In order to estimate the average time spent on the computer terminals per student 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. Answer questions 7 - 9.

7. What is the standard error of the mean?

a. 7.50
b . 0.39
c. 2.00
d. 0.20

8. With a 0.95 probability, the margin of error is approximately

a. 0.39
b 1.96
c. 0.20
d. 1.64

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

a. 7.04 to 10.96 hours
b. 7.36 to 10.64 hours
c. 7.80 to 10.20 hours
1d. 8.61 to 9.39 hours

7) d)

standard error of the mean of one sample of 'n' observation = 0.20

8) a)

The margin of Error = 0.392

9) d

The 95% of confidence intervals are (8.61 , 9.39)

Step-by-step explanation:

7)

solution:-

The Given data sample size 'n' = 81

Given Population standard deviation 'σ' = 1.8 hours

The standard error of the mean of one sample of 'n' observation is

Standard error (SE)

                               =  

                               = σ / √n

                               =

standard error of the mean of one sample of 'n' observation = 0.20

8)

Solution:-

The Given data sample size 'n' = 81

Given Population standard deviation 'σ' = 1.8 hours

Given the probability is 0.95

The z- score = 1.96 at 0.05 level of significance.

The margin of Error   =  

                                   =

                                   =

                                   = 0.392

The margin of Error = 0.392

9)

Solution:-

The 95% of confidence intervals are

(9 - 0.392 , (9 + 0.392)

(8.609 , 9.392)

The 95% of confidence intervals are (8.61 , 9.39)