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

`E = 3.46\ movies`

Explanation:

The formula to find the error is:

`E = z_{(\alpha)/(2)}(\sigma)/(√(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\\\\(\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*(12)/(√(80))\\\\E = 3.46\ movies`