Question

4. Julie visits the local park 90 times, and sees her favorite pigeon 38 times. Let p be the probability that Julie sees the pigeon on any given visit, and assume that whether Julie sees the pigeon on some visit is independent of whether she sees the pigeon on any other visit.

(a) Construct a 90% two-sided confidence interval for p.

(b) Julie believes that there is a 50% chance that she sees the pigeon when she visits the park. Is your interval in part (a) consistent with that belief? Explain your answer.

(c) Test Julie’s claim from part (b) with a hypothesis test. Use a significance level of α = 0.05. Perform the test by comparing a test statistic to a critical value found in the tables in the back of your textbook.

(d) Perform the test in part (c) using a p-value.

Answer 1

a)

Level of Significance,   α =    0.10          
Number of Items of Interest,   x =   38          
Sample Size,   n =    90          
                  
Sample Proportion ,    p̂ = x/n =    0.422          
z -value =   Zα/2 =    1.645   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0521          
margin of error , E = Z*SE =    1.645   *   0.0521   =   0.0856
                  
90%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.422   -   0.0856   =   0.3366
Interval Upper Limit = p̂ + E =   0.422   +   0.0856   =   0.5079
                  
90%   confidence interval is (   33.7%   < p <    50.8%   )

b)

Yes, as we can see our interval contain 50%.

c)

Ho :   p =    0.5                  
H1 :   p ╪   0.5       (Two tail test)          
                          
Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   38                  
Sample Size,   n =    90                  
                          
Sample Proportion ,    p̂ = x/n =    0.4222                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0527                  
Z Test Statistic = ( p̂-p)/SE = (   0.4222   -   0.5   ) /   0.0527   =   -1.4757
                          
critical z value =    ±    1.960   [excel formula =NORMSINV(α/2)]      

As Z stats < Critical Z   ,do not reject null hypothesis

d)

p-Value   =   0.140016503   [excel formula =2*NORMSDIST(z)]              
Decision:   p value>α ,do not reject null hypothesis                       

Please revert back in case of any doubt.

Please upvote. Thanks in advance.