Calculation: The average of means of low A and low B treatment is 0.5(μ2+μ4) and the mean of Placebo treatment is μ1. Therefore, null and alternative hypotheses can be formed as follows:

H0:ψ1=0Ha:ψ1>0

Null hypotheses represent that there is no difference in the Placebo treatment and average of low A and low B treatment, which is represented as below:

H0:μ1=0.5(μ2+μ4)

Hence, on equating ψ1=0, the expression of contrast can be obtained as given below:

μ1−0.5(μ2+μ4)=0, since ψ1 is equal to 0.

The difference of means of low A and high A treatment is (μ3−μ2) and the difference of means of high B and low B treatment is (μ5−μ4). Therefore, null and alternative hypotheses can be formed as follows:

H0:ψ2=0Ha:ψ2>0

Null hypotheses represent that there is no difference in the difference of low A and high A treatment and difference of high B and low B treatment, which is represented as below:

H0:(μ3−μ2)=(μ5−μ4)

Hence, on equating ψ1=0, expression of contrast can be obtained as given below:

(μ3−μ2)−(μ5−μ4)=0, since ψ2=0

The coefficients of contrast for the contrast ψ1 are given below:

a1=1a2=−0.5a3=0a4=−0.5

a5=0

The coefficients of contrast for the contrast ψ2 are given below:

a1=0a2=−1a3=1a4=1

a5=−1

Calculation: The value of contrast c1 can be calculated by using the formula given below:

c1=∑iaix¯i=(1×11.80)+((−0.5)×15.25)+((−0.5)×16.15)=−3.9

The value of contrast c2 can be calculated by using the formula given below:

c2=∑iaix¯i

c2=((−1)×15.25)+(1×18.55)+(1×16.15)+((−1)×17.10)=2.35

Standard error can be calculated by using the formula given below:

SEc1=spooled∑ai2ni

Pooled standard deviation can be calculated as given below:

sp=(n1−1)s12+(n2−1)s22+⋯+(n5−1)s22n1+n2+n3+n4+n5−k=(5−1)17.20+(5−1)13.10+⋯+(5−1)12.50(5+5+5+5+5)−5=3.487

For the first contrast, the standard error SEc1 can be calculated as given below:

SEc1=3.48714+(−0.5)24+(−0.5)24=3.487×0.375=2.1353

For the second contrast, the standard error SEc2 can be calculated as given below:

SEc1=3.48714+14+14+14=3.487×1=3.487

Interpretation: Therefore, it can be concluded that for the first contrast, the estimated value of standard deviation is 2.1353 and for the second contrast, the estimated value of standard deviation is 3.487.

Calculation: Test statistic can be calculated by using following formula, which is mentioned below:

t=cSEc

To calculate the test statistic for first contrast, the value of c1=−3.9 and is substituted in the formula mentioned above:

t=cSEc=−3.92.1353=−1.8264

The total sample size of rats is 25. There are five groups, which are represented by k.

The degree of freedom for contrast can be obtained as shown below:

N−k=25−5=20

The P-value can also be calculated by the software Microsoft Excel by using the following command:

Insert the values of x and degree of freedom, which are –1.8264 and 20, respectively. After this, press enter and the output is shown below in the snapshot:

Hence, the P- value is obtained as 0.0414, which is less than the significance level 0.05, so the decision is to reject the null hypothesis.

To calculate the test statistic for second contrast, the value of c2=2.35 and SEc2=3.487 is substituted in the formula mentioned above:

t=cSEc=2.353.487=0.6739

The P-value can also be calculated by the software Microsoft Excel by using the following command:

Insert the values of x and degree of freedom, which is 0.6739 and 20, respectively. After this, press enter and the output is shown below in the snapshot:

Hence, the P- value is obtained as 0.2540, which is greater than the significance level 0.05, so the decision is to accept the null hypothesis.

Conclusion: Therefore, it can be concluded that the result is significant for the first contrast and the result is not significant for the second contrast.