Question

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. Suppose that you plan to use this data to construct a 99% confident interval. Determine the margin of error. (Use the table of critical values of the Normal distributions. Usc decimal notation. Give your answer to two decimal places.)

margin of error:_____

Answer 1

The margin of error for a 99% confidence interval can be found using the formula: Margin of error = Critical value x Standard deviation. With a 99% confidence level, the critical value is approximately 2.58. Multiplying this by the standard deviation gives us a margin of error of approximately 22.47 hours.

To determine the margin of error for a 99% confidence interval, we need to use the formula:

Margin of error = Critical value x Standard deviation

First, we need to find the critical value from the table of critical values for the normal distribution.

With a 99% confidence level, we want to find the z-score that corresponds to a 0.995 probability.

Looking at the table, we find that the z-score is approximately 2.58.

Next, we multiply the critical value by the standard deviation:

Margin of error = 2.58 x 8.7 = 22.47

Therefore, the margin of error is approximately 22.47 hours.