point(s) possible
K
Here are summary statistics for the weights of Pepsi in randomly selected cans: $n=36, \bar{x}=0.82412 \mathrm{lb}, s=0.00567 \mathrm{lb}$ 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.)
b. Find the margin of error.
\[
E=\square \mathrm{lb}
\]
(Round to five decimal places as needed.)
c. Find the confidence interval estimate of $\mu$.
$\square \mathrm{lb}< \mu< \square \mathrm{lb}$
(Round to five decimal places as needed.)
d. Write a brief statement that interprets the confidence interval. Choose the correct answer below.
Answer
Final Answer: The critical value \(t_{\alpha / 2}\) used for finding the margin of error is approximately \(\boxed{1.69}\).
Step 1 ：Given values are: sample size \(n = 36\), sample mean \(\bar{x} = 0.82412\) lb, sample standard deviation \(s = 0.00567\) lb, and confidence level \(90\% = 0.90\).
Step 2 ：Degrees of freedom is calculated as \(df = n - 1 = 36 - 1 = 35\).
Step 3 ：Alpha level is calculated as \(1 - \text{confidence level} = 1 - 0.90 = 0.10\).
Step 4 ：For a two-tailed test, we use \(\alpha/2 = 0.10/2 = 0.05\).
Step 5 ：The critical value \(t_{\alpha / 2}\) is found using a t-distribution table or a statistical calculator for \(df = 35\) and \(\alpha/2 = 0.05\). The critical value \(t_{\alpha / 2}\) is approximately 1.69.
Step 6 ：Final Answer: The critical value \(t_{\alpha / 2}\) used for finding the margin of error is approximately \(\boxed{1.69}\).
