Question

Q1. After surveying 625 potential voters polled, you find out that 325 support candidate X and 300 support candidate Y.

a. What is the 95% confidence interval for the support rate for X?

b. Based on your survey, what is the chance that the candidate Y will WIN in a general election ?

c. If you want to shrink the standard deviation of your survey to 1.5%, how many people do you need to survey?

Please show work for all

Answer 1

a)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   325          
Sample Size,   n =    625          
                  
Sample Proportion ,    p̂ = x/n =    0.5200          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.019984          
margin of error , E = Z*SE =    1.960   *   0.01998   =   0.0392
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.52000   -   0.03917   =   0.48083
Interval Upper Limit = p̂ + E =   0.52000   +   0.03917   =   0.55917
                  
95%   confidence interval is (   0.4808   < p <    0.5592   )

b)

Standard Error ,    SE = √( p(1-p)/n ) =    0.020                  
Z Test Statistic = ( p̂-p)/SE = (   0.4800   -   0.5   ) /   0.0200   =   -1.0000
                          
      
                          
p-Value   =   0.841   [Excel function =NORMSDIST(-z)