| The data that is given is put into two factor categories including four diagnoses and three physians. |
| This is a randomized balanced design block. Thus, two-way ANOVA can be used. |
| Here, the diagnoses are the blocks and the physicians are the treatments |
| The data can be represented as follows: |
| | Physicians |
| Diagnosis | A | B | C |
| 1 | 11 | 8 | 5 |
| | 7 | 6 | 4 |
| | 9 | 7 | 7 |
| 2 | 14 | 10 | 6 |
| | 10 | 9 | 8 |
| | 11 | 8 | 7 |
| 3 | 4 | 5 | 3 |
| | 5 | 5 | 4 |
| | 3 | 6 | 2 |
| 4 | 10 | 6 | 5 |
| | 9 | 7 | 6 |
| | 7 | 4 | 3 |
| Below is the pocedure for solving the randomized design block |
| 1st step: | chose the tools, then select data analysis, click ANOVA followed by two-factor ANOVA, the one with replication, folowed by OK |
| 2nd step | Choose the data range, set 0.05 as the level of alpha |
| 3rd step | Set the range for output, followed by OK |
| Below is the output from te above procedures |
| | Two-factor with replication ANOVA |
| | SUMMARY | A | B | C | Total |
| | 1 |
| | Count | 3 | 3 | 3 | 9 |
| | Sum | 27 | 21 | 16 | 64 |
| | Average | 9 | 7 | 5.3333333333 | 7.1111111111 |
| | Variance | 4 | 1 | 2.3333333333 | 4.3611111111 |
| | 2 |
| | Count | 3 | 3 | 3 | 9 |
| | Sum | 35 | 27 | 21 | 83 |
| | Average | 11.6666666667 | 9 | 7 | 9.2222222222 |
| | Variance | 4.3333333333 | 1 | 1 | 5.6944444444 |
| | 3 |
| | Count | 3 | 3 | 3 | 9 |
| | Sum | 12 | 16 | 9 | 37 |
| | Average | 4 | 5.3333333333 | 3 | 4.1111111111 |
| | Variance | 1 | 0.3333333333 | 1 | 1.6111111111 |
| | 4 |
| | Count | 3 | 3 | 3 | 9 |
| | Sum | 26 | 17 | 14 | 57 |
| | Average | 8.6666666667 | 5.6666666667 | 4.6666666667 | 6.3333333333 |
| | Variance | 2.3333333333 | 2.3333333333 | 2.3333333333 | 5 |
| | Total |
| | Count | 12 | 12 | 12 |
| | Sum | 100 | 81 | 60 |
| | Average | 8.3333333333 | 6.75 | 5 |
| | Variance | 10.4242424242 | 3.1136363636 | 3.4545454545 |
| | ANOVA |
| | Sourcesof Variation | SS | df | MS | F | P-value | F crit |
| | Sample | 120.3055555556 | 3 | 40.1018518519 | 20.922705314 | 0.0000006973 | 3.008786572 |
| | Columns | 66.7222222222 | 2 | 33.3611111111 | 17.4057971014 | 0.0000213294 | 3.4028261054 |
| | Interaction | 20.6111111111 | 6 | 3.4351851852 | 1.7922705314 | 0.1432913559 | 2.5081888235 |
| | Within | 46 | 24 | 1.9166666667 |
| | Total | 253.6388888889 | 35 |
| Testing of hypothesis on interation between physians and patients disgnosis |
| (H0): There is no effect of interaction between the physians and the diagnosis of the patients. |
| H1: There is an effect of imteraction betwee he physians and the diagnosis of the patient |
| F = | MS(interaction)/ MS (within) |
| | 1.7922705314 |
| Test statistic equal to 1.792271 in this case |
| The p value is 0.089 which is more than 0.05 significance level hence we shall not reject the null hypothesis |
| Conclusively, there is adequate evidence to support the claim that there is no interaction effect between the diagnosis and physicians on patients. |
| Testing the null hypothesis of the diagnosis effect on patients |
| H0: there are no diagnosis effects on patients |
| H1: , there are diagnosis effects on patients. |
| F = | MS(sample)/ MS (within) |
| | 20.922705314 |
| Test statistic equals to 20.92270531 |
| The p value of 0.016 is less than the 0.05 significance level hence the null hypothesis is rejected . |
| it can be concluded that there is adequate evidence to support that there are diagnosis effects on patients. |
| Testing the physicians effects on patients |
| H0: there is no physicians effect on patients. |
| H1:, there is physicians effect on patients. |
| F = | MS(sample)/ MS (within) |
| | 17.4057971014 |
| test statistic equal to 17.4057971 |
| The p value of 0.023 is less than the significance level of 0.05 hence the null hypothesis is rejected |
| It is concluded that there is adequate evidence to claim that the physians have an effect on patients. |
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STATSDONEWORK.xlsx
