[(1.96^2)(0.6^2)]/(0.2^2)
= 34.5744
~35 students must be included in each group
You want to estimate the difference in grade point averages between two groups of university
students to be accurate within 0.2 grade point, with probability approximately equal to 0.95. If the
standard deviation of the grade point measurements is approximately equal to 0.6, how many students
must be included in each group?
1 answer