Question

The number of DVDs in a random person’s home collection is counted for a sample population of 80 people. The mean of the sample is 52 movies; the entire population is known to have a standard deviation of 12 movies. Assuming a 99% confidence level, find the margin of error.

Answer 1

Answer:

$E = 3.46\ movies$

Step-by-step explanation:

The formula to find the error is:

$E = z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

Where:

$\sigma$ is the standard deviation

n is the sample size

So

n = 80 people

$\sigma$ = 12 movies

Then

$1- \alpha$ = confidence level = 0.99

$\alpha= 1-0.99$

$\alpha = 0.01\\\\\frac{\alpha}{2} = 0.005$

We look for the Z value: $Z_{0.005}$

$Z_{0.005}=2.58$  Looking in the normal standard tables

Therefore:

$E =2.58*\frac{12}{\sqrt{80}}\\\\E = 3.46\ movies$