Question

here are summary statistics for the weights of pepsi in randomly selected cans: n = 36, \\( \overline { x } = 0.82409 \\) lb, s = 0.00569 lb. use a confidence level of 90% 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 \(C = 0.90\). The significance level \(\alpha=1 - C=1 - 0.90 = 0.10\).

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 \(\alpha/2=\frac{0.10}{2}=0.05\) and \(df = 35\). Using a \(t\) - distribution table or a calculator (e.g., TI - 84: invT(1 - 0.05,35)), we find \(t_{0.05,35}\approx1.69\).

Answer:

\(1.69\)