A sample of 100 information systems managers had an average hourly income of $40.00 with a standard deviation of $8.00.The value of the margin of error at 95% confidence is

A) 80.
B) 8.
C) .10.
D) 1.568.

To find the margin of error, we use the formula:

Margin of Error = z * (standard deviation/square root of sample size)

We know that our sample size is 100, the standard deviation is $8.00, and we want to find the margin of error at a 95% confidence level. To find the z-score, we can use a z-table or calculator to find the z-score that corresponds to a 95% confidence level, which is 1.96.

So, plugging in the numbers:
Margin of Error = 1.96 * ($8.00/square root of 100)
Margin of Error = 1.96 * $0.80
Margin of Error = $1.568

Therefore, the margin of error at 95% confidence is $1.568, which is answer choice D.