Question

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 =

Answer 1

To find the critical value \( t_{\alpha/2} \) corresponding to a confidence level of 90% for a sample size of 23, we first need to determine the degrees of freedom (df) and the significance level (α).

1. **Degrees of Freedom (df)**: Since the sample size \( n \) is 23, the degrees of freedom is calculated as:
\[
df = n - 1 = 23 - 1 = 22
\]

2. **Significance Level (α)**: For a 90% confidence level, the significance level \( \alpha \) is:
\[
\alpha = 1 - 0.90 = 0.10
\]

3. **Critical value \( t_{\alpha/2} \)**: Since we have a two-tailed test, we need to find \( t_{\alpha/2} \):
\[
\alpha/2 = 0.10/2 = 0.05
\]

Now we need to look up the critical value \( t_{0.05} \) with 22 degrees of freedom. This can typically be found in a t-distribution table or calculated using statistical software.

Using a t-distribution table or calculator, the critical value \( t_{0.05, 22} \) is approximately:

\[
t_{0.05, 22} \approx 1.645
\]

So, rounding to three decimal places, the critical value \( t_{\alpha/2} \) is:

\[
\boxed{1.645}
\]