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A nurse at a local hospital is interested in estimating the birth weight of infants. how large a sample must she select if she desires to be 95 ​% confident that the true mean is within 2 ounces of the sample​ mean? the population standard deviation of the birth weights is known to be 5 ounces.

User Paras Shah
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Final answer:

To be 95% confident the true mean birth weight of infants is within 2 ounces of the sample mean with a known population standard deviation of 5 ounces, the nurse would need to sample 25 infants.

Step-by-step explanation:

The student is interested in determining the size of a sample needed to estimate the true mean birth weight of infants within 2 ounces of the sample mean with 95% confidence when the population standard deviation is known to be 5 ounces. This problem can be solved using the formula for the sample size in a confidence interval estimation for a population mean:

Sample Size (n) = (Z*σ/E)^2

where:

  • Z is the Z-score associated with the desired confidence level,
  • σ (the Greek letter sigma) is the population standard deviation,
  • E is the desired margin of error (the maximum difference allowed between the sample mean and the population mean).

For a 95% confidence level, the Z-score (Z*) is typically 1.96. The population standard deviation (σ) is given as 5 ounces, and the margin of error (E) is 2 ounces.

Plugging these values into the formula:

n = (1.96 * 5 / 2)^2

n = (9.8 / 2)^2

n = (4.9)^2

n = 24.01

Since the sample size must be a whole number, we would round up to the next whole number, which is 25. Thus, the nurse would need a sample of 25 infants to be 95% confident that the true mean birth weight is within 2 ounces of the sample mean.

User Kwabena Berko
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