Question

Of 540 Packers fans surveyed, 215 fans were satisfied with the team's draft picks since 2010. Let p represent the proportion of fans that are satisfied with the Packers draft picks.

Create a 99% confidence interval for satisfied Packer fans and give its lower bound as your answer.

Nationwide, football fans since 2010 on average have been satisfied with their team's draft performance 45% of the time. Matt LaFleur wants to know if Packers fans really are less satisfied than the league average. Using the 540 Packers fans sampled in the previous question, LaFleur tests the hypothesis H₀ : p = 0.45,   H_a : p < 0.45.

Answer 1

Level of Significance,   α =    0.01          
Number of Items of Interest,   x =   215          
Sample Size,   n =    540          
                  
Sample Proportion ,    p̂ = x/n =    0.398          
z -value =   Zα/2 =    2.576   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0211          
margin of error , E = Z*SE =    2.576   *   0.0211   =   0.0543
                  
99%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.398   -   0.0543   =   0.3439

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Ho :   p =    0.45                  
H1 :   p <   0.45       (Left tail test)          
                          
Level of Significance,   α =    0.01                  
Number of Items of Interest,   x =   215                  
Sample Size,   n =    540                  
                          
Sample Proportion ,    p̂ = x/n =    0.3981                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0214                  
Z Test Statistic = ( p̂-p)/SE = (   0.3981   -   0.45   ) /   0.0214   =   -2.4220
                                
                          
p-Value   =   0.0077 [excel function =NORMSDIST(z)]              
Decision:   p-value<α , reject null hypothesis                       
There is enough evidence to say that Packers fans really are less satisfied than the league average

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THANKS

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