A chemist wishes to test the effect of four chemical agents on the strength of a particular type of cloth. Because there might be variability from one bolt to another, the chemist decides to use a randomized block design, with the bolts of cloth considered as blocks. She selects five bolts and applies all four chemicals in random order to each bolt. The resulting tensile strengths follow.
Analyze the data from this experiment (use \(\alpha\)= 0.05) and draw appropriate conclusions.
\(H_0\):\(\tau_i\)=0
\(H_a\):\(\tau_i\) doesn’t equal 0
Where \(\tau_i\) is the chemical treatment
Chemical_1 <- c(73,68,74,71,67)
Chemical_2<- c(73,67,75,72,70)
Chemical_3 <- c(75,68,78,73,68)
Chemical_4<- c(73,71,75,75,69)
observations <-c(Chemical_1,Chemical_2,Chemical_3,Chemical_4)
Chemicals <- as.fixed(c(rep(1,5),rep(2,5),rep(3,5),rep(4,5)))
Bolts <-as.fixed(c(rep(seq(1,5),4)))
Bolt_Chemical_model <- lm(observations~Chemicals+Bolts)
gad(Bolt_Chemical_model)## Analysis of Variance Table
##
## Response: observations
## Df Sum Sq Mean Sq F value Pr(>F)
## Chemicals 3 12.95 4.317 2.3761 0.1211
## Bolts 4 157.00 39.250 21.6055 2.059e-05 ***
## Residual 12 21.80 1.817
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1I concluded that I don’t have enough evidence to reject the null hypothesis that chemicals have the same effect on a piece of cloth.
Assuming that chemical types and bolts are fixed, estimate the model parameters \(\tau_i\) and \(\beta_j\) in Problem 4.3.
Tau_1 <- mean(Chemical_1)-mean(observations)
Tau_2 <- mean(Chemical_2)-mean(observations)
Tau_3 <- mean(Chemical_3)-mean(observations)
Tau_4 <- mean(Chemical_4)-mean(observations)
Beta_1 <- mean(73,73,75,73)-mean(observations)
Beta_2 <- mean(68,67,68,71)-mean(observations)
Beta_3 <- mean(74,75,78,75)- mean(observations)
Beta_4<- mean(71,72,73,75)- mean(observations)
Beta_5 <- mean(67,70,68,69)- mean(observations)
Tau_matrix <- c(Tau_1,Tau_2,Tau_3,Tau_4)
Beta_matrix <- c(Beta_1,Beta_2,Beta_3,Beta_4,Beta_5)
print(Tau_matrix)## [1] -1.15 -0.35 0.65 0.85print(Beta_matrix)## [1] 1.25 -3.75 2.25 -0.75 -4.75\(\tau_1\) = -1.15 \(\tau_2\) = -0.35 \(\tau_3\) = 0.65 \(\tau_4\) = 0.85
\(\beta_1\) = 1.25 \(\beta_2\)=-3.75 \(\beta_3\)=2.25 \(\beta_4\)=-0.75 \(\beta_5\)=-4.75
An industrial engineer is investigating the effect of four assembly methods (A, B, C, D) on the assembly time for a color television component. Four operators are selected for the study. Furthermore, the engineer knows that each assembly method produces such fatigue that the time required for the last assembly may be greater than the time required for the first, regardless of the method. That is, a trend develops in the required assembly time. To account for this source of variability, the engineer uses the Latin square design shown below. Analyze the data from this experiment ( \(\alpha\)=0.5) and draw appropriate conclusions.
( i accidentally did problem 4.23 instead 4.22 but it’s the same method and concepts)
\(H_0\):\(\tau_i\)=0
\(H_a\):\(\tau_i\) doesn’t equal 0
\(\tau_i\) is the assembly time method for a colored tv.
obs <- c(10,14,7,8,
7,18,11,8,
5,10,11,9,
10,10,12,14)
Order_Assembly <-as.factor(c(rep(1,4),rep(2,4),rep(3,4),rep(4,4)))
Operator <-as.factor(c(rep(seq(1,4),4)))
Assembly_Methods <- as.factor(c(3,4,1,2,2,3,4,1,1,2,3,4,4,1,2,3))
IE_model <- aov(obs~Order_Assembly+Operator+Assembly_Methods)
summary(IE_model)## Df Sum Sq Mean Sq F value Pr(>F)
## Order_Assembly 3 18.5 6.167 3.524 0.08852 .
## Operator 3 51.5 17.167 9.810 0.00993 **
## Assembly_Methods 3 72.5 24.167 13.810 0.00421 **
## Residuals 6 10.5 1.750
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1I concluded that I have enough evidence to reject the null hypotheses that assembly methods have the same effect on the assembly time for a color tv.