Question

In a properly validated biotechnology process (i.e., it was proven that the process occurs in the same way each time it is run) three samples of the product are taken during a manufacturing stage to verify the percentage of the active ingredient. The researcher suspects that the first sample influences the result of the second and in turn the second influences the result of the third. For this reason, the researcher decided to compare lots to verify if the samples are independent. Table 1 presents the results of this experiment. Perform an analysis of variance to corroborate the independence between samples. Use a level of significance of 0.05. Carry out and present your calculations by hand (no Excel, no Minitab, etc.).

Table 1: Data Problem 1

Lot	1st Sample	2nd Sample	3rd Sample
A	93.502	92.319	92.451
C	91.177	92.230	92.097
D	87.304	87.496	88.001
D2	81.275	80.564	80.386
G	79.865	79.259	79.209
G2	81.722	80.931	81.351

Answer 1

H0: (The samples are not independent)

HA: (At least one mean is different from other 2 means)

From the given data, the following statistics are calculated:

	1st sample	2nd sample	3rd sample	Total
N	6	6	6	18
	514.845	512.799	513.495	1541.139
Mean	514.845/6 = 85.8075	512.799/6= 85.4665	513.495/6 = 85.5825	1541.13918 = 85.619
	44340.3869	44007.0946	44127.1807	132474.6622
Std. Dev.	5.7066	5.9993	6.0166	5.5526

From the above Table, ANOVA Table is calculated as follows:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Squares	F
Between treatments	0.3607	2	0.3607/2 = 0.1804	0.1804/34.9186 = 0.0052
Within treatments	523.7783	15	523.7783/15 = 34.9186
Total	524.139	17

F = 0.1804/34.9186 = 0.0052

Degrees of Freedom for numerator = 2

Degrees of Freedom for denominator = 15

By Technology, p- value = 0.9949

Since p - value = 0.9949 is greater than = 0.05, the difference is not significant. Fail to reject null hypothesis.

Conclusion:
The data support the claim that the samples are not independent.