A researcher wants to estimate the mean blood cholesterol level of young men ages 15-25 with a 97.59% confidence interval. The blood cholesterol level of young men follows a Normal distribution with standard deviation σ = 15 mg/dl. How large a sample would the researcher need to take to estimate the mean blood cholesterol to within ±9 mg/dl?
To calculate the sample size required to estimate the mean blood cholesterol level within ±9 mg/dl with a 97.59% confidence interval, we can use the sample size formula for estimating population mean:
n = [(Z * σ) / E]^2
Where:
n = Sample size
Z = Z-score corresponding to the desired confidence level
σ = Population standard deviation
E = Desired margin of error
In this case:
Z = Z-score corresponding to a 97.59% confidence level. To find this value, we need to find the Z-score that corresponds to a confidence level of 97.59% and account for the two-tailed nature of the confidence interval.
Using a Z-table or a statistical calculator, we can find that the Z-score for a 97.59% confidence level is approximately 2.57.
σ = Population standard deviation = 15 mg/dl
E = Desired margin of error = ±9 mg/dl
Now, we can substitute these values into the formula:
n = [(Z * σ) / E]^2
n = [(2.57 * 15) / 9]^2
n = (38.55 / 9)^2
n = 4.283^2
n = 18.29
Therefore, the researcher would need to take a sample size of at least 19 (rounded up from 18.29) to estimate the mean blood cholesterol level within ±9 mg/dl with a 97.59% confidence interval.