Answer:
Explanation:
The hypothesis being tested is:
H0: ρ = 0
Ha: ρ ≠ 0
Pearson's r is -0.140.
The critical r-value is 0.404.
Since 0.140 < 0.404, we cannot reject the null hypothesis.
Therefore, we cannot conclude that notebook computer use and grades are related.
Pearson's r is -0.140. It means that there is a weak negative relationship between notebook computer use and grades.
The coefficient of determination is 0.020. 2% of the variation in the model is explained.
The relationship is not significant.
Notebook Computer Use Final Grade for Course
30 86
23 88
6 94
0 56
24 78
36 72
10 80
0 90
0 82
8 60
12 84
18 74
0 78
32 66
36 54
12 98
8 81
18 74
22 70
38 90
5 85
29 93
26 67
10 80
r² = 0.020
r = -0.140
Std. Error = 11.996
n = 24
k = 1
ANOVA table
Source SS df MS F p-value
Regression 63.6652 1 63.6652 0.44 0.5129
Residual 3,165.668 22 143.8940
Total 3,229.333 23