Exhibit 8-5
A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240.
-Refer to Exhibit 8-5. The margin of error at 95% confidence is
A)1.998
B)1400
C)240
D)59.95

Question

Quizplus · Accepted Answer

The margin of error at 95% confidence is calculated as (1.96*standard deviation)/square root of sample size. 
Substituting the given values, we get (1.96*240)/square root of 64 = 59.95. Therefore, the answer is choice D.

Exhibit 8-5 A Random Sample of 64 SAT Scores of Students Applying