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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 136 jugs of detergents for their tests to ensure a 99% confidence level estimation within an error of .04 ounces.

Step by step solution

01

Identify Given Values

The problem gives the following information:\n\n- Population standard deviation (\( \sigma \)): .20 ounces\n- Confidence level: 99% or .99\n- Maximum error (E): .04 ounce\n- The sample size (n) is what we are trying to find
02

Calculate the Z value

The standard advantage for a 99% confidence level is 2.33. This value, known as the Z value, is found on a Z-score table available in most statistics textbooks or online. It corresponds to the number of standard deviations that signifies the 99% confidence level on a normal distribution.
03

Calculate the Sample Size

For determining the needed sample size when the population standard deviation is known, the following formula can be used:\n\n\[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \]\n\nApplying the provided values to this formula:\n\n\[ n = \left( \frac{2.33 \cdot .20}{.04} \right)^2 \]\n\nThis simplifies to:\n\n\[ n = (11.65)^2 = 135.9225 \]
04

Round Up to the Nearest Whole Number

The sample size must be a whole number, as you cannot have a fraction of a sample. Therefore, you always round up to the next whole number even if it's slightly above the decimal. So, the required sample size is 136.

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Most popular questions from this chapter

What assumption(s) must hold true to use the normal distribution to make a confidence interval for the population proportion, \(p\) ?

A random sample of 16 airline passengers at the Bay City airport showed that the mean time spent waiting in line to check in at the ticket counters was 31 minutes with a standard deviation of 7 minutes. Construct a \(99 \%\) confidence interval for the mean time spent waiting in line by all passengers at this airport. Assume that such waiting times for all passengers are normally distributed.

When one is attempting to determine the required sample size for estimating a population mean, and the information on the population standard deviation is not available, it may be feasible to take a small preliminary sample and use the sample standard deviation to estimate the required sample size, \(n\). Suppose that we want to estimate \(\mu\), the mean commuting distance for students at a community college, to a margin of error within 1 mile with a confidence level of \(95 \%\). A random sample of 20 students yields a standard deviation of \(4.1\) miles. Use this value of the sample standard deviation, \(s\), to estimate the required sample size, \(n\). Assume that the corresponding population has a normal distribution.

A company that produces 8 -ounce low-fat yogurt cups wanted to estimate the mean number of calories for such cups. A random sample of 10 such cups produced the following numbers of calories. 147 \(\begin{array}{llllllllll}159 & 153 & 146 & 144 & 148 & 163 & 153 & 143 & 158\end{array}\) Construct a \(99 \%\) confidence interval for the population mean. Assume that the numbers of calories for such cups of yogurt produced by this company have an approximately normal distribution.

A sample selected from a population gave a sample proportion equal to \(.73\) a. Make a \(99 \%\) confidence interval for \(p\) assuming \(n=100\). b. Construct a \(99 \%\) confidence interval for \(p\) assuming \(n=600\). c. Make a \(99 \%\) confidence interval for \(p\) assuming \(n=1500\). d. Does the width of the confidence intervals constructed in parts a through \(\mathrm{c}\) decrease as the sample size increases? If yes, explain why.
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