(1)
For this tudy, we should use One Way ANOVA
(2)
The test statistic for this data = 1.227
(3)
The p - value for this sample = 0.3153
(4)
The p - value is greater than .
(5)
Based on this, we should fail to reject null hypothesis.
(6)
Correct option:
There is insufficient evidence to support the claim that course delivery typeis a fector in final score
Explanation:
From the given data, the following Table is calculated:
Hybrid | Online | Face - to face | Total | |
N | 8 | 7 | 7 | 22 |
522 | 534 | 511 | 1567 | |
Mean | 65.25 | 76.2857 | 73 | 71.227 |
35882 | 41520 | 38467 | 115869 | |
Std. Dev. | 16.1312 | 11.4268 | 13.9284 | 14.2359 |
From the above Table, ANOVA Table is calculated as follows:
Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | Test Statistic F | P - Valu |
Between treatments | 486.9351 | 2 | 486.9351/2=243.4675 | 243.4675/198.3647=1.227 | 0.3153 |
Within treatments | 3768.9286 | 19 | 3768.9286/19=198.3647 | ||
Total | 4255.8636 | 21 |
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