Find the margin of error for a 90% confidence interval when the standard deviation is LaTeX: \sigma= 50????=50 and LaTeX: n = 25????=25. (Round to three decimal places.)

Question

asked Mar 23, 2021 163k views

1 Answer

Gopal Joshi · Answer 1 · 2021-03-28T13:47:51+0000

Answer:

The margin of error for a 90% confidence interval is 16.4

Explanation:

We are given the following in the question:

Sample size, n = 25

Standard deviation = 50

$z_(critical)\text{ at}~\alpha_(0.10) = \pm 1.64$

Margin of error =

$z_(critical)* (\sigma)/(√(n))$

Putting the values, we get,

1.64* (50)/(√(25)) = 16.4

Thus, the margin of error for a 90% confidence interval is 16.4