You are sampling from a population with a known standard deviation of 20 and want to construct a 95% confidence interval with a margin of error of no more than 4. What is the smallest sample that will produce such an interval?
4 > 1.96 (20/sqrt n)
20^2 / ((4/1.96)^2) = 96.04
n>96.04. so at least 97
is this correct?
Yes, your calculation is correct. To determine the smallest sample size needed to construct a 95% confidence interval with a margin of error of no more than 4, you can use the formula:
Margin of Error = Z * (Standard Deviation / √n)
In this case, you want the margin of error to be no more than 4, so you have:
4 = 1.96 * (20 / √n)
By rearranging the formula, you get:
√n = (1.96 * 20) / 4
Squaring both sides, you have:
n = (1.96^2 * 20^2) / 4^2
Simplifying further:
n = (3.8416 * 400) / 16
n = 96.064
Since the sample size must be a whole number, you need to round up to the nearest whole number. Therefore, the smallest sample size needed to construct the desired confidence interval is 97.
Yes, your calculation is correct. The formula for calculating the margin of error in a confidence interval is given by:
Margin of Error = Critical Value * Standard Deviation / (Square Root of Sample Size)
In this case, you want the margin of error to be no more than 4, and the known standard deviation is 20. The critical value for a 95% confidence interval is 1.96. Plugging these values into the formula, we get:
4 = 1.96 * (20 / sqrt(n))
Solving for n, we find:
n > (20^2) / ((4/1.96)^2)
n > 96.04
Therefore, the smallest sample size that will produce a 95% confidence interval with a margin of error no more than 4 is at least 97.