Question

Determine the margin of error for a confidence interval to estimate the population mean with n=18 and s=12.1 for the confidence levels below.

A) The margin of error for an 80% confidence interval is:

B) The margin of error for an 90% confidence interval is:

C) The margin of error for an 99% confidence interval is:

Answer 1

Sample size = n = 18

Standard deviation = s = 12.1

A)

We have to find the margin of error for an 80% confidence interval.

Here E is a margin of error.

Degrees of freedom = n - 1 = 18 - 1 = 17

Level of significance = 0.20

tc = 1.333  ( Using t table)

---------------------------------------------------------------------------------------------------------------------------------

B)

We have to find the margin of error for a 90% confidence interval.

Here E is a margin of error.

Degrees of freedom = n - 1 = 18 - 1 = 17

Level of significance = 0.10

tc = 1.740    ( Using t table)

-------------------------------------------------------------------------------------------------------

We have to find the margin of error for a 99% confidence interval.

Here E is a margin of error.

Degrees of freedom = n - 1 = 18 - 1 = 17

Level of significance = 0.01

tc = 2.898    ( Using t table)

----------------------------------------------------------------------------------

If you have any doubt please comment!