Question

In: Statistics and Probability

3. The following table presents data on the presence and absence of respiratory distress syndrome (RDS)...


3. The following table presents data on the presence and absence of respiratory distress syndrome (RDS) in two groups of infants. Group 1 consisted of 42 infants whose fetal membrances ruptured 24 hours or less before delivery, while group 2 was composed of 22 infants whose membrances ruptured more than 24 hours before delivery. Test the null hypothesis that the two populations are homogeneous. Let α=.05. Incidence of respiratory distress syndrome (RDS) in two groups of infants RDS Group Yes No Total 1 27 15 42 2 7 15 22 Total 34 30 64

Solutions

Expert Solution

Null Hypothesis:Ho:  two populations are homogeneous

Alternate Hypothesis:Ho:  two populations are not homogeneous

degree of freedom(df) =(rows-1)*(columns-1)= 1
for 1 df and 0.05 level , critical value       χ2= 3.841
Decision rule : reject Ho if value of test statistic X2>3.841
Applying chi square test of Homogeneity:
Expected Ei=row total*column total/grand total yes No Total
Group 1 22.31 19.69 42
Group 2 11.69 10.31 22
total 34 30 64
chi square    χ2 =(Oi-Ei)2/Ei yes No Total
Group 1 0.9848 1.1161 2.1008
Group 2 1.8800 2.1307 4.0107
total 2.8648 3.2468 6.1115
test statistic X2 = 6.1115
since test statistic falls in rejection region we reject null hypothesis
we have sufficient evidence to conclude that two populations are not homogeneous

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