Question

58593 people had cardiac arrest in daytime. 11604 survived. 28155 had cardiac arrest at night. 4139 survived. what does this tell is about cardiac arrest hypothesis test

Answer 1

let p1 and p2 be proportion of people who survived after cardiac arrest in day and night time resp.

Ho:   p1 - p2 =   0
Ha:   p1 - p2 >   0

first sample size,     n1=   58593
number of successes, sample 1 =     x1=   11604
proportion success of sample 1 , p̂1=   x1/n1=   0.198044135
      
second sample size,     n2 =    28155
number of successes, sample 2 =     x2 =    4139
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.147007636
      
difference in sample proportions,    p̂1 - p̂2 =     0.051036499
      
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.1815
      
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0028
      
Z-statistic =    (p̂1 - p̂2)/SE =     18.2609

p-value =        0.0000
decision :    p-value<α=0.05,Reject null hypothesis   

so, there is enough evidence to conlcude that true proportion of people who survived after cardiac arrest in day is greater than that of  night time at α=0.05