Question

Computer use.

Answer 1

Answer:

The margin of error will be approximately =

(c.) 1.6

Step-by-step explanation:

Sample of 12th-grade students = 197

Average hours spent on computer per week = 23.5 hours

Total hours spent = 4,625 (197 * 23.5) hours

Standard deviation = 8.7 hours

Confidence interval = 99%

Square root of sample size = $\sqrt{x}$ = 14.04

Standard deviation/square root of sample size = 8.7/14.04 = 0.6196

Z value at 99% confidence = 2.58

Margin of error = 0.6196 * 2.58

= 1.59857

= 1.6