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 97.1%.
(Report answer accurate to three decimal places with appropriate rounding.)

ta/2 =

1 answer

To find the critical value \( t_{\alpha/2} \) corresponding to a confidence level of 97.1% with a sample size of 18, you'll need to follow these steps:

  1. Determine the significance level \( \alpha \): \[ \alpha = 1 - \text{confidence level} = 1 - 0.971 = 0.029 \] Then, \( \alpha/2 \) is: \[ \alpha/2 = 0.029/2 = 0.0145 \]

  2. Determine the degrees of freedom: Since the sample size \( n = 18 \), the degrees of freedom \( df \) is: \[ df = n - 1 = 18 - 1 = 17 \]

  3. Find the critical value \( t_{\alpha/2} \): You will look for the \( t \) value that corresponds to \( 0.0145 \) in the upper tail with \( df = 17 \). This can be obtained from a t-distribution table or using statistical software.

Using statistical tables or software, we find:

  • For \( df = 17 \) and \( \alpha/2 = 0.0145 \), the critical value is approximately \( t_{0.0145} \).

After searching through a t-table or using software, the closest value found for \( t_{0.0145} \) with 17 degrees of freedom is approximately 2.624.

Thus, \[ t_{\alpha/2} \approx 2.624 \]

Final answer rounded to three decimal places: \[ t_{\alpha/2} = 2.624 \]