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The given data on phosphorus concentration in topsoil for four different soil treatments appeared in the article 鈥淔ertilisers for Lotus and Clover Establishment on a Sequence of Acid Soils on the East Otago Uplands鈥 (N. Zeal. J. of Exptl. Ag., 1984: 119鈥129). Use a distributionfree procedure to test the null hypothesis of no difference in true mean phosphorus concentration (mg/g) for the four soil treatments.

I 8.1 5.9 7.0 8.0 9.0

II 11.5 10.9 12.1 10.3 11.9

III 15.3 17.4 16.4 15.8 16.0

IV 23.0 33.0 28.4 24.6 27.7

Short Answer

Expert verified

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

Groups

1

8.1

5.9

7

8

9

2

11.5

10.9

12.1

10.3

11.9

3

15.3

17.4

16.4

15.8

16

4

23

33

28.4

24.6

27.7

i

Ranks

\({r_i}\)

1

4

1

2

3

5

15

2

8

7

10

6

9

40

3

11

15

14

12

13

65

4

16

20

19

17

18

90

03

testing static value:

The test statistic value is,

\(\begin{array}{l}k = \frac{{12}}{{N\left( {N + 1} \right)}}\sum\limits_{i = 1}^I {{J_i}} {\left( {{{\bar R}_i} - \frac{{N + 1}}{2}} \right)^2}\\ = \frac{{12}}{{N\left( {N + 1} \right)}}\sum\limits_{i = 1}^I {\frac{{R_i^2}}{{{J_i}}}} - 3\left( {N + 1} \right)\\ = \frac{{12}}{{20.\left( {20 + 1} \right)}}\left( {\frac{{{{15}^2}}}{5} + \frac{{{{40}^2}}}{5} + \frac{{{{65}^2}}}{5} + {{\frac{{90}}{5}}^2}} \right) - 3 \cdot 21\\ = 17.87\end{array}\)

Degrees of freedom are

\({d_f} = I - 1 = 4 - 1 = 3\)

Critical value at significant level 0.05 is


\(\begin{array}{l}X_{0.05,3}^2 = 7.815\\X_{0.05,3}^2 = 7.815 < 17.87 = K\end{array}\)

Reject null hypothesis

Hence, reject null hypothesis.

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Most popular questions from this chapter

The accompanying data refers to concentration of the radioactive isotope strontium-90 in milk samples obtained from five randomly selected dairies in each of four different regions.

1 6.4 5.8 6.5 7.7 6.1

2 7.1 9.9 11.2 10.5 8.8

3 5.7 5.9 8.2 6.6 5.1

4 9.5 12.1 10.3 12.4 11.7

Test at level .10 to see whether true average strontium-90 concentration differs for at least two of the regions.

Suppose that observations X1, X2,鈥, Xn are made on a process at times 1, 2,鈥, n. On the basis of this data, we wish to test H0: the Xi鈥檚 constitute an independent and identically distributed sequence versus Ha: Xi11 tends to be larger than Xi for i =1,鈥, n (an increasing trend) Suppose the Xi鈥檚 are ranked from 1 to n. Then when Ha is true, larger ranks tend to occur later in the sequence, whereas if H0 is true, large and small ranks tend to be mixed together. Let Ri be the rank of Xi and consider the test statistic D = on i= 1(Ri = i)2.

Then small values of D give support to Ha (e.g., the smallest value is 0 for R1 = 10, R2 = 2,鈥, Rn= n). When H0 is true, any sequence of ranks has probability 1yn!. Use this to determine the P-value in the case n = 4, d= 2. (Hint: List the 4! rank sequences, compute d for each one, and then obtain the null distribution of D. See the Lehmann book (in the chapter bibliography), p. 290, for more information.).

The article 鈥淓ffects of a Rice-Rich Versus Potato-Rich Diet on Glucose, Lipoprotein, and Cholesterol Metabolism in Noninsulin-Dependent Diabetics鈥 (Amer. J. of Clinical Nutr., 1984: 598鈥606) gives the accompanying data on cholesterol synthesis rate for eight diabetic subjects. Subjects were fed a standardized diet with potato or rice as the major carbohydrate source. Participants received both diets for specified periods of time, with cholesterol-synthesis rate (mmol/day) measured at the end of each dietary period. The analysis presented in this article used a distribution-free test. Use such a test with significance level .05 to determine whether the true mean cholesterol-synthesis rate differs significantly for the two sources of carbohydrates.

The sign test is a very simple procedure for testing hypotheses about a population median assuming only that the underlying distribution is continuous. To illustrate, consider the following sample of 20 observations on component lifetime (hr):

1.7 3.3 5.1 6.9 12.6 14.4 16.4

24.6 26.0 26.5 32.1 37.4 40.1 40.5

41.5 72.4 80.1 86.4 87.5 100.2

We wish to test \({H_0}:\tilde \mu = 25.0\) versus \({H_0}:\tilde \mu > 25.0\)The test statistic is Y 5 the number of observations that exceed 25

a. Determine the P-value of the test when Y 5 15. (Hint: Think of a 鈥渟uccess鈥 as a lifetime that exceeds 25.0. Then Y is the number of successes in the sample. What kind of a distribution does Y have when\(\tilde \mu = 25.0\)?)

b. For the given data, should H0 be rejected at significance level .05? (Note: The test statistic is the number of differences \({X_i} - 25\)that have positive signs, hence the name sign test.)

Suppose we wish to test.

: the X and Y distributions are identical

versus

: the X distribution is less spread out than the Y

distribution

The accompanying figure pictures X and Y distributions for which is true. The Wilcoxon rank-sum test is not appropriate in this situation because when is true as pictured, the Y鈥檚 will tend to be at the extreme ends of the combined sample (resulting in small and large Y ranks), so the sum of X ranks will result in a W value that is neither large nor small.

Consider modifying the procedure for assigning ranks as follows: After the combined sample of m + n observations is ordered, the smallest observation is given rank 1, the largest observation is given rank 2, the second smallest is

given rank 3, the second largest is given rank 4, and so on. Then if is true as pictured, the X values will tend to be in the middle of the sample and thus receive large ranks. Let W鈥 denote the sum of the X ranks and consider an uppertailed test based on this test statistic. When is true, every possible set of X ranks has the same probability, so W鈥 has the same distribution as does W when H0 is true. The accompanying data refers to medial muscle thickness for arterioles from the lungs of children who died from sudden infant death syndrome (x鈥檚) and a control group of children (y鈥檚). Carry out the test of versus at level .05.

SIDS 4.0 4.4 4.8 4.9

Control 3.7 4.1 4.3 5.1 5.6

Consult the Lehmann book (in the chapter bibliography) for more information on this test, called the Siegel-Tukey test.

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