Question

A recent headline announced / impeachment causes a dip in a President’s Approval Rating.
This conclusion was based on a University poll of 100 adults. In the poll, 41.1% of respondents approved of the president’s job performance.

(a) Based on this poll, what is the probability that more than 50% of the population approve of the president’s job performance?
(b) Form a 90% confidence interval for this estimate.

(c) Assume we know without sampling error the population support was 42% before the impeachment story broke in September. Test whether the impeachment proceedings have actually caused a dip in approval from the 42% baseline at the .05 significance level.

Answer 1

a),p=   0.411                      
n=   100                      
                          
std error , SE = √( p(1-p)/n ) =    0.0492                      
                          
Z=( p̂ - p )/SE= (   0.5   -   0.411   ) /    0.0492   =   1.809
P ( p̂ >    0.5   ) =P(Z > ( p̂ - p )/SE) =                  
                          
=P(Z >   1.809   ) =    0.0352              

b)

Level of Significance,   α =    0.10          
Sample Size,   n =    100          
                  
Sample Proportion ,    p̂ = 0.4110          
z -value =   Zα/2 =    1.645   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0492          
margin of error , E = Z*SE =    1.645   *   0.0492   =   0.0809
                  
90%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.411   -   0.0809   =   0.3301
Interval Upper Limit = p̂ + E =   0.411   +   0.0809   =   0.4919
                  
90%   confidence interval is (   0.3301   < p <    0.4919   )

c)

Ho :   p =    0.42                  
H1 :   p <   0.42       (Left tail test)          
                          
Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   41.1                  
Sample Size,   n =    100                  
                          
Sample Proportion ,    p̂ = x/n =    0.4110                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0494                  
Z Test Statistic = ( p̂-p)/SE = (   0.4110   -   0.42   ) /   0.0494   =   -0.1823
                          
  
p-Value   =   0.4277   [excel function =NORMSDIST(z)]              
Decision:   p value>α ,do not reject null hypothesis                       
There is not enough evidence that  the impeachment proceedings have actually caused a dip in approval from the 42% baseline at the .05 significance leve