A researcher wants to estimate the mean serum cholesterol level for college age women. In a past study the stndard devition for this measure was 42 units. If the researcher wants to be off in his estimate by no more than three units ( with 99% confidence), how large must his sample be.?
Formula:
n = [(z-value * sd)/E]^2
...where n = sample size, z-value will be 2.575 using a z-table to represent the 99% confidence interval, sd = 42, E = 3, ^2 means squared, and * means to multiply.
Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.
To determine how large the sample size should be in order to estimate the mean serum cholesterol level for college age women with a margin of error no greater than three units, we can use the formula for sample size calculation:
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 measure
E = the desired margin of error
Given:
σ = 42 units (standard deviation)
E = 3 units (margin of error)
Confidence level = 99%, which corresponds to a z-score of approximately 2.576 (obtained from a z-table or statistical software).
Plugging the values into the formula:
n = (2.576 * 42 / 3)^2
n = (108.192 / 3)^2
n = 36.064^2
n ≈ 1299.77
Since we can't have a fraction of participants, we need to round up to the nearest whole number. Therefore, the researcher would need a sample size of at least 1300 college age women to estimate the mean serum cholesterol level with a margin of error no greater than three units at a 99% confidence level.