/*! 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 32 A supermarket chain packages gro... [FREE SOLUTION] | 91Ó°ÊÓ

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A supermarket chain packages ground beef using meat trays of two sizes: one that holds 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 wants the amount of ground beef in the smaller trays to be 1 pound on average. Should the confidence interval in part a concern the quality control department? Explain.

Short Answer

Expert verified
Explain your answer. Answer: The quality control department should not be overly concerned about the confidence interval, as the desired average weight of 1 pound falls within the calculated interval of (0.927, 1.093) pounds. This indicates that the observed sample average of 1.01 pounds is relatively close to the desired average and is not significantly different from the target at a 99% confidence level. However, they should continue to monitor the weight of packages and ensure that the manufacturing process stays within acceptable weight limits.

Step by step solution

01

Identify sample size, mean, and standard deviation

We have the following data available: n = 35, x-bar = 1.01 pounds, and s = 0.18 pounds.
02

Calculate the margin of error

To calculate the margin of error, we use the formula: \(E = t^*_\alpha \cdot (\frac{s}{\sqrt{n}})\) Where \(t^*_\alpha\) is the t-score at the given level of confidence (99% in this case), s is the standard deviation and n is the sample size.
03

Find the t-score

Since the sample size is small (n < 30), we use the t-distribution. For a 99% confidence interval and using df = n-1 = 34, we find the t-score from the t-table as: \(t^*_{0.005} = 2.733\)
04

Calculate Margin of Error

Using the formula, we get: \(E = 2.733 \cdot (\frac{0.18}{\sqrt{35}}) = 0.083\)
05

Construct the confidence interval

We can construct the 99% confidence interval with the margin of error: \((x-bar - E, x-bar + E) = (1.01 - 0.083, 1.01 + 0.083) = (0.927, 1.093)\) So, the 99% confidence interval for the average weight of all packages sold in the smaller meat trays by this supermarket chain is \((0.927, 1.093)\) pounds. #b. What does the phrase \(" 99 \%\) confident" mean?#
06

Definition of Confidence Level

A confidence interval gives an estimated range of values which is likely to include the true population parameter. The confidence level, in this case, 99%, is the probability that the interval will capture the true population mean if we were to repeat the process of taking random samples many times. In other words, if we took 100 different random samples and calculated confidence intervals with 99% confidence, we would expect that 99 out of those 100 intervals would contain the true population mean. #c. Suppose that the quality control department of this supermarket chain wants the amount of ground beef in the smaller trays to be 1 pound on average. Should the confidence interval in part a concern the quality control department? Explain.#
07

Interpret the Confidence Interval

The confidence interval we calculated is \((0.927, 1.093)\) pounds. Notice that the desired average weight of 1 pound falls within this interval. This means that the observed sample average of 1.01 pounds is relatively close to the desired average and is not significantly different from the target at a 99% confidence level. Therefore, the quality control department should not be overly concerned about the confidence interval. However, they should continue to monitor the weight of packages and ensure that the manufacturing process remains within acceptable weight limits.

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Key Concepts

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

Sample Size
Understanding the role of sample size in statistics is vital for interpreting results accurately. When considering confidence intervals, the sample size () is an essential factor because it influences the margin of error and the width of the confidence interval. A larger sample size generally leads to a smaller margin of error, which means the estimates are more precise. In the case of the supermarket chain's packages, the sample size of 35 helps to provide a reasonable estimation for the average weight, but increasing the sample size could yield an even narrower confidence interval, offering a more accurate reflection of the true population mean.

In educational efforts, it's important to emphasize that a properly chosen sample size should balance practical considerations, such as cost and time, with the need for precision. For small sample sizes (<30), statisticians use the t-distribution to account for additional uncertainty in the estimate of the standard deviation. This is because small samples may not provide a reliable estimate of the population standard deviation.
Standard Deviation
The term standard deviation () represents the amount of variability or dispersion in a set of data values. A low standard deviation means that the data points tend to be close to the mean (average) of the set, while a high standard deviation indicates that the data points are spread out over a wider range. In context, the supermarket chain's data had a standard deviation of 0.18 pounds, suggesting that the weights of the ground beef packages were fairly consistent around the sample mean of 1.01 pounds.

When explaining the concept, it's good practice to stress that the standard deviation can greatly affect the margin of error in a confidence interval calculation. A smaller standard deviation would result in a smaller margin of error and a more precise confidence interval, indicating that the sample mean is a reliable estimate of the population mean.
T-Distribution
The t-distribution is a probability distribution that is used when estimating population parameters, such as the mean, when the sample size is small and the population standard deviation is unknown. It resembles the standard normal distribution but has heavier tails, meaning it accounts for the increased uncertainty associated with small sample sizes. The concept of t-distribution is crucial when constructing confidence intervals from small sample sizes, typically defined as less than 30 observations.

In our exercise, because we have a sample size of 35, which is close to 30, the t-distribution is used instead of the Z-distribution to calculate the critical t-value, which factors into the margin of error. As the sample size grows, the t-distribution approaches the normal distribution. Thus, highlighting the difference in distributions based on sample size is important when teaching students about constructing confidence intervals.
Margin of Error
The margin of error reflects the range of values below and above the sample statistic in a confidence interval. It quantifies the level of uncertainty associated with the estimate of a population parameter. The formula to calculate it is dependent on the critical value from the t-distribution (or Z-distribution for large samples), the standard deviation, and the sample size.

In our exercise example, the margin of error is engineered using the t-score and represents how far off the estimated average weight of the meat packages could be from the true average. Understanding that the margin of error affects the width of the confidence interval is fundamental. A smaller margin of error means a more precise estimate but requires a larger sample size or a smaller standard deviation. The example provided a margin of error of 0.083, indicating that the true mean weight of the packages is likely within this range from the sample mean.
Statistical Significance
Lastly, the concept of statistical significance is crucial in determining whether a statistical result reflects a real effect or is due to random chance. It helps in making decisions or inferences about a population based on sample data. Statistical significance is often associated with a probability threshold (the alpha level), such as 0.01 (for a 99% confidence level) or 0.05 (for a 95% confidence level).

For the supermarket's quality control team, statistical significance plays a role in determining whether the average weight of ground beef in the smaller trays sufficiently differs from the target weight of 1 pound. Since the confidence interval of (0.927, 1.093) pounds includes the target value, there might not be statistically significant evidence to suggest that the actual average weight is different from the target. This indicates that any deviation from 1 pound seen in the sample is likely due to natural sample variation, and not an indication of a systemic issue with the packing process.

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Most popular questions from this chapter

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Find the z-values needed to calculate large-sample confidence intervals for the confidence levels given. Confidence coefficient \(1-\alpha=.98\)

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