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

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 = [tex]\sqrt{x}[/tex] = 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

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  = [tex]\bold{(197 \times23.5=4629.5)hours}[/tex]  

Standard deviation = 8.7 hours (Given)

Confidence Interval =99% (Given)

Square root of  random sample size =  = 14.04

[tex]\bold{\dfrac{Standard\: deviation}{Square\: root\: of\: sample\: size} =\dfrac{8.7}{14.04 } = 0.6196}[/tex]

Therefore,

Z value at 99% confidence = 2.58

Margin of error = [tex]0.6196 \times2.58[/tex]  

                          = 1.59857 error

                          = 1.6 error

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

For further details, please refer this link:

https://brainly.com/question/13905583

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