3. In randomized clinical trial 154 women with breast cancer were assigned to re

Question

3. In randomized clinical trial 154 women with breast cancer were assigned to receive chemotherapy. Another 174 women were assigned to receive chemotherapy combined with radiation therapy. Survival data after 15 years are presented below Treatment type Chemo only Chemo+Radiation 15 year Survival Status Died Survived 78 ( 76 ( 66 ( 108 ) 3 total 154 174 Conduct the Chi-square test of the hypothesis that there is no association between the type of treatment and 15 year survival status. Use =.05 . Use p-value method. (place expected counts under null hypothesis in parenthesis next to the observed counts)

Explanation / Answer

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 association b/w treatment and survival status

alternative, H1: exists a association b/w treatment and survival status

level of significance, = 0.05

from standard normal table, chi square value at right tailed, ^2 /2 =3.8415

since our test is right tailed,reject Ho when ^2 o > 3.8415

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

from the table , ^2 o = 5.3656

critical value

the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 2 - 1 ) = 1 * 1 = 1 is 3.8415

we got | ^2| =5.3656 & | ^2 | =3.8415

make decision

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

^2 p_value =0.0205

ANSWERS

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null, Ho: no association b/w treatment and survival status

alternative, H1: exists a association b/w treatment and survival status

test statistic: 5.3656

critical value: 3.8415

p-value:0.0205

decision: reject Ho

exists a association b/w treatment and survival status

MATRIX col1 col2 TOTALS row 1 78 66 144 row 2 76 108 184 TOTALS 154 174 N = 328

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