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A state meat inspector in lowa has been given the assignment of estimating the mean net weight of packages of ground chuck labeled "3 pounds." Of course, he realizes that the weights cannot be precisely 3 pounds. A sample of 36 packages reveals the mean weight to be 3.01 pounds, with a standard deviation of 0.03 pounds. a. What is the estimated population mean? b. Determine a 95 percent confidence interval for the population mean.

Short Answer

Expert verified
a. 3.01 pounds b. [3.0002, 3.0198] pounds

Step by step solution

01

Estimated Population Mean

Since the mean weight of the sample taken by the inspector is 3.01 pounds, this sample mean is considered the best estimate of the population mean. Therefore, the estimated population mean is 3.01 pounds.
02

Determine the Standard Error (SE)

To calculate the confidence interval, we first need to determine the standard error (SE) of the sample mean. The formula for the standard error is given by \( SE = \frac{s}{\sqrt{n}} \), where \( s = 0.03 \) pounds is the standard deviation of the sample and \( n = 36 \) is the sample size. So, \( SE = \frac{0.03}{\sqrt{36}} = \frac{0.03}{6} = 0.005 \) pounds.
03

Identify the Z-score for 95% Confidence

For a 95% confidence interval, we use a Z-score corresponding to a 95% confidence level. The Z-score for 95% confidence is 1.96.
04

Calculate the Margin of Error (ME)

The margin of error can be calculated using the formula: \( ME = Z \times SE \). Substituting the known values, \( ME = 1.96 \times 0.005 = 0.0098 \) pounds.
05

Determine the Confidence Interval

The confidence interval is calculated by adding and subtracting the margin of error from the sample mean. Thus, the lower limit is \( 3.01 - 0.0098 = 3.0002 \) pounds, and the upper limit is \( 3.01 + 0.0098 = 3.0198 \) pounds. Therefore, the 95% confidence interval for the population mean is \( [3.0002, 3.0198] \) pounds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

population mean
The population mean is a crucial statistic that tells us the average value of a specific characteristic in a population. In this exercise, we are interested in the mean weight of all packages of ground chuck labeled "3 pounds." Instead of measuring each package, a sample is taken as a more practical approach.

The state meat inspector took a sample of 36 packages and found that the average weight was 3.01 pounds. This value becomes the best estimate of the population mean. Therefore, even though we don't check every package, we can confidently use 3.01 pounds as our estimate for the average weight of the entire population of packages.
standard error
The standard error (SE) is a measure that tells us how much the sample mean is likely to vary from the actual population mean. It acts as a bridge between the sample data and the wider population.

The standard error is calculated using the formula:
  • SE = \( \frac{s}{\sqrt{n}} \)
where \( s \) is the standard deviation of the sample (0.03 pounds in this case), and \( n \) is the sample size (36 packages).

Plugging in our values, we find:
SE = \( \frac{0.03}{6} = 0.005 \) pounds.

This tells us that, on average, the estimated sample mean (3.01 pounds) could vary by 0.005 pounds above or below the true population mean.
margin of error
The margin of error provides a range of values above and below the sample mean to help capture where the true population mean is likely to lie. It accounts for the uncertainty in sampling and statistical estimations.

To calculate the margin of error, we use the formula:
  • ME = Z \( \times \) SE
The Z-score is a statistical measurement that reflects how many standard errors you need to go from the sample mean to ensure a specific confidence level.

For a 95% confidence interval, the Z-score is 1.96.
So, ME = \( 1.96 \times 0.005 = 0.0098 \) pounds.

This means we can expect the true population mean to fall within 0.0098 pounds of our sample mean of 3.01 pounds.
sample size
Sample size refers to the number of observations in a sample. In this example, the inspector took a sample of 36 packages. The size of the sample is important because it affects the reliability and accuracy of our estimates.

Larger samples tend to provide more reliable estimates because they better represent the entire population. This is due to the law of large numbers, which suggests that as a sample size increases, its mean will tend to get closer to the population mean.

In our exercise, the sample size of 36 allows us to calculate a meaningful confidence interval and make more precise estimates about the population mean of package weights. Additionally, the larger the sample, the smaller the standard error, allowing for narrower confidence intervals and more precise estimates. Thus, choosing an appropriate sample size is key for accurate statistical analysis.

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Most popular questions from this chapter

A recent study of 50 self-service gasoline stations in the Greater Cincinnati- Northern Kentucky metropolitan area revealed that the mean price of unleaded gas was \(\$ 2.799\) per gallon. The sample standard deviation was \(\$ 0.03\) per gallon. a. Determine a 99 percent confidence interval for the population mean price. b. Would it be reasonable to conclude that the population mean was \(\$ 2.50 ?\) Why or why not?

We want to estimate the population mean within 5 , with a 99 percent level of confidence. The population standard deviation is estimated to be \(15 .\) How large a sample is required?

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A processor of carrots cuts the green top off each carrot, washes the carrots, and inserts six to a package. Twenty packages are inserted in a box for shipment. To test the weight of the boxes, a few were checked. The mean weight was 20.4 pounds, the standard deviation 0.5 pounds. How many boxes must the processor sample to be 95 percent confident that the sample mean does not differ from the population mean by more than 0.2 pound?

A sample of 10 observations is selected from a normal population for which the population standard deviation is known to be 5 . The sample mean is 20 . a. Determine the standard error of the mean. b. Explain why we can use formula \((9-1)\) to determine the 95 percent confidence interval even though the sample is less than \(30 .\) c. Determine the 95 percent confidence interval for the population mean.

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