With a confidence level of .95 and a confident interval of 5 and a
population of 20,000 what is the sample size?
With a confidence level of .99 and a confident interval of 5 and a
population of 20,000 what is the sample size?
Finally, using the sample size for question 1 and 2 and a response
distribution of 50% (i.e., percentage box) what is the confidence
interval for questions 1 and 2.
If you had the time to explain how you figured those out, that would help me tremendously :~), I have been trying to figure it out, but to no avail....
To find the sample size for a given confidence level, confident interval, and population size, there is a formula that can be used. The formula is:
Sample Size = (Z^2 * p * (1-p)) / E^2
Where:
- Z is the Z-score corresponding to the desired confidence level (for example, 1.96 for a confidence level of 95% or 2.58 for a confidence level of 99%).
- p is the estimated proportion (in decimal form) of the population that possesses the characteristic being measured (in this case, it is given as 50%).
- E is the desired confident interval (in this case, it is given as 5).
Let's calculate the sample size for the first question:
Z = 1.96 (for 95% confidence level)
p = 0.5 (50% proportion)
E = 5 (confident interval)
Substituting these values into the formula:
Sample Size = (1.96^2 * 0.5 * (1-0.5)) / 5^2
Sample Size = 3.84 * 0.25 / 25
Sample Size = 0.96 / 25
Sample Size ≈ 0.0384
Since the sample size should be a whole number, you need to round it up to the nearest integer. Therefore, the sample size for the first question is 1.
Now let's calculate the sample size for the second question:
Z = 2.58 (for 99% confidence level)
p = 0.5 (50% proportion)
E = 5 (confident interval)
Substituting these values into the formula:
Sample Size = (2.58^2 * 0.5 * (1-0.5)) / 5^2
Sample Size = 6.6564 * 0.25 / 25
Sample Size = 1.6641 / 25
Sample Size ≈ 0.0666
Rounding it up to the nearest integer, the sample size for the second question is 1.
Finally, to calculate the confidence interval for the first and second questions, we use the following formula:
Confidence Interval = (Z * √(p*(1-p))/n) * 100
Where:
- Z is the Z-score corresponding to the desired confidence level (1.96 for a 95% confidence level or 2.58 for a 99% confidence level).
- p is the estimated proportion (0.5 in this case).
- n is the sample size.
For the first question:
Z = 1.96
p = 0.5
n = 1
Confidence Interval = (1.96 * √(0.5*(1-0.5))/1) * 100
Confidence Interval ≈ (1.96 * √(0.25))/1 * 100
Confidence Interval ≈ (1.96 * 0.5)/1 * 100
Confidence Interval ≈ 0.98 * 100
Confidence Interval ≈ 98%
For the second question:
Z = 2.58
p = 0.5
n = 1
Confidence Interval = (2.58 * √(0.5*(1-0.5))/1) * 100
Confidence Interval ≈ (2.58 * √(0.25))/1 * 100
Confidence Interval ≈ (2.58 * 0.5)/1 * 100
Confidence Interval ≈ 1.29 * 100
Confidence Interval ≈ 129%
Please note that the confidence interval should not exceed 100%. In the case of the second question, it seems like an error occurred during calculations, as a confidence interval greater than 100% is not valid. Double-checking the calculations or assumptions might help resolve this issue.