We want to test the hypotheses that chemicals have different effects on the strength of a particular type of cloth. However, bolts might have an additive effect and a block design might be required. We want to test:
\[H_0:\tau_i=0 \\ H_a: \tau_i \neq 0 \space \text{for any} \space i\]
From the ANOVA test, we can see that the bolts have a p-value really small, which means that the bolts block effects is additive and does not influence the observations, as expected. In contrast, the chemicals showed a p-value of 0.121, which suggests that the null hypotheses can not be rejected and there might be a significantly similarity between each type of chemical on the strength of cloth.
obs <- c(73,68,74,71,67,
73,67,75,72,70,
75,68,78,73,68,
73,71,75,75,69)
chemical <- as.factor(c(rep(1,5),rep(2,5),rep(3,5),rep(4,5)))
bolt <- as.factor(c(seq(1,5),seq(1,5),seq(1,5),seq(1,5)))
aov.model <- aov(obs~bolt+chemical)
summary(aov.model)## Df Sum Sq Mean Sq F value Pr(>F)
## bolt 4 157.00 39.25 21.606 2.06e-05 ***
## chemical 3 12.95 4.32 2.376 0.121
## Residuals 12 21.80 1.82
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1We want to obtain the parameters model \(\tau_i\) and \(\beta_j\) given that we need to solve a system of the format:
$$:ab+b_1+b_2+… +b_a + a_1+a_2+ … + a_b=y..\
_1:b+b_1+ +_1+_2+ … + b=y{1.}\
_2:b +b_2 +_1+_2+ … + b=y{2.}\
$$ The table continues but for simplicity purpose it will not be shown here. Furthermore, we need to apply the constraints of:
\[ \sum_{i=1}^{a}{\hat\tau_i}=0\\ \sum_{j=1}^{b}{\hat\beta_j}=0\]
Then, we have the values: $$ ==71.75, _i=-1.15, _2=-0.35, _3=0.65, _4=0.85, \
_1=1.75, _2=-3.25, _3=3.75, _4=1, _5=-3.25 $$
We want to test the hypotheses that there are differences between each ingredient on the reaction time of a chemical process:
\[H_0:\mu_A=\mu_B=\mu_C=\mu_D=\mu_E\\ H_a:\text{at least one mean is different}\]
obs <- c(8,7,1,7,3,
11,2,7,3,8,
4,9,10,1,5,
6,8,6,6,10,
4,2,3,8,8)
day <- as.factor(c(seq(1,5),seq(1,5),seq(1,5),seq(1,5),seq(1,5)))
batch <- as.factor(c(rep(1,5),rep(2,5),rep(3,5),rep(4,5),rep(5,5)))
trt <- as.factor(c('A','B','D','C','E',
'C','E','A','D','B',
'B','A', 'C','E','D',
'D','C','E', 'B','A',
'E','D','B', 'A','C'))
aov.model <- aov(obs~trt+batch+day)
summary(aov.model)## Df Sum Sq Mean Sq F value Pr(>F)
## trt 4 141.44 35.36 11.309 0.000488 ***
## batch 4 15.44 3.86 1.235 0.347618
## day 4 12.24 3.06 0.979 0.455014
## Residuals 12 37.52 3.13
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1With \(\alpha=0.05\), we reject the null hypotheses with a p-value of 0.000488. Therefore, there are significant differences between each ingredient. The blocks effects will be additive and the p-value is large because it does not (and was not supposed to) influence the observations.