Question

here are summary statistics for the weights of pepsi in randomly selected cans: ( n = 36 ), ( overline{x}=0.82414 \text{lb} ), ( s = 0.00572 \text{lb} ). use a confidence level of ( 90% ) to complete parts (a) and (b) below.

a. find the confidence interval estimate of ( mu ).

( square \text{lb}<mu<square \text{lb} )
(round to five decimal places as needed.)

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

( \bigcirc ) a. there is a ( 90% ) 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

( \bigcirc ) b. approximately ( 90% ) of sample mean weights of pepsi in a can will fall between the lower bound and the upper bound

( \bigcirc ) c. one has ( 90% ) confidence that the sample mean weight of pepsi in a can is equal to the population mean

Explanation:

Step1: Determine the critical value

Since \(n = 36\) (\(n>30\)), we can use the normal distribution (approximate \(t -\)distribution). For a \(90\%\) confidence level, the significance level \(\alpha=1 - 0.90=0.10\), and \(\alpha/2 = 0.05\).
The critical value \(z_{\alpha/2}\) from the standard normal distribution table is \(z_{0.05}\approx1.645\)

Step2: Calculate the margin of error

The formula for the margin of error \(E\) is \(E = z_{\alpha/2}\frac{s}{\sqrt{n}}\)
Given \(n = 36\), \(s=0.00572\), \(z_{\alpha/2}=1.645\)
\(E=1.645\times\frac{0.00572}{\sqrt{36}}=1.645\times\frac{0.00572}{6}\)
\(E = 1.645\times0.0009533\approx0.00157\)

Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x} + E\)
Given \(\bar{x}=0.82414\)
\(\bar{x}-E=0.82414- 0.00157=0.82257\)
\(\bar{x}+E=0.82414 + 0.00157=0.82571\)

A confidence interval gives a range of values within which we are confident the population parameter lies. A \(90\%\) confidence interval means that if we were to take many samples and construct confidence intervals in the same way, about \(90\%\) of those intervals would contain the true population parameter. It is not about the probability that the population parameter is in a single - constructed interval (the population parameter is a fixed value, not a random variable), nor is it about sample means (except in the sense of the sampling distribution used to construct the interval).

Answer:

a. \(0.82257\) lb\(<\mu<0.82571\) lb
b. None of the options A, B, C are correct. The correct interpretation is: One has \(90\%\) confidence that the interval from \(0.82257\) lb to \(0.82571\) lb actually contains the true value of the population mean weight of Pepsi in a can.

If we assume there is a typo and we have to choose from the given options (even though they are not perfectly correct in a strict sense):
For option A: The population mean is a fixed value. We don't say there is a \(90\%\) chance (in the frequentist interpretation, the probability is either \(0\) or \(1\) for a single - constructed interval). But if we use the layman's understanding of confidence interval (a common mis - interpretation but widely used in some contexts), option A is the closest.
For option B: It's about the population mean, not sample means.
For option C: The sample mean is just an estimate, and the confidence interval is about the range for the population mean.