Question

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. Assume the population standard deviation is 1.8 hours. With a .95 probability, the margin of error is approximately _________.

Answer 1

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 $(\sigma )$ = 1.8 hours

$\alpha$ = 1 -.95 =.05

$Z_{\frac{\alpha }{2}}$ = $Z_{\frac{.05 }{2}}$ = 1.96           (Using Z table)

With a .95 probability, the margin of error is

Margin of error  =  $Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$

                          = $Z_{\frac{.05 }{2}} \frac{1.8 }{\sqrt{81}}$

                           = $1.96\times 0.2$

Margin of error   = 0.392