Green Spectrophotometric Determination of Organophosphate in Selected Fruits and Vegetables

Table 3

Statistical analysis of the data.

Concentration of Indian sample in mg (X1)	Concentration of local sample in mg (X2)	X1^2	X2^2
1.976	0.473	3.904576	0.223729
0.024	0.003	0.000576	0.000009
1.992	0.007	3.968064	0.000049
2.144	0.029	4.596736	0.000841
1.992	0.007	3.968064	0.000049
1.353	0.013	1.830609	0.000169
∑X1 = 9.481	∑X2 = 0.532	1^2 = 18.26863	∑ X2^2 = 0.224846

S_p² = 1/(n₁ + n₂ − 2) [∑X₁² − (∑X₁₎²/n₁ + ∑X₂² − (∑X₂)²/n₂]. S_p² = 0.346. Again, mean of first sample = ∑X₁/n₁ = 1.58. Mean of second sample = ∑X₂/n₂ = 0.00887. Similarly, independent t-test (t) = (mean of 1^st sample − mean of 2^nd sample)/√ (S_p²/(1/n₁ + 1/n₂) = 4.391244267. Here, d.f = n₁+n₂ − 2 = 6+6 − 2 = 10. The tabulated value of t-test at 5% level of significance at 10 d. f. is 2.228 (two tails). Now, value for the t-test is between 0.002 − 0.001. Let’s call it 0.0015. So, value is lower than 0.05. Thus, calculated value > tabulated value so there is significant greater amount of organophosphate in Indian sample than local sample. Also, the value shows significant difference in concentration.