Given the following data from a repeated-measures design study examining the effect of a treatment by measuring a group of 9 participants before and after they received treatment: Participant Before After
A 8 7
B 7 5
C 6 6
D 7 6
E 9 7
F 8 5
G 5 4
H 9 4
I 7 4
a. Calculate the difference scores and MD.
b. Compute SS, sample variance, and estimated standard error.
c. Is there a significant treatment effect?

MD = 2

SS = 18

SAMPLE VARIANCE = 2.25

STANDARD ERROR = 0.5

Step-by-step explanation:

Given :

A 8 7

B 7 5

C 6 6

D 7 6

E 9 7

F 8 5

G 5 4

H 9 4

I 7 4

Difference, d = Before - After

d

A 8 7 __ 1

B 7 5 __ 2

C 6 6 __ 0

D 7 6 __ 1

E 9 7 __ 2

F 8 5 __ 3

G 5 4 __ 1

H 9 4 __5

I 7 4 ___3

The mean of difference, MD ;

MD = Σd/ n = (1+2+0+1+2+3+1+5+3) / 9 = 18 / 9 = 2

The sum of square, SS ;

(1 - 2)^2 + (2 - 2)^2 + (0 - 2)^2 + (1 - 2)^2 + (2 - 2)^2 + (3 - 2)^2 + (1 - 2)^2 + (5 - 2)^2 + (3 - 2)^2 = 18

Sample variance, S² = SS/(N-1) = 18 / (9 - 1) = 18 / 8 = 2.25

Sample standard deviation, S = √Variance = √2.25 = 1.5

Standard Error, S.E = S / √n = 1.5 / √9 = 0.5

Test statistic : MD / S.E = 2 / 0.5 = 4

We test at α = 0.05 since no α - value is stated in the question.

Critical value at 0.05, df = 8 ;

Critical value = 2.306

Since; Test statistic > Critical value, then result is significant at α = 0.05

i think the answer is b.

step-by-step explanation:

not 100% sure though.

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.