A researcher wants to estimate the mean serum cholesterol level for college age women. In a past study the standard deviation for this measure was 42 units. of the researcher wants to be off in his estimate by no more than three units (with 99% confidence) how large must his sample be?

n = [(z-value * sd)/E]^2

n =[ (2.575 * 42)/3]^2

To determine how large the sample size should be in order for the researcher to be off in their estimate by no more than three units with 99% confidence, we can use the formula for the confidence interval of a population mean:

n = ((Z * σ) / E)²

Where:
n = sample size
Z = Z-score corresponding to the desired confidence level (in this case, 99% confidence level)
σ = standard deviation of the population
E = maximum error tolerance (in this case, three units)

First, we need to find the Z-score associated with a 99% confidence level. Since the confidence level is 99%, we need to find the critical value that leaves 1% in the tails of the standard normal distribution. This can be done using a Z-table or a statistical calculator. The Z-score for a 99% confidence level is approximately 2.57.

Next, we have the given information that the standard deviation (σ) for the measure of serum cholesterol is 42 units, and the maximum error tolerance (E) is three units.

Plugging these values into the formula:

n = ((2.57 * 42) / 3)²
n = (108.54 / 3)²
n = 36.18²
n ≈ 1308.97

Therefore, the researcher would need a sample size of approximately 1309 in order to estimate the mean serum cholesterol level for college-age women with a maximum error of three units and 99% confidence. Since sample sizes must be whole numbers, rounding up to the nearest whole number would give a sample size of 1310.