Question

A certain politician is set to vote on a bill. While the politician is inclined to vote against the bill, if more than 55% of their constituents are in support of the bill the politician will vote for the bill. The politician decides to survey 85 randomly selected constituents to determine if more than 55% of their constituents are in support of the bill.

A.Identify the null and alternative hypothesis.

B.Identify what a Type I and Type II error would be in this problem situation.

C.Compute the power of the test if 60% of the politician’s constituents support the bill and a significance level of 0.05 is used.

D.The politician is especially concerned about voting against the bill if more than 60% of their constituents support the bill. How could the politician reduce the likelihood of this occurring?

Answer 1

a)

Ho :   p =    0.55
H1 :   p >   0.55

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b)

type I error if we conclude that more than 55% of their constituents are in support of the bill.but in actual it is not true

type II errpr if we conclude that not more than 55% of their constituents are in support of the bill but in actual it is more than 55%

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c)

true proportion,   p=   0.6                      
                              
hypothesis proportion,   po=    0.550                      
significance level,   α =    0.05                      
sample size,   n =   85                      
                              
std error of sampling distribution,   σpo = √(po*(1-po)/n) = √ (   0.550   *   0.450   /   85   ) =   0.0540
std error of true proportion,   σp = √(p(1-p)/n) = √ (   0.6   *   0.4   /   85   ) =   0.0531

Zα =       1.645   (right tailed test)      
                  
We will fail to reject the null (commit a Type II error) if we get a Z statistic <                   1.645
this Z-critical value corresponds to X critical value( X critical), such that                  
                  
(p^ - po)/σpo ≤ Zα                  
p^ ≤ Zα*σpo + po                  
p^ ≤    1.645*0.054+0.55       =   0.6388  
                  
now, type II error is ,ß =    P( p^ ≤    0.6388   given that p =   0.6  
                  
   = P ( Z < (p^ - p)/σp )=       P(Z < (0.6388-0.6) / 0.0531)      
   = P ( Z < (   0.729   )      
   =   0.76712          
                  
                  
                  
power =    1 - ß =   0.2329          

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d)

increase sample size

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THANKS

revert back for doubt

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