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Chapter 15: Q7E (page 661)

Use the large-sample version of the Wilcoxon test at significance level .05 on the data of Exercise 37 in Section 9.3 to decide whether the true mean difference between outdoor and indoor concentrations exceeds .20.

Short Answer

Expert verified

Because of the fact that

笔=0.0019&濒迟;0.05=伪

reject null hypothesis.

Step by step solution

01

Wilcoxon rank sum test

The Wilcoxon test is based upon ranking the nA + nB observations of the combined sample. Each observation has a rank: the smallest has rank 1, the 2nd smallest rank 2, and so on. The Wilcoxon rank-sum test statistic is the sum of the ranks for observations from one of the samples.

02

Finding the value of P

The hypothesis of interest are

H0:饾泹=0.2,

Versus alternative

HD:饾泹D>0.2,

The following table represents the value required to compute the test statistic value and corresponding P value:

i

\(x_i^1\)

\(x_i^2\)

\({y_i} = x_i^1 - x_i^2\)

\({y_i} - 0.2\)

\(\left| {{y_i} - 0.2} \right|\)

\(rank({y_i})\)

\(sign({y_i})\)

1

0.29

0.07

0.22

0.02

0.02

2

1

2

0.68

0.08

0.6

0.4

0.4

22

1

3

0.47

0.09

0.38

0.18

0.18

16

1

4

0.54

0.12

0.42

0.22

0.22

18

1

5

0.97

0.12

0.85

0.65

0.65

29

1

6

0.35

0.12

0.23

0.03

0.03

3.5

1

7

0.49

0.13

0.36

0.16

0.16

13

1

8

0.84

0.14

0.7

0.5

0.5

26

1

9

0.86

0.15

0.71

0.51

0.51

27

1

10

0.28

0.15

0.13

-0.07

0.07

7

-1

11

0.32

0.17

0.15

-0.05

0.05

5.5

-1

12

0.32

0.17

0.15

-0.05

0.05

5.5

-1

13

1.55

0.18

1.37

1.17

1.17

32

1

14

0.66

0.18

0.48

0.28

0.28

20

1

15

0.29

0.18

0.11

-0.09

0.09

8

-1

16

0.21

0.18

0.03

-0.17

0.17

14

-1

17

1.02

0.19

0.83

0.63

0.63

28

1

18

1.59

0.2

1.39

1.19

1.19

33

1

19

0.9

0.22

0.68

0.48

0.48

25

1

20

0.52

0.22

0.3

0.1

0.1

10

1

21

0.12

0.23

- 0.11

-0.31

0.31

21

-1

22

0.54

0.23

0.31

0.11

0.11

12

1

23

0.88

0.25

0.63

0.43

0.43

23

1

24

0.49

0.26

0.23

0.03

0.03

3.5

1

25

1.24

0.28

0.96

0.76

0.76

31

1

26

0.48

0.28

0.2

0

0

1

0

27

0.27

0.29

-0.02

-0.22

0.22

17

-1

28

0.37

0.34

0.03

-0.17

0.17

15

-1

29

1.26

0.39

0.87

0.67

0.67

30

1

30

0.7

0.4

0.3

0.1

0.1

9

1

31

0.76

0.45

0.31

0.11

0.11

11

1

32

0.99

0.54

0.45

0.25

0.25

19

1

33

0.36

0.62

-0.26

0.46

0.46

24

-1

The test statistic value is

s+=442.8

For the large sample test, use different test statistic value:

\(\begin{array}{l}Z = \frac{{{S_ + } - \frac{{n(n + 1)}}{4}}}{{\sqrt {\frac{{n(n + 1)(2n + 1)}}{{24}}} }}\\z = 2.89\end{array}\)

The corresponding P-value is (the test is upper tailed)

笔=笔(窜鈮�2.89)=0.0019

Because of the fact that

笔=0.0019&濒迟;0.05=伪

reject null hypothesis.

Hence, the final answer is that it will reject null hypothesis.

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