Significant test using One Way Anova
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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.
Significant test using One Way Anova
- 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.