7.11 A large series of patients with central retinal vein occlusion (CRVO) are d

Question

7.11 A large series of patients with central retinal vein occlusion (CRVO) are di- vided into two primary diagnostic groups: 282 patients who have developed venous stasis retinopathy (VSR) and 78 patients who suffer from hemorrhagic retinopathy (HR). The severity of the disease in each group is: CRVO VSR HR Moderate Marked 51 48 Total 282 78 Mild 89 24 (a) Use a chi-square test to test the null hypothesis that diagnostic group classifi- cation and severity of disease are independent. Use = 0.05 (b) Using the dichotomous grouping of mild versus moderate or marked combined, determine if the proportion with mild severity differs between the two groups

Explanation / Answer

PART A.

Given table data is as below

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calculation formula for E table matrix

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expected frequecies calculated by applying E - table matrix formulae

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calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above

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set up null vs alternative as

null, Ho: no relation b/w X and Y OR X and Y are independent

alternative, H1: exists a relation b/w X and Y OR X and Y are dependent

level of significance, = 0.05

from standard normal table, chi square value at right tailed, ^2 /2 =5.991

since our test is right tailed,reject Ho when ^2 o > 5.991

we use test statistic ^2 o = (Oi-Ei)^2/Ei

from the table , ^2 o = 60.226

critical value

the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 3 - 1 ) = 1 * 2 = 2 is 5.991

we got | ^2| =60.226 & | ^2 | =5.991

make decision

hence value of | ^2 o | > | ^2 | and here we reject Ho

^2 p_value =0

ANSWERS

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null, Ho: no relation b/w X and Y OR X and Y are independent

alternative, H1: exists a relation b/w X and Y OR X and Y are dependent

test statistic: 60.226

critical value: 5.991

p-value:0

decision: reject Ho

PART B.

set up null vs alternative as
null hypothesis is that all proportions are equal
H0: p1 = p2 = p3.
alternative hypothesis is that atleast one proportions are unequal
H0: p1 = p2 = p3.
level of significance, = 0.05
from standard normal table, chi square value at right tailed, ^2 /2 =5.991
since our test is right tailed,reject Ho when ^2 o > 5.991
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 60.226
critical value
the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 3 - 1 ) = 1 * 2 = 2 is 5.991
we got | ^2| =60.226 & | ^2 | =5.991
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho

we have evidence that the proportion with mild severity differs between the two groups

col1 col2 col3 row 1 89 142 51 282 row 2 6 24 48 78 TOTALS 95 166 99 N = 360

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