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Chapter 8: Problem 23

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 sample size should be at least 27.

Step by step solution

01

Identify the Confidence Level, Margin of Error, and Standard Deviation

The confidence level is \(99\%\) which corresponds to a critical value or z-score of approximately \(2.58\) in a Normal distribution. This is because we use a two-tail test for this type of estimation. The margin of error (E) is \(.04\) ounce, and the standard deviation (蟽) is \(.20\) ounces.
02

Use the Sample Size Formula

The Sample Size formula is \(n = \frac{Z^2*\sigma^2}{E^2}\), where n is the sample size, Z is the z-value corresponding to the level of confidence, 蟽 is the population standard deviation, and E is the margin of error.
03

Substitute the values into the formula

Insert the identified values into the formula. \[n = \frac{(2.58)^2*(.20)^2}{(.04)^2}\]
04

Calculate the Sample Size, n

Now perform the mathematical calculations. Resulting value for n is \(26.01\).
05

Round up to the Nearest Whole Number

Since sample size cannot be a fraction, round up to guarantee that the desired confidence level is achieved. Therefore, the sample size should be 27.

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Most popular questions from this chapter

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