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. Calculate the margin of error (to two decimals) for a confidence interval with a confidence coefficient of 0.95.

Answer 1

Answer:

Margin of error  for a 95% of confidence intervals is 0.261

Step-by-step explanation:

Step1:-

 Sample n = 81 business students over a one-week period.

 Given the population standard deviation is 1.2 hours

 Confidence level of significance = 0.95

 Zₐ = 1.96

Margin of error (M.E) = $\frac{Z_{\alpha }S.D }{\sqrt{n} }$

Given n=81 , σ =1.2 and  Zₐ = 1.96

Step2:-

$Margin of error (M.E) = \frac{Z_{\alpha }S.D }{\sqrt{n} }$

$Margin of error (M.E) = \frac{1.96(1.2) }{\sqrt{81} }$

On calculating , we get

Margin of error = 0.261

Conclusion:-

Margin of error  for a 95% of confidence intervals is 0.261