9) Water samples were taken to assess the concentration of trace metals in a bod
ID: 3364426 • Letter: 9
Question
9) Water samples were taken to assess the concentration of trace metals in a body of water; in this case, the concentration of zinc was of interest. Water samples were taken at ten different locations. Does the data suggest differences in the concentration at the bottom (top row) and surface (bottom row)? (Are these samples dependent or independent?) (2)
L 1
L 2
L 3
L 4
L 5
L 6
L 7
L 8
L 9
L 10
.43
.266
.567
.531
.707
.716
.651
.589
.469
.723
.415
.238
.390
.410
.605
.609
.632
.523
.411
.612
L 1
L 2
L 3
L 4
L 5
L 6
L 7
L 8
L 9
L 10
.43
.266
.567
.531
.707
.716
.651
.589
.469
.723
.415
.238
.390
.410
.605
.609
.632
.523
.411
.612
Explanation / Answer
An independent variable sometimes called an experimental or predictor variable, is a variable that is being manipulated in an experiment in order to observe the effect on a dependent variable, sometimes called an outcome variable
given two samples differences in the concentration at the bottom (top row) and surface (bottom row).so that it is independent variables
Given that,
mean(x)=0.5649
standard deviation , s.d1=0.1468
number(n1)=10
y(mean)=0.4845
standard deviation, s.d2 =0.1312
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.262
since our test is two-tailed
reject Ho, if to < -2.262 OR if to > 2.262
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =0.5649-0.4845/sqrt((0.02155/10)+(0.01721/10))
to =1.291
| to | =1.291
critical value
the value of |t | with min (n1-1, n2-1) i.e 9 d.f is 2.262
we got |to| = 1.29135 & | t | = 2.262
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.2913 ) = 0.229
hence value of p0.05 < 0.229,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 1.291
critical value: -2.262 , 2.262
decision: do not reject Ho
p-value: 0.229
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