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Chapter 53: Problem 3

Compute a mean value correctly. A researcher finds that the mean vitamin concentration in three replicate samples designated \(A, B\) and \(C\) is \(3.0,2.5\) and \(2.0 \mathrm{mg}\), respectively. He computes the mean vitamin concentration as \(2.5 \mathrm{mg}\), but forgets that the sample sizes were 24,37 and 6, respectively. What is the true mean vitamin concentration? (Answer to three significant figures.)

Short Answer

Expert verified
The true mean vitamin concentration is 2.633mg.

Step by step solution

01

Identify the sample sizes and corresponding mean values

The sample sizes of A, B, and C are 24, 37, and 6, respectively. The corresponding mean vitamin concentrations for A, B, and C are 3.0mg, 2.5mg, and 2.0mg, respectively.
02

Multiply each mean concentration by its sample size

For sample A: 24 samples * 3.0mg/sample = 72mg\nFor sample B: 37 samples * 2.5mg/sample = 92.5mg\nFor sample C: 6 samples * 2.0mg/sample = 12mg
03

Sum up the total amount

Total amount of Vitamin = Amount from A + Amount from B + Amount from C = 72mg + 92.5mg + 12mg = 176.5mg
04

Sum up the total sample size

Total sample size = Size of A + Size of B + Size of C = 24 + 37 + 6 = 67 samples
05

Compute the true mean

Mean = Total amount / Total number = 176.5mg / 67 samples = 2.633mg

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Most popular questions from this chapter

Calculate and interpret standard errors. Two samples, \(A\) and \(B\), gave the following descriptive statistics (measured in the same units): Sample \(A\), mean \(=16.2\), standard deviation \(=12.7\), number of data values \(=12\); Sample B, mean \(=\) 13.2, standard deviation 14.4, number of data values \(=20\). Which has the lower standard error in absolute terms and in proportion to the sample mean? (Express answers to three significant figures.)
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