Answer to Question #173588 in Statistics and Probability for ANJU JAYACHANDRAN

Question

4. a) Plating of gold on wrist watches requires a chemical called . AiCI 3

Concentration of

the chemical that is added to the solution is an important factor. The objective is to

choose that concentration for which plating is uniform. Three concentration of AiCI 3

chosen were 5%, 15% and 20%. These were added in the solution and plating was

done and thickness (in units) measured on 5 samples is gives below.

Sample

→

↓

Conc. of AiCI3

1 2 3 4 5

5% 3 2 4 2 4

15% 4 4 3 2 4

25% 3 3 4 4 2

Use ANOVA to comment on whether the concentration of AiCI gives same result or 3

not. (Use 05.0 α = )

Answer 1

The given data is-

Let

$H_o$ ​: The concentartion of $Alcl_3$ gives same same results.

$H_a$ ​:The concentartion of $Alcl_3$ does not gives same same results.

The variance table is given by-

The combines sample size $n=n_1+n_2+n_3+n_4$

$=3+3+3+3+3\\=15$

The number of samples $K=5$

The mean of the combined sample-

$\bar{x}=\dfrac{x_1+x_2+x_3+x_4+x_5}{5}$

$=\dfrac{3.33+3+3.66+2.66+3.33}{5}$

$=\dfrac{14.792}{5}=2.96$

The Mean square for treatment(MST)-

$=\dfrac{n_1(x_1-\bar{x})^2+n_2(x_2-\bar{x_2})^2+n_3(x_3-\bar{x_3})^2+n_4(x_4-\bar{x_4})^2+n_5(x_5-\bar{x_5})^2}{k-1}$

$=\dfrac{3(3.33-2.96)^2+3(3-2.96)^2+3(3.66-2.96)^2+3(2.66-2.96)+3(3.33-2.96)^2}{5-1}$

$=\dfrac{3(0.1369+0.0016+0.49+0.09+0.1369)}{5}=0.51324$

The Mean square for Error(MSE)-

$=\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2+(n_3-1)s_3^2+(n_4-1)s_4^2+(n_5-1)s_{5}^2}{n-K}$

$=\dfrac{(3-1)(0.0456)+(3-1)(0.00053)+(3-1)(0.163)+(3-1)(0.03)+(3-1)(0.0456)}{15-5}$

$=\dfrac{2(0.0456+0.0053+0.163+0.03+0.0456)}{10}$

$=0.0579$

Then, $F=\dfrac{MST}{MSE}$

$=\dfrac{0.51324}{0.0579}=8.8642$

For Annova test the degree of freedom $df_1=n_1-1=3-1=2$ and $df_2=n-K=15-5=10$

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 $H_o$ is rejected. Hence The concentartion of $Alcl_3$ does not gives same same results