Question

A study investigated survival rates for in-hospital patients who suffered cardiac arrest. Among 58,583 patients who has cardiac arrest during the day, 11,604 survived. Among 28,155 patients who suffered cardiac arrest during the night, 5350 survived. A physician claims that the survival rate of those who suffer cardiac arrest during the day is higher than the survival rate of those who suffer cardiac arrest during the night.

a) Perform a complete hypothesis test to test their claim at the 5% level of significance. Use the P-value method.

b) Construct a 90% confidence interval on the difference of proportions. Explain how your CI comes to the same conclusion as part a).

Answer 1

P1 :- Proportion of patients who has cardiac arrest during the day

P2 :- Proportion of patients who has cardiac arrest during the night

To Test :-

H0 :- P1 <= P2

H1 :- P1 > P2

p̂1 = 11604 / 58583 = 0.1981
p̂2 = 5350 / 28155 = 0.19

Test Statistic :-
Z = ( p̂1 - p̂2 ) / √( p̂ * q̂ * (1/n1 + 1/n2) ))
p̂ is the pooled estimate of the proportion P
p̂ = ( x1 + x2) / ( n1 + n2)
p̂ = ( 11604 + 5350 ) / ( 58583 + 28155 )
p̂ = 0.1955
q̂ = 1 - p̂ = 0.8045
Z = ( 0.1981 - 0.19) / √( 0.1955 * 0.8045 * (1/58583 + 1/28155) )
Z = 2.8022

Test Criteria :-
Reject null hypothesis if Z > Z(α)
Z(α) = Z(0.05) = 1.6449
Z > Z(α) = 2.8022 > 1.6449, hence we reject the null hypothesis
Conclusion :- We Reject H0

Decision based on P value
P value = P ( Z > 2.8022 )
P value = 0.0025
Reject null hypothesis if P value < α = 0.05
Since P value = 0.0025 < 0.05, hence we reject the null hypothesis
Conclusion :- We Reject H0

There is sufficient evidence to support the claim that  the survival rate of those who suffer cardiac arrest during the day is higher than the survival rate of those who suffer cardiac arrest during the night.

Part b)

p̂1 = 0.1981
p̂2 = 0.19
q̂1 = 1 - p̂1 = 0.8019
q̂2 = 1 - p̂2 = 0.81
n1 = 58583
n2 = 28155
(p̂1 - p̂2) ± Z(α/2) * √( ((p̂1 * q̂1)/ n1) + ((p̂2 * q̂2)/ n2) )
Z(α/2) = Z(0.1 /2) = 1.645
Lower Limit = ( 0.198078 - 0.19002 )- Z(0.1/2) * √(((0.198078 * 0.801922 )/ 58583 ) + ((0.19002 * 0.80998 )/ 28155 ) = 0.0034
upper Limit = ( 0.198078 - 0.19002 )+ Z(0.1/2) * √(((0.198078 * 0.801922 )/ 58583 ) + ((0.19002 * 0.80998 )/ 28155 )) = 0.0128

90% Confidence interval is ( 0.0034 , 0.0128 )
( 0.0034 < ( P1 - P2 ) < 0.0128 )

Since 0 does not lie in the interval, hence we conclude to reject null hypothesis.