Question #121051
A sample of 150 people from a certain industrial community showed that 80 people suffered from a lung disease. A sample of 100 people from a rural community showed that 30 suffered from the same lung disease. At α =0.05, is there a difference between the proportions of people who suffer from the disease in the two communities?
1
Expert's answer
2020-06-09T18:05:48-0400

Let the proportion of industrial community is  p1and the proportion of rural community is  p2,p^1=80150=0.533p^2=30100=0.3p^p=80+30150+100=0.44H0:p1=p2H1:p1p2Let  α=0.05    Zα2=Z0.025=1.96,Z=p^1p^2p^p(1p^p)(1n1+1n2)Z=0.5330.30.44(0.56)(1150+1100)=3.64the rejection region:  z>1.96,z<1.96The decision is to reject:  H0this means that there is a difference between  the proportions of people who suffer from the disease in the two communities \text{Let the proportion of industrial community is}\; p_{1}\\ \text{and the proportion of rural community is}\; p_{2},\\ \hat p_{1}=\frac{80}{150}=0.533\\ \hat p_{2}=\frac{30}{100}=0.3\\ \hat p_{p}=\frac{80+30}{150+100}=0.44\\ H_0: p_{1}= p_{2}\\ H_1:p_{1}\neq p_{2}\\ Let \; \alpha =0.05\implies Z_{\frac{α}{2}}=Z_{0.025}=1.96,\\ Z=\frac{\hat p_{1} -\hat p_{2}}{\sqrt{\hat p_{p}(1-\hat p_{p})(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}\\ Z=\frac{0.533 -0.3}{\sqrt{0.44(0.56)(\frac{1}{150}+\frac{1}{100})}}=3.64\\ \text{the rejection region} :\;z>1.96, z<-1.96\\ \text{The decision is to reject}:\; H_0\\ \text{this means that there is a difference between }\\ \text{ the proportions of people who suffer from }\\ \text{the disease in the two communities }


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