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A doctor 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 3 ounces of the sample mean? The standard deviation of the birth weights is known to be 7 ounces.

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Final answer:

To be 95% confident that the true mean is within 3 ounces of the sample mean, the doctor must select a sample size of 21. The formula for the margin of error is E = Z * (σ / sqrt(n)).

Step-by-step explanation:

To determine how large a sample must be selected, we can use the formula for the margin of error (E): E = Z * (σ / sqrt(n)), where Z is the z-score corresponding to the desired confidence level, σ is the standard deviation of the birth weights, and n is the sample size. We know that E = 3 ounces and σ = 7 ounces. Since the desired confidence level is 95%, we can use a Z-score of 1.96, which corresponds to a 95% confidence interval.

Substituting these values into the formula, we can solve for n:

3 = 1.96 * (7 / sqrt(n))

Squaring both sides and solving for n, we get:

n = (1.96^2 * 7^2) / 3^2

n ≈ 20.56

Since the sample size must be a whole number, we round up to the nearest whole number and conclude that she must select a sample size of 21 to be 95% confident that the true mean is within 3 ounces of the sample mean.

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