- #1
aptiva
- 5
- 0
Hi all:
Sample 1: M = 62.77, SD = 6.52
Sample 2: M = 72.91, SD = 6.22
Sample 3: M = 74.61, SD = 6.60
Now if I wanted to calculate what proportion of students in each population would be expected to receive a grade <50 and receive a grade >80, would I perform a Z-test using Z = (X-mu)/Sd
For example, in sample 1:
(For <50)
Z = 50 - 62.77 / 6.52 = -1.959, P = 0.0202 (according to Z-table)
So, .5 - .0202 = .4798 = approximately 48% of the students in the first population would receive a score <50...is that correct?
(For >80)
Z = 80 - 62.77 / 6.52 = 2.65, P = 0.4965(according to Z-table)
So, .5 - .4965 = .0035 = approximately .35% of the students in the first population would receive a score >80...?
Homework Statement
Sample 1: M = 62.77, SD = 6.52
Sample 2: M = 72.91, SD = 6.22
Sample 3: M = 74.61, SD = 6.60
Homework Equations
Now if I wanted to calculate what proportion of students in each population would be expected to receive a grade <50 and receive a grade >80, would I perform a Z-test using Z = (X-mu)/Sd
The Attempt at a Solution
For example, in sample 1:
(For <50)
Z = 50 - 62.77 / 6.52 = -1.959, P = 0.0202 (according to Z-table)
So, .5 - .0202 = .4798 = approximately 48% of the students in the first population would receive a score <50...is that correct?
(For >80)
Z = 80 - 62.77 / 6.52 = 2.65, P = 0.4965(according to Z-table)
So, .5 - .4965 = .0035 = approximately .35% of the students in the first population would receive a score >80...?