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 98% confident that the true mean is within 2 ounces of the sample mean? The standard deviation of the birth weights is known to be 5 ounces.

A) 34
B) 5
C) 33
D) 6

To determine the sample size needed for estimating the birth weight of infants with a 98% confidence level and a margin of error of 2 ounces, you can use the formula for sample size determination:

n = (Z * σ / E)^2

Where:
n = sample size needed
Z = z-score corresponding to the desired confidence level (in this case, 98% confidence level corresponds to a z-score of 2.33, as per the standard normal distribution table)
σ = standard deviation of the birth weights (5 ounces)
E = margin of error (2 ounces)

Plugging in the values into the formula:

n = (2.33 * 5 / 2)^2
n = (11.65 / 2)^2
n = 5.825^2
n ≈ 34

Therefore, the doctor must select a sample size of approximately 34.

The answer is A) 34

To determine the sample size needed to estimate the mean birth weight with a desired level of confidence and a given margin of error, you can use the formula for sample size estimation:

n = (Z * σ / E)^2

Where:
n = the required sample size
Z = the Z-score corresponding to the desired confidence level
σ = the standard deviation of the population
E = the desired margin of error

In this case, the doctor wants to be 98% confident (which corresponds to a Z-score of 2.33 for a two-tailed test), the standard deviation is known to be 5 ounces, and the desired margin of error is 2 ounces.

Plugging these values into the formula:

n = (2.33 * 5 / 2)^2
n ≈ 34.23

Rounding up to the nearest whole number, the required sample size is 35.

So, the correct answer is A) 34.

3.2

You want mean ± 2.33 SEm for 98% confidence interval.

SEm = SD/√n

I'll leave the rest to you.