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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.

Respuesta :

luisejr77

Answer:

[tex]E = 3.46\ movies[/tex]

Step-by-step explanation:

The formula to find the error is:

[tex]E = z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}[/tex]

Where:

[tex]\sigma[/tex] is the standard deviation

n is the sample size

So

n = 80 people

[tex]\sigma[/tex] = 12 movies

Then

[tex]1- \alpha[/tex] = confidence level = 0.99

[tex]\alpha= 1-0.99[/tex]

[tex]\alpha = 0.01\\\\\frac{\alpha}{2} = 0.005[/tex]

We look for the Z value: [tex]Z_{0.005}[/tex]

[tex]Z_{0.005}=2.58[/tex]  Looking in the normal standard tables

Therefore:

[tex]E =2.58*\frac{12}{\sqrt{80}}\\\\E = 3.46\ movies[/tex]

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