Response Quantitative Quant Cat Quant Explanatory categorical Quant Cat Analysis Oneway ANOVA Etest Regression (dumany variables) Edest Two-sample @test Wilcoxon Rank-Sum Test Regression Etest Pearson correlation (Edest Spearman correlation 2 X test Two-sample proportion -test - One-sample Prest Sign test Signed-rank test Cat - (2) test One-sample proportion Betest Cat Quant Quant Quant Cat × Cat Quant x Cat Strong Logistic regression (E-test Two-way ANOVA (Ftest Multiple regression Atest Moderate 0.05 Weak Inconclusive 0 0.01 0.1 1 If the question does not say increase or decrease, then use two sided tests.
Sensitive questions: Drop coins for two times, if the first one is head than answer honestly if the first one tail, second one head than answer ves if the second one is tail than answer no. p .. Yes Calculate P: H 0.5 0.5 T 1-p 0.5 0.5 · No · Yes · No Assume we get a probability of Yes is 0. 717 Than P(Yes) = P + 0.5*0.5 = 0.717 P = 0.934
ANOVA Assumption of using ANOVA: Groups are independent. · The residuals have a normal distribution. · The variability of the residuals does not depend on the group. . Homogeneity of variance. SST (Sum of squares total) SSR (Sum of squares residual) SSG (Sum of squares explained by groups) = SST - SSR DFT (Degrees of freedom total) = n - 1 DFR (Degrees of freedom residual) = n - k, k = number of groups DFG (Degrees of freedom explained by groups) = DFT - DFR R2 = SSG/SST, if R2 = 0.66 then it explains 66% of the overall variability observed If there was no association, then we would expect SST is close to SSR so that SSG is close to 0. F = MSG/MSR, MSG = SSG/DFG, MSR = SSR/DFR MS (Mean sum of squares) = SS/DF Researchers conducted an observational study of green tea and coffee consumption from 537 men and women at two workplaces in Japan (Pham et al., 2013). Subjects were classified into three groups for green tea consumption: 's 1 cup/day', '2-3 cups/day' and '? 4 cups/day'. A one-way analysis of variance for comparing the mean body mass index (BMI) between the three green tea groups found a group sum of squares of 64.42 and a residual sum of squares of 5809.64. These give a start for the ANOVA table: Source DF SS MS F P GreenTea 64.42 Residuals 5809.64 Total DF (Total) = 537 - 1 = 536 DF (Green Tea) = 3 (groups) - 1 = 2 DF (Residuals) = 536 - 2 = 534 SS (Total) = 64.42 + 5809.64 = 5874.06 MS = SS/DF MS (Green Tea) = 64.42/2 = 32.21 MS (Residuals) = 5809.64/534 = 10.879 MS (Total) = 32.21 + 10.897 = 43.089 F = MSG/MSR