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 =

1 answer

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:

\[ \boxed{1.643} \]