Question

here are summary statistics for the weights of pepsi in randomly selected cans: ( n = 36 ), ( overline{x}=0.82412 \text{lb} ), ( s = 0.00572 \text{lb} ). use a confidence level of ( 95% ) 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: Calculate the significance level

The confidence level is \(C = 0.95\). The significance level \(\alpha=1 - C=1 - 0.95 = 0.05\).

Step2: Calculate the degrees of freedom

The sample size is \(n = 36\). The degrees of freedom \(df=n - 1=36 - 1 = 35\).

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

We have \(\alpha/2=\frac{0.05}{2}=0.025\). Using a \(t\) - distribution table or a calculator with a \(t\) - distribution function (e.g., in Excel: T.INV.2T(0.05,35) or in R: qt(0.025,35,lower.tail = FALSE)), we find that \(t_{0.025,35}\approx 2.03\).

Answer:

\(2.03\)