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]

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 confiden

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