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 her estimate is within 2 ounces of the true mean? Assume that s=7 ounces based on earlier studies.
To determine the sample size needed, we can use the formula for sample size estimation for means:
n = (Z * σ / E)^2
Where:
n = required sample size
Z = Z-score for the desired level of confidence
σ = population standard deviation
E = margin of error
In this case, the doctor wants to be 98% confident, so the Z-score for a 98% confidence level is 2.33 (you can find this value in a Z-table or use a statistical calculator).
The desired margin of error is 2 ounces.
The population standard deviation (σ) is given as 7 ounces.
Let's plug these values into the formula:
n = (2.33 * 7 / 2)^2
n = (16.31 / 2)^2
n = 8.16^2
n ≈ 66.63
Since the sample size must be a whole number, we round up to the nearest whole number.
Therefore, the doctor needs to select a sample size of at least 67 infants to be 98% confident that her estimate of birth weight is within 2 ounces of the true mean.
To determine the sample size required for estimating the birth weight of infants, we can use the formula for the confidence interval:
n = (Z * σ / E)^2
Where:
n = sample size
Z = Z-score for the desired confidence level (98% = 2.33)
σ = population standard deviation (s = 7 ounces)
E = maximum error tolerance (2 ounces)
Plugging in the values:
n = (2.33 * 7 / 2)^2
n = (16.31 / 2)^2
n = 8.155^2
n ≈ 66.61
Therefore, the doctor must select a sample size of at least 67 infants to be 98% confident that her estimate is within 2 ounces of the true mean.