You intend to estimate a population mean with a confidence interval. You believe the population to have a normal distribution. Your sample size is 79.

While it is an uncommon confidence level, find the critical value that corresponds to a confidence level of 82.6%.
(Report answer accurate to three decimal places with appropriate rounding.)

ta/2 =

1 answer

To find the critical value \( t_{\alpha/2} \) that corresponds to a confidence level of 82.6%, we first need to calculate \( \alpha \):

\[ \alpha = 1 - 0.826 = 0.174 \]

Since we are looking for the value \( t_{\alpha/2} \), we need to divide \( \alpha \) by 2:

\[ \alpha/2 = 0.174 / 2 = 0.087 \]

Next, we need to find the degrees of freedom, which is computed as:

\[ df = n - 1 = 79 - 1 = 78 \]

Now, we look up the critical value \( t_{\alpha/2} \) for \( \alpha/2 = 0.087 \) with 78 degrees of freedom using a t-table or a calculator.

Using a t-table or calculator, we find the critical value corresponding to \( t_{0.087} \) and \( df = 78 \).

Using statistical software or tables, we find:

\[ t_{0.087} \approx 1.375 \]

Thus, rounding to three decimal places, the value is:

\[ t_{\alpha/2} \approx 1.375 \]

So, the final answer is:

\[ \boxed{1.375} \]