Question 1

Hypothesis:

Null Hypothesis:

\(H_o : \mu_1 = \mu_2 = \mu_3 =\mu_i\)
Alternative Hypothesis:

\(Ha\) : atleast one \(\mu_i\) differs

As we know it has fixed effects hence we can write our hypothesis as,

Null Hypothesis:

\(H_o : \tau_i=0\) for all i
Alternative Hypothesis:

\(H_a\) : \(\tau_i \neq 0\) for some i

Linear Effects:

\(y_{ij} = \mu + \tau_i + \beta_j + \epsilon_{ij}\)

Where \(\mu\) is the grand mean

Where \(\tau_i\) is the fixed effects for treatment i

Where \(\beta_j\) is the block effect for j

Where \(\epsilon_{ij}\) is the random error

chemical <- c(rep(1,5),rep(2,5),rep(3,5),rep(4,5))
bolts <- c(rep(seq(1,5),4))
obs <- c(73,68,74,71,67,73,67,75,72,70,75,68,78,73,68,73,71,75,75,69)

chemical <- as.fixed(chemical)
bolts <- as.fixed(bolts)
model <- lm(obs~chemical+bolts)
gad(model)

## Analysis of Variance Table
## 
## Response: obs
##          Df Sum Sq Mean Sq F value    Pr(>F)    
## chemical  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 ' ' 1

P value is 0.1211 < 0.15 hence we reject Null Hypothesis

Question 2

Hypothesis:

Null Hypothesis

\(H_o : \mu_1 = \mu_2 = \mu_3 =\mu_i\)
Alternative Hypothesis:

\(Ha\) : atleast one \(\mu_i\) differs

As we know it has fixed effects hence we can write our hypothesis as,

Null Hypothesis:

\(H_o : \tau_i=0\) for all i
Alternative Hypothesis:

\(H_a\) : \(\tau_i \neq 0\) for some i

Linear Effects:

\(y_{ij} = \mu + \tau_i + \epsilon_{ij}\)

Where \(\mu\) is the grand mean

Where \(\tau_i\) is the fixed effects for treatment i

Where \(\epsilon_{ij}\) is the random error

chemical <- c(rep(1,5),rep(2,5),rep(3,5),rep(4,5))
replication <- c(rep(seq(1,5),4))
obs <- c(73,68,74,71,67,73,67,75,72,70,75,68,78,73,68,73,71,75,75,69)

chemical <- as.fixed(chemical)
model2 <- lm(obs~chemical)
gad(model2)

## Analysis of Variance Table
## 
## Response: obs
##          Df Sum Sq Mean Sq F value Pr(>F)
## chemical  3  12.95  4.3167  0.3863 0.7644
## Residual 16 178.80 11.1750

P value is 0.7644 > 0.15 hence we fail to reject Null Hypothesis

Question 3 Comparing results of question 1 and question 2

Answer :- We can see in first question that we blocked bolts , and we got our p value as 0.1211 which was less than 0.15 , hence we reject Null Hypothesis. But on the other hand in Q2) we did not consider blocking on bolts(replication) which gave us a p value of 0.7644 which is greater than 0.15, hence we fail to reject Null Hypothesis. This tells us that when we block we included nuisance variability and it compelled us to reject hypothesis which is actually what we must do when we know there is a nuisance variability. But when we ignore(which must not be a standard practice we accepted the NULL Hypothesis. We can state that if we do not block it can significantly affect our decisions. So for question on to block or not to block we must do it wherever it is necessary, so that we do not make wrong decision.

Yes, i do think so that it represented a significant amount of nuisance variability as it changed our decision of rejecting Null hypothesis while blocking and accepting without blocking.

DOE Flipped Assignment - 11

Thakur , Siddiq , Mustoor

Question 1

Hypothesis:

Null Hypothesis:

\(H_o : \mu_1 = \mu_2 = \mu_3 =\mu_i\)

Alternative Hypothesis:

\(Ha\) : atleast one \(\mu_i\) differs

As we know it has fixed effects hence we can write our hypothesis as,

Null Hypothesis:

\(H_o : \tau_i=0\) for all i

Alternative Hypothesis:

\(H_a\) : \(\tau_i \neq 0\) for some i

Linear Effects:

\(y_{ij} = \mu + \tau_i + \beta_j + \epsilon_{ij}\)

Where \(\mu\) is the grand mean

Where \(\tau_i\) is the fixed effects for treatment i

Where \(\beta_j\) is the block effect for j

Where \(\epsilon_{ij}\) is the random error

P value is 0.1211 < 0.15 hence we reject Null Hypothesis

Question 2

Hypothesis:

Null Hypothesis

\(H_o : \mu_1 = \mu_2 = \mu_3 =\mu_i\)

Alternative Hypothesis:

\(Ha\) : atleast one \(\mu_i\) differs

As we know it has fixed effects hence we can write our hypothesis as,

Null Hypothesis:

\(H_o : \tau_i=0\) for all i

Alternative Hypothesis:

\(H_a\) : \(\tau_i \neq 0\) for some i

Linear Effects:

\(y_{ij} = \mu + \tau_i + \epsilon_{ij}\)

Where \(\mu\) is the grand mean

Where \(\tau_i\) is the fixed effects for treatment i

Where \(\epsilon_{ij}\) is the random error

P value is 0.7644 > 0.15 hence we fail to reject Null Hypothesis

Question 3 Comparing results of question 1 and question 2

Yes, i do think so that it represented a significant amount of nuisance variability as it changed our decision of rejecting Null hypothesis while blocking and accepting without blocking.