Question

Unusually high concentration of metals in drinking water can pose a health hazard. Twenty couples of data were taken from different locations of Kuş Cenneti Lake measuring zinc concentration in bottom water and surface water.

Table 1. Zinc Concentration (mg/l) in bottom and surface water in different regions and areas of Karacakaçan Lake

Location	Region	Area	Zinc_Bottom	Zinc_Surface	Zinc_Mean	TZ3_Mean
Loc 1	North	Area A	0.51	0.49	0.50	2.67
Loc 2	North	Area A	0.46	0.45	0.46	2.45
Loc 3	North	Area A	0.52	0.52	0.52	2.66
Loc 4	North	Area A	0.51	0.50	0.50	2.56
Loc 5	North	Area A	0.58	0.57	0.58	3.03
Loc 6	North	Area A	0.57	0.55	0.56	3.21
Loc 7	North	Area A	0.59	0.58	0.59	3.50
Loc 8	North	Area B	0.46	0.44	0.45	2.40
Loc 9	North	Area B	0.46	0.44	0.45	2.35
Loc 10	North	Area B	0.45	0.43	0.44	2.56
Loc11	East	Area B	0.41	0.40	0.41	1.33
Loc 12	East	Area B	0.41	0.39	0.40	1.56
Loc 13	East	Area B	0.41	0.39	0.40	1.23
Loc 14	East	Area B	0.47	0.45	0.46	1.12
Loc 15	East	Area C	0.41	0.39	0.40	1.23
Loc 16	East	Area C	0.37	0.36	0.37	1.10
Loc 17	East	Area C	0.41	0.39	0.40	1.20
Loc 18	East	Area C	0.40	0.39	0.39	1.10
Loc 19	East	Area C	0.39	0.37	0.38	1.12
Loc 20	East	Area C	0.46	0.44	0.45	1.30

Does the data suggest that the true average concentration in the north region is higher than that of east region? Formulating your hypothesis and providing an explanation on the reason of your choice, conduct the appropriate test and interprete your results (α=0.05)

Answer 1

From the given scenario, the null and alterate hypotheses can be stated as follows: Null hypothesis: The true average concentration in the north region is less than or equal to that of east region. H: 12 Alternate hypothesis: The true average concentration in the north region is higher than that of east region. H:14 > 113 Calculate the test statistic as follows: 5 n 0.5+...+0.44 10 5.050 10 = 0.505 (x-7) ni (0.5-0.505) +...+(0.44 -0.505)? 10-1 = 0.003 n2 0.41+ ... +0.45 10 4.060 10 = 0.406 (41;-) ni (0.41 -0.406)2 +...+(0.45-0.406)? 10-1 = 0.001 si 57

n 72 0.505 -0.406 0.003 0.001 + 10 10 0.099 0.020 = 4.97 P=T DIST.RT (t.df) Use Excel function T DIST RT (t.df) to get the probability value for right tailed test.

= 1 + n df s m-1 m 1-1 m2 0.003 0.001 + 10 10 1 0.003 + 10-11 10 10-1 0.00000016 0.0000000118 =13 P=T DIST RT(4.97.13) =0.0001 1 0.001 10 The P-value of the test 0.0001 is less than the level of significance 0.05. Thus, the researcher can reject the null hypothesis. Conclusion: The true average concentration in the north region is higher than that of east region.