p>0.10 n a Midwestern city, the association between racial dfferences in the inc
ID: 3198795 • Letter: P
Question
p>0.10 n a Midwestern city, the association between racial dfferences in the incidence of cardiac arrest and subsequent survival was studied in 6117 cases of non-traumatic, out of hospital cardiac arrest . During a 12-month period, less than 24 of 2910 survived an arrest to hospital discharge (24 of 2910), compared to 84 of 3207 Caucasians, Is there an and survival in cases of non-traumatic out-of-hospital cardiac arrest? association between race (African- Let a 0.05 Race African 6009 total 1 pts D Question 7 What is the test statistic used?Explanation / Answer
Solution
Q7
Test statistic: ?2 = ?(i = 1 to r, j = 1 to c){(oij - eij)2/eij}
Q8
Test statistic value = 28.328
Q9
Critical value = 3.841
Q10
p-value = 1.02E – 07
Q11
There is evidence to suggest that there is association between race and survival rate.
DONE
Back-up Theory and Details of calculations
Back-up Theory
Suppose we have a contingency table consisting of r rows and c columns, i.e., (r x c) cells, each row representing a particular level of one attribute and each column representing a particular level of another attribute. Let oij be the observed frequency in the cell at ith row-jth column and eij be the corresponding expected frequency.
H0: The two attributes are independent. vs H1: H0 is false.
Test Statistic
?2 = ?(i = 1 to r, j = 1 to c){(oij - eij)2/eij}. [This can also be simplified as ?2 = ?(i = 1 to r, j = 1 to c)(o2ij/eij) – n, where n = total frequency = ?(i = 1 to r, j = 1 to c)(oij) which is also equal to ?(i = 1 to r, j = 1 to c)(eij).]
Under H0, ?2 is distributed as Chi-square distribution with degrees of freedom = (r - 1)(c - 1).
Under H0, eij = (oi. . o.j)/n where oi. = ?(j = 1 to c)(oij) = sum of all observed frequencies in the ith row and o.j = ?(i = 1 to r)(oij) = sum of all observed frequencies in the jth column.
Decision Criterion
Reject H0 if ?2cal > ?2(r - 1)(c - 1), ?. i.e. reject H0 if ?2cal > ?2tab (i.e., calculated value of ?2 is greater than the upper ?% point of Chi-square distribution with degrees of freedom = (r - 1)(c - 1), which can be read off from standard table).
Details of Excel Calculations
?2cal =
28.32729
p-value
?2crit =
3.841459
for DF =
OIJ (given)
Oi1
Oi2
Oi.
O1j
24
84
108
O2j
2886
3123
6009
O.j
2910
3207
6117
Eij
Ei1
Ei2
Total
E1j
51.37813
56.62187347
108
E2j
2858.622
3150.378127
6009
Total
2910
3207
6117
CHECK: respective row totals, column totals and grand total of
both Oij and Eij must be equal
(Oij - Eij)^2/Eij
1
2
1
14.58912
13.23802563
27.82715
2
0.262211
0.23792757
0.500138
14.85133
13.4759532
28.32729
DONE
?2cal =
28.32729
p-value
?2crit =
3.841459
for DF =
OIJ (given)
Oi1
Oi2
Oi.
O1j
24
84
108
O2j
2886
3123
6009
O.j
2910
3207
6117
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