Computer use. A random sample of 197 12th-grade students from across the United States was surveyed, and it was observed that these students spent an average of 23.5 hours on the computer per week, with a standard deviation of 8.7 hours. If we plan to use these data to construct a 99% confident interval, the margin of error will be approximately.

(a.) 0.07
(b.) 0.62
(c.) 1.6
(d.) 8.7

This question can be determined by the  concept of standard deviation. The margin of 1.6 error will be approximately.

Given:

Random sample of 12th-grade students = 197

Average hours spent on computer per week = 23.5 hours

Therefore,

Total hours spent by 12th grade students  =  

Standard deviation = 8.7 hours (Given)

Confidence Interval =99% (Given)

Square root of  random sample size =  = 14.04

Therefore,

Z value at 99% confidence = 2.58

Margin of error =  

                          = 1.59857 error

                          = 1.6 error

Thus, the correct option is (c) 1.6 error.

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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 = = 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

good luck on your studies! : )

david edward.

(23*2)-12=x     x=34

step-by-step explanation: