A pioneer study andomly assigned pregnant women to either a daily multivitamin s

Question

A pioneer study andomly assigned pregnant women to either a daily multivitamin supplement including folic acid or a placebo. O the 2107 pregnancies witn the vitamin regimen, there were cases of congenital malformation. Among the 2046 pregnancies with the placebo, 46 cases of congenital malformation were recorded. Note: Round your answer only after completing the entire calculations. Do not use rounded values in your calculation (a) The difference of the two sample proportions of congenital malformation cases (placebo minus the multivitamin) is 0.0101 (Round to 4 decimal places) (b) The difference computed in part (a) was small. However, this inequality seems more substantial when viewed by their ratio, which is a statistic commonly reported in medical journals. Find and interpret the relative risk of congenital malformation between the two groups. (Select all that apply. Hint: there are two correct answers) 1.82; those who took multivitamin supplement were 1.82 times as likely to have congenital malformation cases as those who took placebo. 0.55; those who took placebo were 0.55 times as likely to have congenital malformation cases as those who took multivitamin supplement. 1.82; those who took placebo were 1.82 times as likely to have congenital malformation cases as those who took multivitamin supplement. 0.55; those who took multivitamin supplement were 0.55 times as likely to have congenital malformation cases as those who took placebo.

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 relation b/w extent of sleep disorder breathing and death

alternative, H1: exists a relation b/w extent of sleep disorder breathing and death

level of significance, = 0.05

from standard normal table, chi square value at right tailed, ^2 /2 =7.815

since our test is right tailed,reject Ho when ^2 o > 7.815

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

from the table , ^2 o = 61.84

critical value

the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 4 - 1 ) = 1 * 3 = 3 is 7.815

we got | ^2| =61.84 & | ^2 | =7.815

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 extent of sleep disorder breathing and death

alternative, H1: exists a relation b/w extent of sleep disorder breathing and death

expected cell count that from the severe sleep disordered breathing and death cell is 57.57

test statistic: 61.84

critical value: 7.815

p-value:0

decision: reject Ho

MATRIX col1 col2 col3 col4 TOTALS row 1 476 320 165 90 1051 row 2 2953 1477 562 255 5247 TOTALS 3429 1797 727 345 N = 6298

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