Determine the margin of error for a 90% confidence interval to estimate the population mean when s = 40 for the sample sizes below. a) n=14
b) n = 28
c) n = 45
a) The margin of error for a 90% confidence interval when n = 14 is
(Round to two decimal places as needed.)
b) The margin of error for a 90% confidence interval when n=28 is
(Round to two decimal places as needed.)
c) The margin of error for a 90% confidence interval when n = 45 is
(Round to two decimal places as needed.)

a) The margin of error for a 90% confidence interval when n = 14 is 18.93.

b) The margin of error for a 90% confidence interval when n=28 is 12.88.

c) The margin of error for a 90% confidence interval when n = 45 is 10.02.

Step-by-step explanation:

The t-distribution is used to solve this question:

a) n = 14

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 14 - 1 = 13

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 13 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.7709

The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.

The margin of error for a 90% confidence interval when n = 14 is 18.93.

b) n = 28

27 df, T = 1.7033

The margin of error for a 90% confidence interval when n=28 is 12.88.

c) The margin of error for a 90% confidence interval when n = 45 is

44 df, T = 1.6802

The margin of error for a 90% confidence interval when n = 45 is 10.02.

the answer is a

step-by-step explanation:

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