Question

here are summary statistics for the weights of pepsi in randomly selected cans: ( n = 36 ), ( \bar{x} = 0.82412 ) lb, ( s = 0.00567 ) lb. use a confidence level of 99% to complete parts (a) through (d) below.

a. identify the critical value ( t_{alpha/2} ) used for finding the margin of error.

( t_{alpha/2} = square )

(round to two decimal places as needed.)

Explanation:

Step1: Determine degrees of freedom

Degrees of freedom \( df = n - 1 \). Given \( n = 36 \), so \( df = 36 - 1 = 35 \).

Step2: Determine confidence level and significance level

Confidence level is 90%, so significance level \( \alpha = 1 - 0.90 = 0.10 \). Then \( \alpha/2 = 0.05 \).

Step3: Find critical value \( t_{\alpha/2} \)

Using t - distribution table or calculator with \( df = 35 \) and \( \alpha/2 = 0.05 \), we find \( t_{0.05, 35} \approx 1.69 \) (rounded to two decimal places).

Answer:

\( 1.69 \)