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anova [2018/10/19 08:06]
hkimscil [Post hoc test]
anova [2020/05/20 15:55] (current)
hkimscil [Example 2]
Line 518: Line 518:
  
 ====== Example ====== ====== Example ======
 +가설. 단어맞히기 게임에서 첫글자를 힌트로 주거나, 마지막 글자를 힌트로 주거나, 힌트를 주지 않은 세 그룹 간에 틀린 단어의 숫자에 차이가 있을 것이다. 
 +
 |  |First Letter \\ Condition 1 \\ X<sub>1</sub>|Last Letter \\ Condition 2 \\ X<sub>2</sub>|No Letter \\ Condition 3 \\ X<sub>3</sub> | |  |First Letter \\ Condition 1 \\ X<sub>1</sub>|Last Letter \\ Condition 2 \\ X<sub>2</sub>|No Letter \\ Condition 3 \\ X<sub>3</sub> |
 | | 15 | 21 | 28 | | | 15 | 21 | 28 |
Line 530: Line 532:
 | | 41 | 21 | 15 | | | 41 | 21 | 15 |
 | T  | 180  | 240  | 270  | | T  | 180  | 240  | 270  |
-| n  | 10  |  10  | 10  | +| n  | 10  | 10  | 10  | 
-| Mean  | $\overline{X_1}$ = 18  | $\overline{X_2}$ = 24  | $\overline{X_3}$ = 27  | +| Mean  | 18  | 24  | 27  | 
-Total  $ T_1 = 180 $  $ T_2 = 240 $ $ T_3 270 $  | +Mean%%^%%2  324  576  729  | 
-| $\sum{(X_i)^2}$  | 3888  | 5868  | 7614  |+| b $n*(mean)^2 | 3240  | 5760  | 7290  | 
 +a = $\sum{(X_i)^2}$  | 3888  | 5868  | 7614  
 +| a-b = Var.within  | 648  | 108  | 324  | $f = $1080   |
  
 +\begin{eqnarray*}
 +a & = & \sum (X_i)^2 = 17370 \\
 +b & = & \overline{X_{total}} = 23 \\
 +c & = & \overline{X}^2 = 529 \\
 +d & = & N = 30 \\
 +e & = & SS_{total} = \sum {(X_i)^2} = a - (d * c) = 1500 \\
 +f & = & SS_{within} = 648 + 108+ 324 = 1080 \\
 +g & = & SS_{between} = e - f = 420  
 +\end{eqnarray*}
  
-$\sum (X_i)^2 17370 $ \+| m.total = Mean.total $= \$ 23  ||| 
-$\overline{X_{total}} = 23 $ \\ +| m.within = Mean.each.group  | 18  | 24  | 27  | 
-$\overline{X}^2 = 529 \\+| o = m.total - m.within  | 5   | -1  | -4  | 
 +| o%%^%% | 25  | 1  | 16  | 
 +| n  | 10  | 10  | 10  | 
 +| o%%^%%2 * n | 250  | 10  | 160  | 
 +| sum  | $g \; 420  |||
  
-$ G = ? $ \\ 
-$ G^2 = ? $ \\ 
-$ \sum {(X_i)^2} = 17370 $ \\  
-\\ 
-$ k =  $ \\ 
-$ n =  $ \\ 
-$ N =  $ \\ 
-\\ 
-$ df_{total} = $ \\  
-$ df_{between} = $ \\ 
-$ df_{within} = $ \\ 
  
