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Chapter 8: Problem 31

A company that produces detergents wants to estimate the mean amount of detergent in 64 -ounce jugs at a \(99 \%\) confidence level. The company knows that the standard deviation of the amounts of detergent in all such jugs is \(.20\) ounce. How large a sample should the company select so that the estimate is within \(.04\) ounce of the population mean?

Short Answer

Expert verified
The company should select a sample size of 1035 jugs to estimate the mean amount within .04 ounce of the population mean at the 99% confidence level.

Step by step solution

01

Understand the Problem and Given Information

In this problem, we are asked to calculate the sample size. The given information includes: \n 1. Confidence level: \(99\%\) \n 2. Standard deviation (\(\sigma\)): \(.20\) ounce \n 3. Margin of Error (\(E\)): \(.04\) ounce \n\nWe use the confidence level to find the Z value from the standard normal distribution table. For a \(99\%\) confidence level, \(Z = 2.57 \).
02

Formulate the Sample Size Formula

The sample size formula in statistics is derived on the basis that the sample mean follows a normal distribution. It is given by: \n\n\[n = \left( \frac{Z \cdot \sigma}{E} \right)^2\] \nWhere: \n \(n\) = required sample size \n \(Z\) = standard normal value corresponding to the desired confidence level \n \(\sigma\) = population standard deviation \n \(E\) = desired margin of error
03

Substitute Given Values into the Formula

Substitute the given values into the formula: \n\[n = \left( \frac{2.57 \cdot .20}{.04} \right)^2\]
04

Calculate The Sample Size

Solve the mathematical expression to obtain the value of \(n\). Make sure to round up the result as it's impossible to have a fraction of a sample. It results in \(n = 1034.42\), but we round to \(1035\) because the number of samples must be a whole number.

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Key Concepts

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

Confidence Interval
A confidence interval is a range of values that estimates the true value of a population parameter. It gives us an upper and lower limit around an estimate. The width of this interval depends on three factors: the level of confidence, the standard deviation, and the sample size.

Confidence intervals allow you to say, "I am X% confident that the true mean falls within this range." In this scenario involving detergent production, the company seeks a 99% confidence interval. This high confidence level means they require a narrow interval, leading to a larger sample size to ensure accuracy.
  • The interval gives a sense of reliability of the estimation.
  • 99% confidence means there is only a 1% chance that the true mean does not lie within this range.
  • Greater confidence levels necessitate more data, hence larger samples.
Margin of Error
The margin of error represents the amount by which your sample estimate may differ from the actual population value. Lowering the margin of error provides a more precise estimate. In the context of the detergent company, the margin of error is set at 0.04 ounces.

A smaller margin of error results in a higher level of precision in the results but usually requires a larger sample size.
  • This is the maximum allowable difference from the sample mean to the population mean.
  • Reducing the margin of error often necessitates increasing the sample size.
  • It's crucial in ensuring that results are within an acceptable range.
Z-value
The Z-value, or Z-score, is a statistical measurement representing the number of standard deviations a data point is from the mean. It is essential in determining the size of the confidence interval. The Z-value correlates directly with the confidence level and can be found in a standard normal distribution table.

For a 99% confidence level, the Z-value is 2.57. This number reflects the area under a normal distribution curve.
  • The Z-value affects the width of the confidence interval.
  • Higher Z-values correspond to wider confidence intervals and more confidence in the estimate.
  • Identifying the correct Z-value is crucial for accurate sample size calculation.
Population Standard Deviation
The population standard deviation (\(\sigma\)) measures the amount of variation or dispersion of a set of data values in the entire population. It is a critical element in the formula for sample size calculation.

In our case, the population standard deviation for the detergent jugs is 0.20 ounces. This value influences how wide or narrow the confidence interval will be.

A higher standard deviation indicates more spread among the data points.
  • It's a fixed quantity that represents variability within the population.
  • Standard deviation alters the margin of error and, consequently, the required sample size.
  • The larger the standard deviation, the larger the sample size required for a given confidence and error margin.

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Most popular questions from this chapter

A Centers for Disease Control and Prevention survey about cell phone use noted that \(14.7 \%\) of U.S. households are wireless-only, which means that the household members use only cell phones and do not have a landline. Suppose that this percentage is based on a random sample of 855 U.S. households (Source: http://www.cdc.gov/nchs/data/nhsr/nhsr014.pdf). a. Construct a \(95 \%\) confidence interval for the proportion of all U.S. households that are wireless-only. b. Explain why we need to construct a confidence interval. Why can we not simply say that \(14.7 \%\) of all U.S. households are wireless-only?

You want to estimate the proportion of students at your college who hold off- campus (part-time or full-time) jobs. Briefly explain how you will make such an estimate. Collect data from 40 students at your college on whether or not they hold off-campus jobs. Then calculate the proportion of students in this sample who hold off-campus jobs. Using this information, estimate the population proportion. Select your own confidence level.

a. A sample of 1100 observations taken from a population produced a sample proportion of \(.32 .\) Make a \(90 \%\) confidence interval for \(p\). b. Another sample of 1100 observations taken from the same population produced a sample proportion of .36. Make a \(90 \%\) confidence interval for \(p\). c. A third sample of 1100 observations taken from the same population produced a sample proportion of .30. Make a \(90 \%\) confidence interval for \(p\). d. The true population proportion for this population is \(.34 .\) Which of the confidence intervals constructed in parts a through c cover this population proportion and which do not?

A group of veterinarians wants to test a new canine vaccine for Lyme disease. (Lyme disease is transmitted by the bite of an infected deer tick.) In an area that has a high incidence of Lyme disease, 100 dogs are randomly selected (with their owners' permission) to receive the vaccine. Over a 12 -month period, these dogs are periodically examined by veterinarians for symptoms of Lyme disease. At the end of 12 months, 10 of these 100 dogs are diagnosed with the disease. During the same 12 -month period, \(18 \%\) of the unvaccinated dogs in the area have been found to have Lyme disease. Let \(p\) be the proportion of all potential vaccinated dogs who would contract Lyme disease in this area. a. Find a \(95 \%\) confidence interval for \(p\). b. Does \(18 \%\) lie within your confidence interval of part a? Does this suggest the vaccine might or might not be effective to some degree? c. Write a brief critique of this experiment, pointing out anything that may have distorted the results or conclusions.

A mail-order company promises its customers that the products ordered will be mailed within 72 hours after an order is placed. The quality control department at the company checks from time to time to see if this promise is fulfilled. Recently the quality control department took a sample of 50 orders and found that 35 of them were mailed within 72 hours of the placement of the orders. a. Construct a \(98 \%\) confidence interval for the percentage of all orders that are mailed within 72 hours of their placement. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?
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