contestada

Internet Viewing A researcher wishes to estimate the average number of minutes per day a person spends on the Internet. How olo
large a sample must she select if she wishes to be 99% confident that the population mean is within 10 minutes of the sample
mean? Assume the population standard deviation is 38 minutes. Round your final answer up to the next whole number.

Respuesta :

Using the z-distribution, as we have the population standard deviation, it is found that a sample of 96 people must be taken.

What is the margin of error for the z-distribution?

It is given by:

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which:

  • z is the critical value.
  • [tex]\sigma[/tex] is te population standard deviation.
  • n is the sample size.

In this problem, we have a 99% confidence level, hence[tex]\alpha = 0.99[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.99}{2} = 0.995[/tex], so the critical value is z = 2.575.

Additionally, we have that [tex]\sigma = 38, M = 10[/tex], hence we solve for n to find the minimum sample size needed.

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]10 = 2.575\frac{38}{\sqrt{n}}[/tex]

[tex]10\sqrt{n} = 38 \times 2.575[/tex]

[tex]\sqrt{n} = 3.8 \times 2.575[/tex]

[tex](\sqrt{n})^2 = (3.8 \times 2.575)^2[/tex]

n = 95.7

Rounding up, a sample of 96 people must be taken.

More can be learned about the z-distribution at https://brainly.com/question/25890103

Otras preguntas