Here are summary statistics for randomly selected weights of newborn girls: \( \mathrm{n}=36, \overline{\mathrm{x}}=3197.2 \mathrm{~g}, \mathrm{~s}=692.6 \mathrm{~g} \). Use a confidence level of \( 90 \% \) to complete parts (a) through (d) below. a. Identify the critical value \( \mathrm{t}_{\alpha / 2} \) used for finding the margin of error. \( \mathrm{t}_{\alpha / 2}=\square \) (Round to two decimal places as needed.)

To find the critical value \( t_{\alpha / 2} \) needed for a 90% confidence interval, we need to determine \( \alpha \) first. For a 90% confidence level, \( \alpha = 1 - 0.90 = 0.10 \). Since we are looking for \( t_{\alpha / 2} \), we divide \( \alpha \) by 2, yielding \( \alpha / 2 = 0.05 \). Next, we need the degrees of freedom (df) for our t-distribution, which is calculated as \( df = n - 1 = 36 - 1 = 35 \). Using a t-distribution table or calculator, we locate the critical value corresponding to \( df = 35 \) and \( \alpha / 2 = 0.05\). This gives us the critical value: \( t_{\alpha / 2} \approx 1.690 \) So, the answer is: \( t_{\alpha / 2} = 1.69 \) (rounded to two decimal places).