Correlation Homework

Question

A researcher is interested in how self-efficacy in statistics (how well you feel you can do statistics) is related to life enjoyment. He has his participants take a self-efficacy questionnaire and then a life enjoyment questionnaire. Both measures are on an 11-point scale. The data is as follows:

Participant	Stats self-efficacy	Life enjoyment
A	10	11
B	9	10
C	10	11
D	11	11
E	2	3
F	10	10
G	11	10
H	10	9
I	9	8
J	10	11

a. Who are the groups being tested?

b. What are the null and research hypotheses?

c. What is the mean of X?

What is the mean of Y?

Find the deviances for both groups. (X – My) and (Y – My)

Multiply the deviances like in the lecture

Sum these.

What number do you get? ___________________________

This is the numerator

Square the deviances in X and Y

What is the Sum of Squares of X?

What is the Sum of Squares of Y?

What do you get when you multiply SSx and SSy? _______________________________

Take the square root of that ____________________________

This is your denominator

d. What is r? ______________________

e. What is the critical value? ___________________________________

f. What is your decision? ______________________________

5. What are the three magnitudes of strength for correlations?

6. Which correlation is strongest? A. 0        B. .56             C. -0.65          D. 1.01

Answer 1

a. Groups Being Tested

The groups being tested are participants' self-efficacy in statistics and their life enjoyment levels.

b. Hypotheses

1. Null Hypothesis (H0): There is no relationship between self-efficacy in statistics and life enjoyment (r = 0).
2. Research Hypothesis (H1): There is a relationship between self-efficacy in statistics and life enjoyment (r ≠ 0).

c. Calculate Means

1. Mean of X (Stats Self-Efficacy)
2. Mean of X=10+9+10+11+2+10+11+10+9+1010=9210=9.2\text{Mean of X} = \frac{10 + 9 + 10 + 11 + 2 + 10 + 11 + 10 + 9 + 10}{10} = \frac{92}{10} = 9.2Mean of X=1010+9+10+11+2+10+11+10+9+10​=1092​=9.2
3. Mean of Y (Life Enjoyment)
4. Mean of Y=11+10+11+11+3+10+10+9+8+1110=10210=10.2\text{Mean of Y} = \frac{11 + 10 + 11 + 11 + 3 + 10 + 10 + 9 + 8 + 11}{10} = \frac{102}{10} = 10.2Mean of Y=1011+10+11+11+3+10+10+9+8+11​=10102​=10.2

d. Calculate Deviances

1. Deviances for X (X - Mean of X)
2. 10−9.2=0.810 - 9.2 = 0.810−9.2=0.8
3. 9−9.2=−0.29 - 9.2 = -0.29−9.2=−0.2
4. 10−9.2=0.810 - 9.2 = 0.810−9.2=0.8
5. 11−9.2=1.811 - 9.2 = 1.811−9.2=1.8
6. 2−9.2=−7.22 - 9.2 = -7.22−9.2=−7.2
7. 10−9.2=0.810 - 9.2 = 0.810−9.2=0.8
8. 11−9.2=1.811 - 9.2 = 1.811−9.2=1.8
9. 10−9.2=0.810 - 9.2 = 0.810−9.2=0.8
10. 9−9.2=−0.29 - 9.2 = -0.29−9.2=−0.2
11. 10−9.2=0.810 - 9.2 = 0.810−9.2=0.8
12. Deviances for Y (Y - Mean of Y)
13. 11−10.2=0.811 - 10.2 = 0.811−10.2=0.8
14. 10−10.2=−0.210 - 10.2 = -0.210−10.2=−0.2
15. 11−10.2=0.811 - 10.2 = 0.811−10.2=0.8
16. 11−10.2=0.811 - 10.2 = 0.811−10.2=0.8
17. 3−10.2=−7.23 - 10.2 = -7.23−10.2=−7.2
18. 10−10.2=−0.210 - 10.2 = -0.210−10.2=−0.2
19. 10−10.2=−0.210 - 10.2 = -0.210−10.2=−0.2
20. 9−10.2=−1.29 - 10.2 = -1.29−10.2=−1.2
21. 8−10.2=−2.28 - 10.2 = -2.28−10.2=−2.2
22. 11−10.2=0.811 - 10.2 = 0.811−10.2=0.8

e. Multiply Deviances

Now, multiply the corresponding deviances:

(0.8×0.8)=0.64=0.64=1.44=51.84=−0.16=−0.36=−0.96=0.44=0.64​

(−0.2×−0.2)=0.04

(0.8×0.8)=.64

(1.8×0.8)=1.44

(−7.2×−7.2)=51.84

(0.8×−0.2)=-.16

(1.8×−0.2)=-.36

(0.8×−1.2)=-.96

(−0.2×−2.2) =.44

(0.8×0.8)​=.64

f. Sum of Products

0.64+0.04+0.64+1.44+51.84−0.16−0.36−0.96+0.44+0.64=53.600.64 + 0.04 + 0.64 + 1.44 + 51.84 - 0.16 - 0.36 - 0.96 + 0.44 + 0.64 = 53.600.64+0.04+0.64+1.44+51.84−0.16−0.36−0.96+0.44+0.64=53.60

g. Calculate Sum of Squares

1. Sum of Squares for X (SSx)
2. SSx=0.64+0.04+0.64+3.24+51.84+0.64+3.24+0.64+0.04+0.64=62.80SSx = 0.64 + 0.04 + 0.64 + 3.24 + 51.84 + 0.64 + 3.24 + 0.64 + 0.04 + 0.64 = 62.80SSx=0.64+0.04+0.64+3.24+51.84+0.64+3.24+0.64+0.04+0.64=62.80
3. Sum of Squares for Y (SSy)
4. SSy=0.64+0.04+0.64+0.64+51.84+0.04+0.04+1.44+4.84+0.64=60.64SSy = 0.64 + 0.04 + 0.64 + 0.64 + 51.84 + 0.04 + 0.04 + 1.44 + 4.84 + 0.64 = 60.64SSy=0.64+0.04+0.64+0.64+51.84+0.04+0.04+1.44+4.84+0.64=60.64

h. Multiply SSx and SSy

SSx×SSy=62.80×60.64≈3804.032SSx \times SSy = 62.80 \times 60.64 \approx 3804.032SSx×SSy=62.80×60.64≈3804.032

i. Take the Square Root

3804.032≈61.69\sqrt{3804.032} \approx 61.693804.032​≈61.69

j. Calculate Correlation Coefficient (r)

r=53.6061.69≈0.87r = \frac{53.60}{61.69} \approx 0.87r=61.6953.60​≈0.87

k. Critical Value

For α = 0.05 and degrees of freedom df=n−2=10−2=8df = n - 2 = 10 - 2 = 8df=n−2=10−2=8:

1. The critical value is approximately ±2.306 (check a t-table).

l. Decision

Since r≈0.87r \approx 0.87r≈0.87 is strong, you would reject the null hypothesis, indicating a significant relationship.

m. Magnitudes of Strength for Correlations

1. Weak: 0.0 to 0.3
2. Moderate: 0.3 to 0.7
3. Strong: 0.7 to 1.0

n. Strongest Correlation

The strongest correlation from the options is C. -0.65 (since -0.65 is closer to -1 than the others are to 1).