Given: x1= 80 , n1=150,
x2= 30, n2=100
Step1: state the hypothesis
H0: "p1=p2"
H1: "p1\\neq p2"
Step2: Name of the test:
2 sample proportion test
Step 3: test statistic:
"\\widehat{p1}=\\frac{x_{1}}{n_{1}}=\\frac{80}{150}=0.5333,\n\n\n \\widehat{p2}=\\frac{x_{2}}{n_{2}}=\\frac{30}{100}=0.3,\n\n\\widehat{p}=\\frac{x_{1}+x_{2}}{n_{1+}n_{2}}=\\frac{80+30}{150+100}=0.44"
Test statistic is calculated as follows:
"z=\\frac{\\widehat{p1}-\\widehat{p2}}{\\sqrt{\\widehat{p}(1-\\widehat{p})(\\frac{1}{n_{1}}+\\frac{1}{n_{2}})}}"
"z=\\frac{0.5333-0.3}{\\sqrt{0.44(1-0.44)(\\frac{1}{150}+\\frac{1}{100})}}"
=3.641
Step 4: P value Using z table : We get a z-score of 3.641. Because this is a two-sided test, it is not enough to just look at the right tail. We also have to look at the equivalent of the left tail, or a -3.641. that is P(Z<-3.641)+P(Z>=3.641)=0.000136+0.000136=0.000272=0.0003
Step 5: conclusion: since P value is less than alpha , null hypothesis is rejected.
Hence,there is enough evidence to claim that there is difference between proportions of people suffer from the disease in two communities.
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