/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 The meat department of a local s... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 33

The meat department of a local supermarket chain packages ground beef using meat trays of two sizes: one designed to hold approximately 1 pound of meat, and one that holds approximately 3 pounds. A random sample of 35 packages in the smaller meat trays produced weight measurements with an average of 1.01 pounds and a standard deviation of .18 pound. a. Construct a \(99 \%\) confidence interval for the average weight of all packages sold in the smaller meat trays by this supermarket chain. b. What does the phrase "99\% confident" mean? c. Suppose that the quality control department of this supermarket chain intends that the amount of ground beef in the smaller trays should be 1 pound on average. Should the confidence interval in part a concern the quality control department? Explain.

Short Answer

Expert verified
Answer: The 99% confidence interval for the average weight of smaller meat trays is approximately (0.927, 1.093) pounds. Although the interval includes the desired average weight of 1 pound, the quality control department may want to investigate further since the lower limit is close to 1 pound.

Step by step solution

01

Identify the given information

The problem provides the following information: Sample size (n) = 35 Average weight (\(\bar{x}\)) = 1.01 pounds Standard deviation (s) = 0.18 pound
02

Calculate the t-score for a 99% confidence interval

To calculate the t-score for a 99% confidence interval, we use the t-distribution and the formula: \[t_{\alpha/2, n-1}\] where - \(\alpha\) is the significance level (0.01 for a 99% confidence interval) - n is the sample size Use a t-table or online calculator to find the t-score for 34 degrees of freedom and a 99% confidence interval, which is approximately 2.733.
03

Calculate the margin of error

To calculate the margin of error, use the formula: \[E = t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}}\] where - \(E\) is the margin of error - \(t_{\alpha/2, n-1}\) is the t-score from Step 2 - s is the standard deviation - n is the sample size \[E = 2.733 \times \frac{0.18}{\sqrt{35}} 鈮� 0.083\]
04

Calculate the confidence interval

To calculate the 99% confidence interval, use the formula: \[(\bar{x} - E, \bar{x} + E)\] \[(1.01 - 0.083, 1.01 + 0.083) 鈮� (0.927, 1.093)\]
05

Answer part b

The phrase "99% confident" means that, if we were to take many samples of the same size and calculate confidence intervals for each sample, about 99% of those intervals would contain the true average weight of all packages.
06

Answer part c

Since the supermarket chain intends for the average weight of the smaller meat trays to be 1 pound, the quality control department should examine the confidence interval found in part a. The interval (0.927, 1.093) includes 1 pound, which means that there is a possibility that the average weight of packages is actually 1 pound. However, since the lower limit is close to 1 pound, the quality control department may want to investigate further to ensure that the true average is not below 1 pound.

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