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. If the sample mean is 9 hours, then the 95% confidence interval is

(8.608, 9.392)

Step-by-step explanation:

We have the following information

Population standard deviation = 1.8

Sample mean = 9 hours

Sample n = 81

C I = 95%

So level of significance

Alpha = 1-0.95

= 0.05

Z critical at 0.05/2

Z(0.025) = 1.96

The 95% c.i =

9+-(1.96)(1.8/√81)

9+-(1.96)(0.2)

(9-0.392)(9+0.392)

(8.608, 9.392)

This is the confidence interval at 95%.

I hope you find my solution useful. Good luck!!!

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17

rx and sx

step-by-step explanation: