You intend to estimate a population mean with a confidence interval. You believe the population to have a normal distribution. Your sample size is 23.

Find the critical value that corresponds to a confidence level of 90%.
(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 90% with a sample size of 23, you'll first need to determine the degrees of freedom.

The degrees of freedom (df) is calculated as:

\[ \text{df} = n - 1 \]

where \( n \) is the sample size. For a sample size of 23:

\[ \text{df} = 23 - 1 = 22 \]

Next, for a 90% confidence level, the significance level \( \alpha \) is:

\[ \alpha = 1 - 0.90 = 0.10 \]

Since we need \( t_{\alpha/2} \), we divide \( \alpha \) by 2:

\[ \alpha/2 = 0.10 / 2 = 0.05 \]

Now, we will look up the critical value in the t-distribution table for 22 degrees of freedom and \( \alpha/2 = 0.05 \).

From the t-distribution table, the critical value \( t_{0.05} \) for 22 degrees of freedom is approximately:

\[ t_{0.05} \approx 1.645 \]

Therefore, the value of \( t_{\alpha/2} \) is:

\[ t_{\alpha/2} = 1.645 \]

Rounded to three decimal places, it is:

\[ \boxed{1.645} \]