Significant test using student T test
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This document describes performing a t-test to analyze the difference in average IB grades between two classes. Data on average scores was collected from both classes and tested for normal distribution. A t-test was performed using the data in Excel. The results showed a t-value less than the critical value and a p-value greater than 0.05, so the null hypothesis that there is no significant difference between the classes was accepted.
Significant test using student T test
- 1. Significant test (T-test) on 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 T test 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 t value < t critical, accept null hypothesis Perform t test and if P < 0.05 or t value > t critical, reject 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 the formulas given for the student t-test. 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 formula Detail Calculation using Excel 1. Follow instruction shown below and choose t test equal variance 2. For variable 1 – choose range class 1, variable 2 – choose class 2
- 5. 3. Display results from Excel t value = 0.82, t critical = 2.04 t value < t critical (0.82 < 2.04) Do no reject null hypothesis p value = 0.41 > 0.05 Conclusion and Evaluation The tvalue is less than the tcrit (0.82 < 2.04), or P value > 0.05, this shows that there is no significant difference between the two IB classes. Thus, the null hypothesis is accepted (fail to reject) over the alternative hypothesis. Significance was determined by t test and P > 0.05, there is not enough evidence to reject the Null Hypothesis, (There is no significant difference between both classes)