Correlation Calculation | Academic Papers Help

Correlation Calculation

QUESTION 1

a)
H0: ? = 50
HA: ? ? 50
Sample mean = 55
Standard deviation = 10
Standard error of mean = s / ? n
Standard error of mean = 10 / ? 25
SE = 10/5
Standard error of mean 2 ——–sM
t = (xbar- ? ) / SE
t = (55-50) / 2
t = 2.5 ————– t

The critical value t with 24 degrees of freedom with ? = .05 = 2.064 in a two tailed test
Since t when computed exceeds critical t, the null hypothesis is rejected.

b)
H0: ? = 50
HA: ? ? 50
Sample mean = 55
Standard deviation = 20
Standard error of mean = s / ? n
Standard error of mean = 20 / ? 25
SE = 20/5
Standard error of mean 4 —————–sM
t = (xbar- ? ) / SE
t = (55-50) / 4
t = 1.25 ———————- t
Since the computed value of t = 1.25 does not exceed the critical value of t, it is not possible to reject the null hypothesis.
c)
The larger the variance the lower the standard error, which lowers the likelihood of rejection of the null hypothesis.
QUESTION 2
For subject A:
Difference scores: -10, 2,-15, -9
Sum of D= -36
MD = -9
For subject B:
Difference scores: -10, 2, -15, -9
Sum of D = -32
MD = -8
For subject C:
Difference scores (D): 10, -2, 15, 9
Sum of D = 32
MD = +8
For subject D:
Difference scores: 10, -2, 15, 9.
Sum of D = 36
MD = +9
QUESTION 3
For treatment 1, M = 15/5 = 3
SS = 10
For treatment 2, M = 60/5 = 12
SS = 80
The pooled variance = (10 + 80)/ 8 = 11.25
standard error = s/?n = 2.2361/1.055 = 2.12
t(8) = -9/2.12 = ?4.25 Therefore the null hypothesis (H0) should be rejected.

QUESTION 4

Source SS df MS F
Between 26 2 13 6.50
Within 18 9 2
Total 44 11
With df = 2, 9 the critical value = 4.26. The null hypothesis is therefore rejected and one can conclude that there are significant differences between the treatments.

QUESTION 7
X Y XY X2 Y2
7 3 21 49 9
3 1 3 9 1
6 5 30 36 25
4 4 16 16 16
5 2 10 25 4
25 15 80 135 55

R = [5(80)-(25)(15)]/?([675-625][275-225] = 25/0 = ?

QUESTION 8
SSX = ({0-1}2 + {1-1}2 + {2-1}2) = 2
SP = 2
MX = 3/3 = 1
MY = 27/3 = 9
The regression equation is: Y8 = (1)X + 8

 

QUESTION 9
Ice Cream Flavor
A B C D
12 18 28 22

Mean = 20
Variance = 64 + 4 + 64 + 4 = 136/4 = 34
Standard deviation = 5.83
Standard error of mean = 5.83/8.94 = 0.652
A: T =12-20/0.652 = -12.26
B: T= 18-20/0.652 = -3.06
C: T = 8/0.652 = 12.26
D T = 2/0.652 = 3.06
All the preferences are therefore significant, with A and B showing negative significance and C as well as D showing positive significance.

QUESTION 10
X Y Rank 1 Rank 2 d d2
2 5 2 5 -3 9
4 1 4 1 3 9
3 2 3 2 1 1
1 4 1 4 -3 9
5 3 5 3 2 4

?d2 = 32

Spearman Correlation = 1- (192/120) = -0.6
The data set therefore have a negative correlation