Question

here are summary statistics for the weights of pepsi in randomly selected cans. n = 36, x̄ = 0.82408 lb, s = 0.00568 lb. use a confidence level of 95% to complete parts (a) and (b) below

a. find the confidence interval estimate of μ.

□ lb < μ < □ lb
(round to five decimal places as needed)

b. write a brief statement that interprets the confidence interval. choose the correct answer below

○ a. one has 95% confidence that the interval from the lower bound to the upper bound contains the true value of the population mean weight of pepsi in a can
○ b. approximately 95% of sample mean weights of pepsi in a can will fall between the lower bound and the upper bound
○ c. there is a 95% chance that the true value of the population mean weight of pepsi in a can will fall between the lower bound and the upper bound
○ d. one has 95% confidence that the sample mean weight of pepsi in a can is equal to the population mean weight of pepsi in a can

Explanation:

Step1: Determine the critical value

Since \(n = 36\) (\(n>30\)), we can use the normal distribution. For a 95% confidence level, the significance level \(\alpha=1 - 0.95=0.05\), and \(\alpha/2=0.025\). The critical value \(z_{\alpha/2}\) is such that \(P(Z>z_{\alpha/2}) = 0.025\). From the standard - normal table, \(z_{\alpha/2}=1.96\).

Step2: Calculate the margin of error \(E\)

The formula for the margin of error \(E\) when the population standard deviation \(\sigma\) is unknown (we use the sample standard deviation \(s\) as an estimate since \(n\) is large) is \(E = z_{\alpha/2}\frac{s}{\sqrt{n}}\).
Given \(n = 36\), \(s=0.00568\), and \(z_{\alpha/2}=1.96\)

$$
LATEXBLOCK0
$$

Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x} + E\). Given \(\bar{x}=0.82408\)

$$
LATEXBLOCK1
$$

Answer:

a. \(0.82222\) lb\(<\mu<0.82594\) lb
b. A. One has 95% confidence that the interval from the lower bound to the upper bound contains the true value of the population mean weight of Pepsi in a can.