Question

In: Statistics and Probability

Pollsters at Quinnipiac University surveyed 751 likely New York voters and found out that 368 of...

Pollsters at Quinnipiac University surveyed 751 likely New York voters and found out that 368 of 751 are going to vote for Candidate A rather than Candidate B.

  1. At the 95 percent confidence level, can a winner be predicted?
  2. What is the margin of error?
  3. What is the probability of error?
  4. What sample size is necessary in order to change the margin of error to 2 percent while leaving the probability of error the same?

Solutions

Expert Solution

a)

Ho :   p =    0.5                  
H1 :   p ╪   0.5       (Two tail test)          
                          
Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   368                  
Sample Size,   n =    751                  
                          
Sample Proportion ,    p̂ = x/n =    0.4900                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0182                  
Z Test Statistic = ( p̂-p)/SE = (   0.4900   -   0.5   ) /   0.0182   =   -0.5474
                                
                          
p-Value   =   0.5841   [excel formula =2*NORMSDIST(z)]              
Decision:   p value>α ,do not reject null hypothesis                       
There is not enough evidence to pridicc a winner

.................

b)

z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0182          
margin of error , E = Z*SE =    1.960   *   0.0182   =   0.0358

..

c)

probability of type 1 error = 0.05

...................

d)

sample proportion ,   p̂ =    0.490013316                          
sampling error ,    E =   0.05                          
Confidence Level ,   CL=   0.95                          
                                  
alpha =   1-CL =   0.05                          
Z value =    Zα/2 =    1.960   [excel formula =normsinv(α/2)]                      
                                  
Sample Size,n = (Z / E)² * p̂ * (1-p̂) = (   1.960   /   0.05   ) ² *   0.49   * ( 1 -   0.49   ) =    384.0
                                  
                                  
so,Sample Size required=       384                          

..................

Please revert back in case of any doubt.

Please upvote. Thanks in advance.

  


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