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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.

A) 59.4671

B) 2.6296

C) 38.6536

D) 3.4547​

Sagot :

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

The formula to find the error is:

Where:

is the standard deviation

n is the sample size

So

n = 80 people

= 12 movies

Then

= confidence level = 0.99

We look for the Z value:

 Looking in the normal standard tables

Therefore: