A doctor wants to estimate the HDL cholesterol of all 20-29 year old females. How many subjects are needed to estimate the HDL cholesterol within 2 points with 99%confidence assuming s=17.8 based on earlier studies? Suppose the doctor would be content with 90% confidence. How does the decrease in cofidence affect the sample size required?
A: A 99% confidence level requires ___ subjects
B: A 90% confidence level requires ___ subjects
A doctor wants to estimate the mean HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the mean HDL cholesterol within 3 points with 99 % confidence assuming s equals 16.7 based on earlier studies? Suppose the doctor would be content with 90 % confidence. How does the decrease in confidence affect the sample size required?
A 99% confidence level requires "206" Subject
A 99% confidence level requires "206" subject
A doctor wants to estimate the mean HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the mean HDL cholesterol within 44 points with 99 %99% confidence assuming s equals 11.2s=11.2 based on earlier studies? Suppose the doctor would be content with 90 %90% confidence. How does the decrease in confidence affect the sample size required?
To estimate the required sample size, we can use the formula for sample size estimation for estimating a mean:
n = (Z * s / E)²
Where:
n = required sample size
Z = Z-score corresponding to the desired confidence level
s = standard deviation of the population (17.8 in this case)
E = desired margin of error (2 points in this case)
A: For a 99% confidence level, we need to find the corresponding Z-score. The Z-score for a 99% confidence level is the value that leaves 0.5% of the area (or probability) in each tail of the distribution. Looking up this value in a standard normal distribution table, the Z-score is approximately 2.576. Substituting this value along with s=17.8 and E=2 into the formula, we get:
n = (2.576 * 17.8 / 2)²
n ≈ 319.52
Therefore, a 99% confidence level requires approximately 320 subjects.
B: For a 90% confidence level, we need to find the corresponding Z-score. The Z-score for a 90% confidence level is the value that leaves 5% of the area (or probability) in each tail of the distribution. Looking up this value in a standard normal distribution table, the Z-score is approximately 1.645. Substituting this value along with s=17.8 and E=2 into the formula, we get:
n = (1.645 * 17.8 / 2)²
n ≈ 78.94
Therefore, a 90% confidence level requires approximately 79 subjects.
The decrease in confidence level from 99% to 90% reduces the required sample size. This happens because as the confidence level decreases, the margin of error becomes larger, allowing for a smaller sample size to achieve the same level of precision. In other words, with a lower confidence level, we can have a wider range of estimates, which requires a smaller sample size.
2.58 = 99% confidence interval
1.645 = 90% confidence interval
Use a formula to find sample size.
Here is one:
n = [(z-value * sd)/E]^2
...where n = sample size, z-value = values above for the intervals needed, sd = 17.8, E = .02, ^2 means squared, and * means to multiply.
Plug the values into the formula and finish the calculations. Round your answers to the next highest whole number.
I'll let you take it from here.