You intend to estimate a population mean with a confidence interval. You believe the population to have a normal distribution. Your sample size is 8.

Find the critical value that corresponds to a confidence level of 99%.
(Report answer accurate to three decimal places with appropriate rounding.)

ta/2 =

1 answer

To find the critical value \( t_{\alpha/2} \) for a 99% confidence level with a sample size of 8, we need to use the t-distribution. The critical value can be found by looking up the appropriate value in a t-table or using statistical software or calculators.

  1. Determine \( \alpha \): For a 99% confidence level, \( \alpha \) (the total level of significance) is: \[ \alpha = 1 - 0.99 = 0.01 \] Therefore, \( \alpha/2 = 0.005 \).

  2. Degrees of Freedom: The degrees of freedom for the t-distribution is calculated as: \[ \text{df} = n - 1 = 8 - 1 = 7 \]

  3. Find \( t_{\alpha/2} \): We are looking for \( t_{0.005} \) with 7 degrees of freedom. You can reference a t-table or use statistical software to find this value.

Looking up \( t_{0.005} \) for 7 degrees of freedom, you find:

\[ t_{0.005} \approx 3.499 \]

Thus, the critical value that corresponds to a 99% confidence level with a sample size of 8 is:

\[ t_{\alpha/2} \approx 3.499 \]

Final Answer: \[ t_{\alpha/2} = 3.499 \]