Question

Starting salaries of 125 college graduates who have taken a statistics course have a mean of $44,687. suppose the distribution of this population is normal and has a standard deviation of $8,401.

Using an 98% confidence level, find both of the following:
(NOTE: Do not use commas nor dollar signs in your answers.)

(a) The margin of error E:

Answer 1

The margin of error (E) for a 98% confidence level is approximately $1,755.9.

Step-by-step explanation:

To find the margin of error, we can use the critical value for a 98% confidence level multiplied by the standard deviation divided by the square root of the sample size. In this case, the critical value for a 98% confidence level is 2.33.

So, the margin of error (E) is calculated as follows:

E = (critical value) * (standard deviation / sqrt(sample size))

E = 2.33 * (8401 / sqrt(125))

E = 2.33 * 752.610

E ≈ 1755.9