In: Statistics and Probability
A petrochemical plant was built by a river. To verify whether the plant was contributing to pollution of the river water with benzo(a)pyrene, two sets of data were obtained by analyzing water samples from upstream ( mean conc. = 0.95ppb; n=5; s=0.05 ppb) and downstream (mean conc.= 1.10 ppb; n=6; s=0.08ppb) with respect to the plant. Based on this data, one would reach this conclusion at a 95% confidence level as to whether the plant is making significant contribution of benzo(a)pyrene pollution of the river water.
A) The contribution is significant.
B) The contribution is not.
C) More data is needed to reach a conclusion
D) none of the above
C) 1.18 x 10^-12 mol/L
D) None of the above
Ho : µ1 - µ2 = 0
Ha : µ1-µ2 < 0
Level of Significance , α =
0.05
Sample #1 ----> 1
mean of sample 1, x̅1= 0.950
standard deviation of sample 1, s1 =
0.050
size of sample 1, n1= 5
Sample #2 ----> 2
mean of sample 2, x̅2= 1.100
standard deviation of sample 2, s2 =
0.080
size of sample 2, n2= 6
difference in sample means = x̅1-x̅2 =
0.9500 - 1.1 =
-0.15
pooled std dev , Sp= √([(n1 - 1)s1² + (n2 -
1)s2²]/(n1+n2-2)) = 0.0683
std error , SE = Sp*√(1/n1+1/n2) =
0.0414
t-statistic = ((x̅1-x̅2)-µd)/SE = ( -0.1500
- 0 ) / 0.04
= -3.6262
Degree of freedom, DF= n1+n2-2 =
9
p-value = 0.0028 [ excel function:
=T.DIST(t stat,df) ] 52
Conclusion: p-value <α , Reject null
hypothesis
A) The contribution is significant.
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