Respuesta :
Answer:
With a .95 probability, the margin of error is = 0.392
Step-by-step explanation:
Given -
In order to estimate the average time spent per student on the computer terminals at a local university, data were collected for a sample of 81 business students over a one-week period.
Sample size ( n ) =81
Standard deviation [tex](\sigma )[/tex] = 1.8 hours
[tex]\alpha[/tex] = 1 -.95 =.05
[tex]Z_{\frac{\alpha }{2}}[/tex] = [tex]Z_{\frac{.05 }{2}}[/tex] = 1.96 (Using Z table)
With a .95 probability, the margin of error is
Margin of error = [tex]Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}[/tex]
= [tex]Z_{\frac{.05 }{2}} \frac{1.8 }{\sqrt{81}}[/tex]
= [tex]1.96\times 0.2[/tex]
Margin of error = 0.392
