Population is normally distributed and there is not

population is normally distributed, and there is not at least 30 participants.
Step 2 - Null Hypothesis: H
0
: µ
1
- µ
2
= 0 Research Hypothesis: H
1
: µ
1
-
Step 3 -
X X - M ( X - M ) 2
10 -0.33 0.11
8 -2.33 5.43
13 2.67 7.13
s 2
X
= ∑( X - M ) 2 / ( N -1) = (0.11+5.43+7.13)/(3-1) = 12.67/2 = 6.34
Y Y - M ( Y - M ) 2
3 -5.33 28.41
15 6.67 44.49
7 -1.33 1.77
s
2
Y
= ∑( Y - M )
2
/ ( N -1) = (28.41+44.49+1.77)/(3-1) = 74.67/2 = 37.34
df
X
= N-1 = 3-1 = 2
Y
= N-1 = 3-1 = 2
µ
2
≠ 0
df
2

PS 307, Final Exam Part 2
df
total
= df
X
+ df
Y
= 4
s 2
pooled
= ( df
X
/ df
total
) s 2
Y
+( df
Y
/ df
total
) s 2
Y
= (2/4)6.34 + (2/4)37.34 = 3.17+18.67 = 21.84
s 2
Mx
= s 2
pooled
/ N
X
= 21.84/3 = 7.28
s
2
My
= s
2
pooled
/ N
Y
= 21.84/3 = 7.28
s 2
difference
= s 2
Mx
+ s 2
My
= 14.56
s
difference
= √ s 2
difference
= √14.56 = 3.82
Step 4 - Critical Value for a two-tailed test with p value of 0.05 and df
total
of 4 = 2.365
Step 5 - t = ( M
X
- M
Y
)/ s
difference
= (10.33-8.33)/3.82 = 0.52
Step 6 - t (11) = 0.52, p < 0.05 ; fail to reject null hypothesis
2 . t (11) = 0.52, p < 0.05, d = =19.84
Cohen’s d = ( M
X
- M
Y
)/ s
pooled
= (10.33-8.33)/21.84 = -19.84
Question 3 - 60 Points
You wish to compare how a low-fat diet, low-fat diet and a brisk four-mile
walk each day, and a low-fat diet and a four-mile jog each day affect
cholesterol levels. You randomly assign six subjects to each group. The
scores in the table below show the reduction in cholesterol count after
two months.
1. Conduct all 6 steps of the appropriate statistic. Be sure to label
all six steps.
2. Calculate and interpret the effect size for this test.
3. Calculate and interpret any post-hoc test if necessary.
Low-Fat Diet Diet and Walk Diet and Run
10 14 19
8 12 15
15 18 14
12 16 16
9 13 18
6 17 20
1.
Step 1 - The populations are (1) individuals on a low-fat diet only, (2) individuals who are on a low-fat
diet and walk four miles every day, and (3) individuals who are on a low-fat diet and run four miles every day.
The comparison distribution will be an F distribution. The hypothesis test will be a one-way between-groups
ANOVA. This study meets one of the three assumptions and may meet the others: (1) The data is randomly
3

PS 307, Final Exam Part 2
selected, (2) we don’t know whether the population is normally distributed, and (3) We will check
homoscedasticity.
Step 2 - H
0
: µ
1
= µ
2
= µ
2
Research Hypothesis: H
1
: At least one µ is different from another µ .
Step 3 - df
between
= N
groups
– 1 = 3-1 = 2
df
1
= n
1
– 1 = 6-1 = 5
df
2
= n
2
– 1 = 6-1 = 5
df
3
= n
3
– 1 = 6-1 = 5
df
within
= df
1
+ df
2
+ df
3
= 5+5+5 = 15
Step 4