

2. **Find \( \alpha/2 \)**: Since we want the area in each tail:
\[
\frac{\alpha}{2} = \frac{0.134}{2} = 0.067
\]

3. **Degrees of freedom**: The degrees of freedom for the t-distribution is given by:
\[
df = n - 1 = 18 - 1 = 17
\]

4. **Find the critical value using a t-table or calculator**: You can look up \( t_{0.067} \) for 17 degrees of freedom. Alternatively, you can use a calculator or statistical software.

Using a t-distribution calculator or table:

For \( df = 17 \) and \( \alpha/2 = 0.067 \), we find:

\[
t_{0.067} \approx 1.643
\]

Thus, the critical value \( t_{\alpha/2} \) is:

\[
\boxed{1.643}
\]