The given data is-

Let
Ho​ ​: The concentartion of Alcl3​ gives same same results.
Ha​ ​:The concentartion of Alcl3​ does not gives same same results.
The variance table is given by-

The combines sample size n=n1​+n2​+n3​+n4​
=3+3+3+3+3=15
The number of samples K=5
The mean of the combined sample-
xˉ=5x1​+x2​+x3​+x4​+x5​​
=53.33+3+3.66+2.66+3.33​
=514.792​=2.96
The Mean square for treatment(MST)-
=k−1n1​(x1​−xˉ)2+n2​(x2​−x2​ˉ​)2+n3​(x3​−x3​ˉ​)2+n4​(x4​−x4​ˉ​)2+n5​(x5​−x5​ˉ​)2​
=5−13(3.33−2.96)2+3(3−2.96)2+3(3.66−2.96)2+3(2.66−2.96)+3(3.33−2.96)2​
=53(0.1369+0.0016+0.49+0.09+0.1369)​=0.51324
The Mean square for Error(MSE)-
=n−K(n1​−1)s12​+(n2​−1)s22​+(n3​−1)s32​+(n4​−1)s42​+(n5​−1)s52​​
=15−5(3−1)(0.0456)+(3−1)(0.00053)+(3−1)(0.163)+(3−1)(0.03)+(3−1)(0.0456)​
=102(0.0456+0.0053+0.163+0.03+0.0456)​
=0.0579
Then, F=MSEMST​
=0.05790.51324​=8.8642
For Annova test the degree of freedom df1​=n1​−1=3−1=2 and df2​=n−K=15−5=10
The Tabulated value of F at degree of freedome 2 and 10 at 0.05 level of significance is 4.10
Conclusion: As The calculated value of F is greater than the tabulated F at 95% level of significance.
So Ho​ is rejected. Hence The concentartion of Alcl3​ does not gives same same results