Question

here are summary statistics for the weights of pepsi in randomly selected cans: ( n = 36 ), ( overline{x}=0.82407 ) 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 ( mu ).
( square mathrm{lb}<mu<square mathrm{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 sample mean weight of pepsi in a can is equal to the population mean weight of pepsi in a can.
b. 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.
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. approximately ( 95 % ) of sample mean weights of pepsi in a can will fall between the lower bound and the upper bound.

Explanation:

Step1: Determine the critical value

Since the sample size \(n = 36\) (\(n>30\)), we can use the normal distribution (approximate \(t -\)distribution with large \(n\)). For a \(95\%\) confidence level, \(\alpha=1 - 0.95=0.05\), and \(\alpha/2=0.025\). The critical value \(z_{\alpha/2}\) from the standard normal distribution table is \(z_{0.025}=1.96\)

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

The formula for the margin of error \(E\) is \(E = z_{\alpha/2}\times\frac{s}{\sqrt{n}}\)
Given \(s = 0.00568\), \(n = 36\), and \(z_{\alpha/2}=1.96\)
\(E=1.96\times\frac{0.00568}{\sqrt{36}}=1.96\times\frac{0.00568}{6}\)
\(E = 1.96\times0.0009467\approx0.001855\)

Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x} + E\)
Given \(\bar{x}=0.82407\)
\(\bar{x}-E=0.82407- 0.001855=0.822215\)
\(\bar{x}+E=0.82407 + 0.001855=0.825925\)

A confidence interval gives a range of values within which we are confident the population parameter lies. A \(95\%\) confidence interval means that if we were to take many samples and construct confidence intervals in the same way, approximately \(95\%\) of those intervals would contain the true population parameter.

For option A: The sample mean is just a single - point estimate. A confidence interval is about the population mean, not stating that the sample mean is equal to the population mean.
For option B: A \(95\%\) confidence interval implies that we are \(95\%\) confident that the interval (calculated using the formula \(\bar{x}\pm E\)) contains the True value of the population mean.
For option C: It's not about a \(95\%\) chance for a single interval. The confidence level refers to the long - run frequency of intervals that contain the True parameter.
For option D: A confidence interval for the population mean is not about sample means. It's about the population mean.

Answer:

a. \(0.82222<\mu<0.82593\)
b. B. 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.