A random sample with n20 provided a sample mean of 18 and a
A random sample with n=20 provided a sample mean of 18 and a sample deviation of 5. What is the margin of error for a 95% confidence interval? (Not the confidence interval itself)
A random sample with n=20 provided a sample mean of 18 and a sample deviation of 5. What is the margin of error for a 95% confidence interval? (Not the confidence interval itself)
Solution
a)
Note that
margin of error = t*s/sqrt(n)
Here, as df = n - 1 = 19, then for a 95% confidence interval,
t = 2.093024054
Thus,
E = t*s/sqrt(n) = 2.093024054*5/sqrt(20)
= 2.340072032 [ANSWER]
