Question

Starting salaries of 90 college graduates who have taken a statistics course have a mean of $43,993 and a standard deviation of $9,144. Using 99% confidence level, find the following: A. The margin of error E: _______________________________________ B. The confidence interval for the mean μ: _______________

Answer 1

Answer:

a) E = 2482.90

b) 99% Confidence interval: (41510.1,46475.9)

Step-by-step explanation:

We are given the following in the question:

Sample mean, $\bar{x}$ =$43,993

Sample size, n = 90

Alpha, α = 0.051

Sample standard deviation, s = $9,144

a) The margin of error

Formula:

$z_{critical}\text{ at}~\alpha_{0.01} = 2.576$

$E =z_{critical}\times \displaystyle\frac{s}{\sqrt{n}}\\\\E = 2.576\times \frac{9144}{\sqrt{90}}\\\\E = 2482.90$

b) 99% Confidence interval:  

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$43993 \pm 2482.90 = (41510.1,46475.9)$