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b. The probability is 90% that the sample percentage will be oontained within what symmetrical limits of the population perce
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Answer #1

a)

                       population proportion ,p=   0.7                      
                       n=   100                      
                                                  
                       std error , SE = √( p(1-p)/n ) =    0.0458                      
                                                  
                       we need to compute probability for                           
                       0.62   < p̂ <   0.79                  
                                                  
                       Z1 =( p̂1 - p )/SE= (   0.62   -   0.7   ) /    0.0458   =   -1.746
                       Z2 =( p̂2 - p )/SE= (   0.79   -   0.7   ) /    0.0458   =   1.964
P(   0.62   < p̂ <   0.79   ) =    P[( p̂1-p )/SE< Z <(p̂2-p)/SE ]    =P(    -1.746   < Z <   1.964   )          
                                                  
= P ( Z <   1.964   ) - P (    -1.746   ) =    0.9752   -   0.040   =   0.9348  

b)

population proportion ,p=   0.7                      
n=   100                      
                          
std error , SE = √( p(1-p)/n ) =    0.0458                      
we need to compute probability for                           
P(p̂1<p̂<p̂2)=   0.9

                   
proportion left    0.10   is equally distributed both left and right side of normal curve                   
P(p̂ < p̂ 1) =   0.05   and P(p̂ > p̂2)=   0.95              
Z value at    0.05   =   -1.645 (excel formula =NORMSINV(   0.10   / 2 ) )  
Z value at    0.95   =   1.645 (excel formula =NORMSINV(   0.10   / 2 ) )  
                          
Z =( p̂ - p )/SE   
so, p̂ = Z*SE+p                          
                          
p̂ 1 = Z*SE+p =   -1.645   *   0.0458   +   0.7   =   0.6246
p̂ 2 = Z*SE+p =   1.645   *   0.0458   +   0.7   =   0.7754

so, answer : above 62.46% and below 77.54%

c)

population proportion ,p=   0.7                      
n=   100                      
                          
std error , SE = √( p(1-p)/n ) =    0.0458                      
we need to compute probability for                           
P(p̂1<p̂<p̂2)=   0.91                      
proportion left    0.09   is equally distributed both left and right side of normal curve                   
P(p̂ < p̂ 1) =   0.045   and P(p̂ > p̂2)=   0.955              
Z value at    0.045   =   -1.695   (excel formula =NORMSINV(   0.09   / 2 ) )  
Z value at    0.955   =   1.695   (excel formula =NORMSINV(   0.09   / 2 ) )  
                          
Z =( p̂ - p )/SE=                          
so, p̂ = Z*SE+p                          
                          
p̂ 1 = Z*SE+p =   -1.695   *   0.0458   +   0.7   =   0.6223
p̂ 2 = Z*SE+p =   1.695   *   0.0458   +   0.7   =   0.7777

above 62.23% and below 77.77%

d)

option d)

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