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The article 鈥淧roduction of Gaseous Nitrogen in Human Steady-State Conditions鈥 (J. of Applied Physiology, 1972: 155鈥159) reports the following observations on the amount of nitrogen expired (in liters) under four dietary regimens: (1) fasting, (2) 23% protein, (3) 32% protein, and (4) 67% protein. Use the Kruskal-Wallis test at level .05 to test equality of the corresponding \[{\mu _I}'s\].

1 4.079 4.859 3.540 5.047 3.298

2 4.368 5.668 3.572 5.848 3.802

3 4.169 5.709 4.416 5.666 4.123

4 4.928 5.608 4.940 5.291 4.674

1 4.679 2.870 4.648 3.847

2 4.844 3.578 5.393 4.374

3 5.059 4.403 4.496 4.688

4 5.038 4.905 5.208 4.806

Short Answer

Expert verified

Do not reject null hypothesis

Step by step solution

01

Kruskal- Wallis test:

The kruskal- wallis test

Test for testing equality of the \[{\mu _i}'s\]. The test static is

\[\begin{array}{l}K = \frac{{12}}{{N\left( {N + 1} \right)}}\sum\limits_{i = 1}^I {{J_i}} {\left( {{{\bar R}_i} - \frac{{N + 1}}{2}} \right)^2}\\ = \frac{{12}}{{N\left( {N + 1} \right)}}\sum\limits_{i = 1}^I {\frac{{R_i^2}}{{{J_i}}} - 3} \left( {N + 1} \right)\end{array}\]

When the null hypothesis is true, and either

\[\begin{array}{l}I = 3,{J_i} \ge 6\left( {i = 1,2,3} \right)\\or,\\I > 3,{J_i} \ge 5\left( {i = 1,2,I} \right)\end{array}\]

Test statistic K has chi-squared distribution with I-1 degrees of freedom. The P-value is corresponding are to the right of k under the \[X_{I - 1}^2\]curve

02

solving further:

Two table goes together- point\[{x_{ij}}\]in the first table corresponds to the\[{r_{ij}}\]rank in the second table.

i

Amount of Nitrogen Expired

1

4.079

4.859

3.54

5.047

3.298

4.679

2.87

4.648

3.847

2

4.368

5.668

3.752

5.848

3.802

4.844

3.578

5.393

4.374

3

4.169

5.709

4.416

5.666

4.123

5.059

4.403

4.496

4.688

4

4.928

5.608

4.94

5.291

4.674

5.038

4.905

5.208

4.806

i

Ranks

\[{r_i}\]

1

8

22

3

27

2

18

1

16

7

104

2

11

34

5

36

6

21

4

31

12

160

3

10

35

14

33

9

28

13

15

19

179

4

24

32

25

30

17

26

23

29

23

226

03

testing static value:

The test statistic value is,

\[\begin{array}{l}k = \frac{{12}}{{N\left( {N + 1} \right)}}\sum\limits_{i = 1}^I {{J_i}} {\left( {{{\bar R}_i} - \frac{{N + 1}}{2}} \right)^2}\\ = \frac{{12}}{{N\left( {N + 1} \right)}}\sum\limits_{i = 1}^I {\frac{{R_i^2}}{{{J_i}}}} - 3\left( {N + 1} \right)\\ = \frac{{12}}{{36.\left( {36 + 1} \right)}}\left( {\frac{{{{104}^2} + {{160}^2} + {{176}^2} + {{226}^2}}}{9}} \right) - 3 \cdot 37\\ = 7.587\end{array}\]

Degrees of freedom are

\[{d_f} = I - 1 = 4 - 1 = 3\]

Critical value at significant level 0.5 is


\[\begin{array}{l}X_{0,0.5,3}^2 = 7.815\\X_{0,0.5,3}^2 = 7.815 > 7.587 = K\end{array}\]

Do not reject null hypothesis

Hence, do not reject null hypothesis.

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