ANOVA: One-Way Analysis of Variance - Statistics Project Example

Assignment 3 (Part 3) – Statistical Analysis of Data – ANOVA (One Way Analysis of Variance) A. Research Question: Is there a difference in the % water content between the groups of kidney exposed to varying times of ischemia? B. Null Hypothesis: There is no difference in the % water content between the group of kidneys exposed to varying times of ischemia. In other words, there is no effect of varying times of ischemia on % content of water in the group of kidneys. In symbolic form, H0: µ1 = µ2 = µ3 = µ4 (all the means are equal) (1-Control, 2-30 min ischemia, 3-60-min ischemia, 4-120 min ischemia) The alternate hypothesis will be H1: Not all the means are equal (at least one mean is different) C. Summary of Results of Statistical Analysis Table 1: The % Water Content (Data) Control (no ischemia) 30 min ischemia 60 min ischemia 120 min ischemia 78.5 83.0 81.8 83.5 79.2 81.9 81.1 83.5 78.9 83.0 82.3 81.4 79.7 79.8 79.7 81.5 78.6 79.0 Table 2: Descriptive Statistics Ischemia n Mean (X) SEM St. Dev Minimum Maximum Q1 Median Q3 Control 7 79.09 0.182 0.481 78.5 79.7 78.60 79.0 79.70 30 min 3 82.63 0.367 0.635 81.9 83.0 81.90 83.0 83.00 60 min 5 81.30 0.423 0.946 79.8 82.3 80.45 81.5 82.05 120 min 3 82.80 0.700 1.212 81.4 83.5 81.40 83.5 83.50 Table 3: Analysis of Variance Table Source SS df MS F η2* FCrit p Significant Between 43.683 3 14.561 23.39 .83 3.34 Read More