Question

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.)

Answer 1

Answer:

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 $1 - \frac{1 - 0.9}{2} = 0.95$. So we have T = 1.7709

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.7709\frac{40}{\sqrt{14}} = 18.93$

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

$M = T\frac{s}{\sqrt{n}} = 1.7033\frac{40}{\sqrt{28}} = 12.88$

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

$M = T\frac{s}{\sqrt{n}} = 1.6802\frac{40}{\sqrt{45}} = 10.02$

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