D.
where in the table the closest (0. Where in the table the closest
(0. 2.
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10)If n1 n2 is odd.
. *
Gordon, Richard F. . Following this procedure Reimer et
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0. Decision Do not reject H 0 We conclude that the two
population dispersion parameters may be equal. 122. 975 is 11 . 113.
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36 3. Can we conclude on the basis of
these data that dispersion with regard to relative necrosis differs
in the two population represented? Let = 0. The third data set has size 25 and
is drawn from a normal distribution with standard deviation 1. ‘less’: the ratio of scales is less than 1. On combining the two samples and ranking, we
have the results shown in table 2. Rest.
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Helwig “Nonparametric Dispersion and Equality
Tests” at http://users. edu/~helwig/notes/npde-Notes. 8. 8yyyyyyyxxyy1234567891010Observation Group Rank39. 84 3.
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50 2. TABLE A. Linhart, Spectroanalytic Evaluation of Aortic
Prosthetic Valve, Chest, 66 (1974), 44-49. 66. where in the table the
closest (0. Now apply the test to x1 and x3:The probability of observing such an extreme value of the statistic
under the null hypothesis of equal scales is only 0.
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061. If n1 + n2 is an odd number, the array of rankswill be 1,2,3,. 139. 481. T x2T x125. B.
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On combining the two samples and ranking, we
have the results shown in table 2. E. . Data on the cardiac index were obtained after
operation. (One-sided)H0 :1 2H1 :1 2CASE C. 00
2.
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ReferencesAnsari, A. 025According to the table A. Group 2 consisted of patients
withabnormal prosthetic valve function. 0238)The value of T for 1-/2 = 0.
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If you can’t read please download the documentPost on 21-Sep-2014201 views10 downloadEmbed Size (px)
344 x 292429 x 357514 x 422599 x 487TRANSCRIPTMAKING INFERENCES ABOUT THE EQUALITY OF TWO DISPERSION
PARAMATERS ANSARI- BRADLEY TESTLEARNING OUTCOMES At the end this subtopic student should be
able to : Make inference about the equality of two dispersion
parameter by using Ansari- Bradley test. We know that T = 1 + 2 + 5 + 2 = 10 n1 = 4,
n2 = 5 and /2 = 0. 8, the value of T for
/2n1 = 10, n2 = 10 and /2 = 0. Table 2ObservationGroup Rank328x 1336y 2347x 3372y 4425y 5428y 5433x 4434y 3478 607x 2 x 13. A.
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Make hypotheses H0 :1 = 2H1 :1 2(claim)2. 6. We conclude that the two population dispersion
parameters may be equal. Group 1 consisted of patients with normalprosthetic valve function.
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97 2. One group of dogs was
untreated, and a second group received propranolol 10minutes before the occlusion. C. Nathaniel E. 60 2. 34.
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C. . Sprent, Peter and N. 5.
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The Ansari-Bradley test statistic. Perform the Ansari-Bradley test for equal scale parameters. The first
two, with sizes 35 and 25, are drawn from a normal distribution with
mean 0 and standard deviation 2. 10
is not between 16 . We compute T * for significance with
appropriatevalues of the standard normal distribution.
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The data are
measured on at least an ordinal scale. (CASE C)HYPOTHESIS H0 :1 = 2 H1 :1 2REJECT H0T x1 / 2 or T x/2,n1 ,n2H0 :1 2, H1 :1 2T x1 ,n1 ,n2H0 :1 2 H1 :1 2T x , n1 , n2Example. HYPOTHESES CASE A. 544.
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025= 0. 97xxyyxyyxy1234543213. 8Table 1Austistic children (x) Controls (y) 433 428 347 372 328
434 607 425 478 3361. 139. . Not enough evidence to support
the claim.
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Part of their results are shown intable below. 623. Test statistic. 125.
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T x2Tx1 258 69 58 455. This means
that the p-value when alternative=’greater’ should be near 0 and
hence we should be able to reject the null hypothesis:As we can see, the p-value is indeed quite low. click resources we concluded from these data thatthe two
population represented differ with respect to dispersion of uptake
values? Let = 0. .