Question

hi can you please help me to find the claim

1) tho said that they attend a religious service at least st the given hypothesis. Assume that the samples are indepe me that the samples are independent and that they have been randomly selected 1) A researcher finds that of 1,000 people who said that they attend a religious se once a week, 31 stopped to help a person with car trouble. Of 1,200 people inter who had not attended a religious service at least once a month, 22 stopped to with car trouble. At the 0.05 significance level, test the claim that the two proporco equal. 1) claim. once a month, 22 stopped to help a person - 1000 Do: 1200 Xr= 31 X 22 2) Ho :P, = P₂ 3) Hi :P. 7P2 2 tails 5) Z= 1.929 [Start, test, 6/2-prop 2 test, Enter] 1.93 6) Meject Ho -CASE Xi 2) 7 of 8,500 people vaccinated against a certain disease later developed the disease. 18 of 10,000 people vaccinated with a placebo later developed the disease. Test the claim that the vaccine is effective in lowering the incidence of the disease. Use a significance level of 0.02. Claim:P, LP₂ -> CASe 3.org 7-16 2) Ho: P;= P2 3) Hi P, LP2 lett Tail DF:6 0.07 3.173 9) Z=-5.96 6) Deject HO case -> 3 7) Sutricient evident to reject the claim of PILP2

Answer 1

1) claim: p1 = p2
sample #1   ----->              
first sample size,     n1=   1000          
number of successes, sample 1 =     x1=   31          
proportion success of sample 1 , p̂1=   x1/n1=   0.0310          
                  
sample #2   ----->              
second sample size,     n2 =    1200          
number of successes, sample 2 =     x2 =    22          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.0183          
                  
difference in sample proportions, p̂1 - p̂2 =     0.0310   -   0.0183   =   0.0127
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.0241          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0066          
Z-statistic = (p̂1 - p̂2)/SE = (   0.013   /   0.0066   ) =   1.9293
                  
p-value =        0.0537   [excel formula =2*NORMSDIST(z)]      
decision :    p-value>α,Don't reject null hypothesis               
conclusion: there is not enough evidence to reject the claim

2)

claim: p1 - p2 <   0

Ho:   p1 - p2 =   0          
Ha:   p1 - p2 <   0          
                  
sample #1   ----->              
first sample size,     n1=   8500          
number of successes, sample 1 =     x1=   7          
proportion success of sample 1 , p̂1=   x1/n1=   0.0008          
                  
sample #2   ----->              
second sample size,     n2 =    10000          
number of successes, sample 2 =     x2 =    18          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.0018          
                  
difference in sample proportions, p̂1 - p̂2 =     0.0008   -   0.0018   =   -0.0010
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.0014          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0005          
Z-statistic = (p̂1 - p̂2)/SE = (   -0.001   /   0.0005   ) =   -1.8017
                  
p-value =        0.0358   [Excel function =NORMSDIST(z)      
decision :    p-value>α,Don't reject null hypothesis               
conclusion: there is not enough evidence to support the claim.