Question

In order to estimate the average time spent on the computer terminals per student at a local university, data were collected from a sample of 81 business students over a one-week period. Assume the population standard deviation is 1.2 hours. Refer to Exhibit 8-1. With a .95 probability, the margin of error is approximately_____.

Answer 1

Answer:

Margin of error = $\pm 0.2613$

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 81

Alpha, α = 0.05

Population standard deviation, σ = 1.2 hours

Margin of error formula:

$z_{critical}\times \dfrac{\sigma}{\sqrt{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting the values, we get

$M.E = 1.96(\dfrac{1.2}{\sqrt{81}} ) = \pm 0.2613$