-<code>x1 <- c(15, 20, 14, 13, 18, 16, 13, 12, 18, 41)+\begin{eqnarray*} 
 +k & = & 3 \\ 
 +n & = & 10 \\ 
 +N & = & 30 \\ 
 +\end{eqnarray*} 
 + 
 +\begin{eqnarray*} 
 +e' & = & df_{between} =  \\ 
 +f' & = & df_{within} =   \\ 
 +g' & = & df_{total} =  \\  
 +\end{eqnarray*} 
 + 
 +|          | SS   | df    | MS  | F   | 
 +| between  | $e = \;$ 420   | $e' = \;$ 2 | 210  |  210/40 = 5.25  | 
 +| within   | $f = \;$ 1080  | $f' = \;$ 27 | 40  |   | 
 +| total    | $g = \;$ 1500  | $g' = \;$ 29 |     | 
 + 
 +F<sub>crit</sub>(2, 27) =  3.35 
 +F<sub>cal</sub> = 5.25 
 + 
 +F<sub>cal</sub> > F<sub>crit</sub> 이므로 3집단 간의 평균은 통계학적으로 의미가 있는 차이를 가지고 있다. 
 +====== Example 2 ====== 
 +<code> 
 +x1 <- c(15, 20, 14, 13, 18, 16, 13, 12, 18, 41)
 x2 <- c(21, 25, 29, 18, 26, 22, 26, 24, 28, 21) x2 <- c(21, 25, 29, 18, 26, 22, 26, 24, 28, 21)
 x3 <- c(28, 30, 32, 28, 26, 30, 25, 36, 20, 15) x3 <- c(28, 30, 32, 28, 26, 30, 25, 36, 20, 15)
 </code> </code>
  
-<code>> data.frame(x1,x2,x3)+<code>> xc <- data.frame(x1,x2,x3)
    x1 x2 x3    x1 x2 x3
 1  15 21 28 1  15 21 28
Line 606: Line 635:
  
 <code> <code>
-> colnames(xs[1]) <- "wrong" +> colnames(xs)  <- c("wrong""condition")
-> colnames(xs[2]) <- "cond"+
 </code> </code>
  
-<code># cf+<code> 
 +# cf
 # lengthofelements <- length(x1) # lengthofelements <- length(x1)
 # varofvariable <- var(x1)</code> # varofvariable <- var(x1)</code>
  
 <code> <code>
-df_x1 +df.total <- length(xs$wrong) - 1 
-df_x2 +ss.total <- var(xs$wrong)*df_tot 
-df_x3+var.total <- ss.total/df.total 
 +var.total.r <- var(xs$wrong)
  
-ss_x1 +df.x1 <- length(x1)-1 
-ss_x2 +df.x2 <- length(x2)-1 
-ss_x3+df.x3 <- length(x3)-1 
 +ss.x1 <- var(x1)*df.x1 
 +ss.x2 <- var(x2)*df.x2 
 +ss.x3 <- var(x3)*df.x3
  
-df_bet +ss.within <- ss.x1 + ss.x2 + ss.x3 
-ss_bet+df.within <- df.x1 + df.x2 + df.x3 
 +ss.between <- ss.total - ss.within 
 +df.between <- df.total - df.within
  
-df_tot +ms.between <- ss.between/df.between 
-ss_tot+ms.within <- ss.within/df.within 
 +f.value <- ms.between/ms.within
  
-df_with +ss.between 
-ss_with+df.between
  
-df_bet +ss.within 
-ss_bet+df.within
  
-</code>+ms.between 
 +ms.within
  
-<code> +f.value 
-df_tot <- length(xs$ind) - +[1] 5.25 
-ss_tot <- var(xs$values)*df_tot + 
-var_tot <- var(xs$values)+f.crit <- qf(.95, df1=2, df2=27## p=.05 level에서 F(2,27)의 값은 qf 펑션으로 구합니다. 
 +f.crit 
 +[1] 3.354131 
 + 
 +#################################### 
 +## f.value가 f.crit 값보다 크므로  
 +## 세 그룹 간에 차이가 있다는 가설을  
 +## 받아들인다. (세 그룹 간에 차이가  
 +## 없다는 영가설을 부정한다) 
 +####################################
  
-df_x1 <- length(x1)-1 
-df_x2 <- length(x2)-1 
-df_x3 <- length(x3)-1 
-ss_x1 <- var(x1)*df_x1 
-ss_x2 <- var(x2)*df_x2 
-ss_x3 <- var(x3)*df_x3 
 </code> </code>
 ====== E.G. 1 (R) ====== ====== E.G. 1 (R) ======
anova.1539903998.txt.gz · Last modified: 2018/10/19 08:06 by hkimscil