Question

As a part of a study, a chemical plant sampled the amount of chemicals in the soil surrounding their building. Samples were taken in front of the building (sample 1) and samples were taken in the back of the building (sample 2). The results in ppm are listed below.

Sample1	0	1.8	1.5	0.4	3.5	1.7	3.9
	5.4	2.5	2.7	3.1	4.7	2.5	4.2
Sample 2	5.5	3.9	5.9	2.6	6.8	1.8	4.5
	2.5	3.9	3.9	2.9	4.8	3.1	3.8

(a) Please test if there is a significant difference in the amount of chemicals in the soil in the front vs the back of the chemical plant. List the 5 steps of the hypothesis test.

(b) What is the 95% confidence interval for the difference in the amount of chemicals between the two soil sources?  

Answer 1

X :- Represent sample 1

Y :- Represent sample 2

Mean X̅ = Σ Xi / n
X̅ = 37.9 / 14 = 2.7071
Sample Standard deviation SX = √ ( (Xi - X̅ )2 / n - 1 )
SX = √ ( 31.6895 / 14 -1 ) = 1.5613

Mean Y̅ = ΣYi / n
Y̅ = 55.9 / 14 = 3.9929
Sample Standard deviation SY = √ ( (Yi - Y̅ )2 / n - 1 )
SY = √ ( 25.7293 / 14 -1) = 1.4068

To Test :-

H0 :- µ1 = µ2
H1 :- µ1 ≠ µ2

Test Statistic :-
t = (X̅1 - X̅2) / SP √ ( ( 1 / n1) + (1 / n2))



t = ( 2.7071 - 3.9929) / 1.4861 √ ( ( 1 / 14) + (1 / 14 ))
t = -2.2892

Test Criteria :-
Reject null hypothesis if | t | > t(α/2, n1 + n2 - 2)
Critical value t(α/2, n1 + n1 - 2) = t(0.05 /2, 14 + 14 - 2) = 2.056
| t | > t(α/2, n1 + n2 - 2) = 2.2892 > 2.056
Result :- Reject Null Hypothesis

There is a significant difference in the amount of chemicals in the soil in the front vs the back of the chemical plant.

Part b)

Confidence interval is :-
( X̅1 - X̅2 ) ± t( α/2 , n1+n2-2) SP √( (1/n1) + (1/n2))
t(α/2, n1 + n1 - 2) = t( 0.05/2, 14 + 14 - 2) = 2.056
( 2.7071 - 3.9929 ) ± t(0.05/2 , 14 + 14 -2) 1.4861 √ ( (1/14) + (1/14))
Lower Limit = ( 2.7071 - 3.9929 ) - t(0.05/2 , 14 + 14 -2) 1.4861 √( (1/14) + (1/14))
Lower Limit = -2.4406
Upper Limit = ( 2.7071 - 3.9929 ) + t(0.05/2 , 14 + 14 -2) 1.4861 √( (1/14) + (1/14))
Upper Limit = -0.131
95% Confidence Interval is ( -2.4406 , -0.131 )