Use the Fisher LSD method to determine which means are different. In the Fisher

Question

Use the Fisher LSD method to determine which means are different. In the Fisher LSD Caclulation, what is N-a (the 2nd DOF for t)??

An article in the Journal of Agricultural Engineering Research (1992, Vol. 52, pp. 53-76) described an experiment to investigate the effect of drying temperature of wheat grain on baking quality bread. Three temperature levels were used, and the response variable measured was the volume of the loaf of bread produced. The data are as follows: Temperature (oco Volume (CC) 70.0 1245 1235 1285 1245 1235 1235 1240 1200 1220 1210 75.0 80.0 1225 1200 1170 1155 1095

Explanation / Answer

Using R

mix_tech <- rep(c(70,75,80),each=5)
mix_tech <- factor(mix_tech)
ten_str <- c(1245,1235,1285,1245,1235,1235,1240,1200,1220,1210,1225,1200,1170,1155,1095)
model <- lm(ten_str~mix_tech)
ANOVA <- anova(model)
ANOVA

# Fisher Least Significant Difference (LSD) method (to compare pairwise Means)

data <- matrix(ten_str,nrow=3,byrow = TRUE)
data <- as.data.frame(data)
Mean_data <- apply(data, 1, mean)

N <- length(ten_str)                         # Total number of Observations
a <- length(unique(mix_tech))                # Level of temperature
n <- dim(data)[2]                            # Number of observations in each level (temperature)
alpha <- 0.05
error_df <- ANOVA$Df[2]                      # error DF (N-a+1)
mix_tech_df <- ANOVA$Df[1]                   # Treatment DF (a-1)

abs_diff <- c()
for(i in 1:(a-1))
{
for(j in (i+1):a){
    if(i>1)
      abs_diff[i+(j-1)] <- abs(Mean_data[i]-Mean_data[j])
    else
      abs_diff[i+(j-2)] <- abs(Mean_data[i]-Mean_data[j])
}
}

abs_diff

MSE <- ANOVA$`Mean Sq`[2]
t <- qt(1-alpha/2,N-a)
t
LSD <- t*sqrt(2*MSE/n)
LSD
# If Absolute Difference > Fisher LSD then Reject H0 (corrousponding means Differs Significantly)
# N-a is (N is total observation)- (a is number of row)
#hypothesis, H0:u1=u2=u3 vs H1:u1=!u2=!u3 (u is mean of corresponding level of temperature)

USING minitab:-

MTB > OneWay;

SUBC>   Response 'volume';

SUBC>   Categorical 'temperature';

SUBC>   IType 0;

SUBC>   Fisher 5;

SUBC>   GMCI;

SUBC>   TANOVA;

SUBC>   Nodefault.

One-way ANOVA: volume versus temperature

Method

Null hypothesis         All means are equal

Alternative hypothesis At least one mean is different

Significance level      = 0.05

Analysis of Variance

Source       DF Adj SS Adj MS F-Value P-Value

temperature   2   16480    8240     7.84    0.007

Error        12   12610    1051

Total        14   29090

Fisher Pairwise Comparisons

Grouping Information Using the Fisher LSD Method and 95% Confidence

temperature N     Mean Grouping

70           5 1249.00 A

75           5 1221.00 A

80           5   1169.0    B

Means that do not share a letter are significantly different.

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.