The document describes conducting a one-way ANOVA to analyze the average IB grades between two different IB classes. It provides the hypotheses, data collection process, data tables for each class, a normal distribution check, and the ANOVA calculation and results in Excel. The ANOVA analysis found no significant difference between the average IB grades of the two classes, as the p-value was greater than 0.05 and the F-value was less than the critical F-value.

1. One way anova on the average IB grades between diff IB classes.
Data independent of each other.
 Showing normal distribution
 Minimum sample size of 15 for each data set
 Similar variance
Hypotheses
Null Hypothesis (H0) – No significant difference between different IB classes.
Alternate Hypothesis (HA) – A significant difference between different IB classes.
Steps for One way anova
1. State the null hypothesis and alternative hypothesis based on your
research question
2. Set critical P level at P = 0.05 (5% by chance)
3. Perform t test and if P > 0.05 or F value < F critical, accept null hypothesis
4. Write your conclusion is there a significant difference between mean of
two samples
Procedure :
1. The average IB scores from two different IB classes were collected
2. The scores are then tabulated according to the table below.
3. The data is then processed using excel.
4. The processed data is analyzed and a conclusion is then found.

2. Data Collection
Average Scores for IB Class 3
Student Score(x) Sorted Value Class Frequency Score (x2)
A 4.8 4.1 4.1 – 4.5 | | 23.04
B 5.6 4.1 31.36
C 4.7 4.7 4.6 – 5.0 | | | | | | 22.09
D 5.5 4.7 30.25
E 5.3 4.8 5.1 – 5.5 | | | | 28.09
F 6.0 4.8 36.00
G 4.1 5.0 5.6 – 6.0 | | | 16.81
H 5.1 5.0 26.01
I 5.1 5.1 6.1 – 6.5 | 26.01
J 4.8 5.1 23.04
K 5.0 5.3 25.00
L 6.0 5.5 36.00
M 4.7 5.6 22.09
N 5.0 6.0 25.00
O 4.1 6.0 16.81
P 6.3 6.3 39.69
n = 16 Σx = 82.1 Σx2 = 427.29
Average Scores for IB Class 2
Student Score(x) Sorted Value Class Frequency Score (x2)
A 5.4 1.4 1.0 – 1.5 | 29.16
B 4.2 3.2 17.64
C 5.5 3.4 3.1 – 3.5 | | | 30.25
D 1.4 3.5 1.96
E 5.3 4.2 4.1 – 4.5 | | | 28.09
F 5.3 4.5 28.09
G 4.5 4.5 5.1 – 5.5 | | | | 20.25
H 3.2 5.3 10.24
I 7.0 5.3 5.6 – 6.0 | | | 49.00
J 6.4 5.4 40.96
K 5.8 5.5 6.1 -6.5 | 33.64
L 3.4 5.7 11.56
M 5.7 5.8 6.6 – 7.0 | 32.49
N 3.5 5.9 12.25
O 4.5 6.4 20.25
P 5.9 7.0 34.81
n = 16 Σx = 77.4 Σx2 = 400.64

3. Showing Normal Distribution for both Classes

4. Detail Calculation using Excel
1. Follow instruction shown below and choose one way anova
2. For input data – select the data, including labels.

5. 3. Display results from Excel
Anova:Single
Factor
SUMMARY
Groups Count Sum Average Variance
a 16 82.1 5.13125 0.400958333
b 16 77 4.8125 2.005166667
ANOVA
Source of Variation SS df MS F P-value F crit
BetweenGroups 0.812813 1 0.812813 0.675619513 0.417584 4.170877
WithinGroups 36.09188 30 1.203063
Total 36.90469 31
P value > 0.05 or F value < F critical. There is no significant difference between
the two sample.
Conclusion and Evaluation
The Fvalue less than Fcrit (0.657 < 4.17), or P (0.417) value > 0.05, this proves that
there is no significant difference between the two IB classes. Thus, the null
hypothesis is accepted over the alternative/research hypothesis.