In: Statistics and Probability
Student | Family Goal | Work Goal |
1 | 7 | 5 |
2 | 6 | 4 |
3 | 8 | 2 |
4 | 3 | 9 |
5 | 4 | 1 |
6 | 10 | 3 |
7 | 5 | 4 |
8 | 7 | 3 |
9 | 2 | 7 |
10 | 6 | 9 |
Independent variable (X) : Family Goal
Dependent variable (Y): Work Goal
Following table shows the calculations:
X | Y | X^2 | Y^2 | XY | |
7 | 5 | 49 | 25 | 35 | |
6 | 4 | 36 | 16 | 24 | |
8 | 2 | 64 | 4 | 16 | |
3 | 9 | 9 | 81 | 27 | |
4 | 1 | 16 | 1 | 4 | |
10 | 3 | 100 | 9 | 30 | |
5 | 4 | 25 | 16 | 20 | |
7 | 3 | 49 | 9 | 21 | |
2 | 7 | 4 | 49 | 14 | |
6 | 9 | 36 | 81 | 54 | |
Total | 58 | 47 | 388 | 291 | 245 |
Sample size: n=10
Now,
The coefficient of correlation is :
Test statistics:
Degree of freedom: df=n-1=8
Critical value using excel function "=TINV(0.05,8)" are: +/- 2.306
Critical value using excel function "=TINV(0.01,8)" are: +/- 3.355
Since test statistics lies between critical values for both level of significance so we fail to reject the null hypothesis at both levels .05 and .01 .
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The p-value using excel function "=TDIST(1.461,8,2)" is 0.1821
Since p-value is greater than 0.05 so we fail to reject the null hypothesis.
The results in APA format is
t(8) = -1.461, p > 0.05
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The r^2 is
It shows that 21.06% of variation in dependent variable is explained by independent variable.