21.6 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.

(c) 1.6

Step-by-step explanation:

Calculation for what the margin of error will be approximately

First step is to calculate Total hours spent

Total hours spent = (197 * 23.5 hours)

Total hours spent =4,625 hours

Second step is to find the square root of the sample size

Sample size=√197

Sample size=14.04

Third step is to divide the Standard deviation of 8.7 hours by square root of sample size of 14.04

= 8.7 hours /14.04

= 0.6196

Fourth step is to use the Z table to find Z value at 99% confidence

Z value at 99% confidence = 2.58

Last step is to calculate the Margin of error

Margin of error = 0.6196 * 2.58

Margin of error = 1.59857

Margin of error= 1.6 (Approximately)

Therefore what the margin of error will be approximately is 1.6

the answer is c

step-by-step explanation: i took a test with this question on it