Question

In order to estimate the average time spent on the computer lab terminals per student at a local university, data were collected for a sample of 98 students over a one-week period. Assume the population standard deviation is 1.8 hours.

If the sample mean is 9, then the lower limit for a 90% confidence interval is

(Round your answer to three decimal places.)

Answer 1

Solution :

Given that,
sample mean = = 9

Population standard deviation =    =1.8

Sample size = n =98

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

Z= Z_{0.10 = 1.282}( Using z table ( see the 0.10 value in standard normal   (z) table corresponding z value is 1.282 )   

Margin of error = E = Z/2 * ( /n)

= 1.282 * ( 1.8/  98 )

= 0.233
At 90%lower confidence interval
is,

- E
9 - 0.233

(8.767)

lower limit for a 90% confidence interval is 8.767

answered by: anonymous