Question

A con artist created a coin to swindle people from their money. If the people playing against the con artist gets heads, they win. A police officer is guessing whether the person is a con artist or not. After playing 100 times (assume this is large enough), the policeman wins 40 times.

a) Construct a hypothesis test of your choice and argue whether the coin toss game is fair for the policeman (fair meaning p=0.5). Choose a significance level of 0.05, make it a one-sided test, and calculate the p-value for your sample estimate. State clearly the conclusion of your test based on the context of the problem.

b) What is the probability that your test gives you the correct conclusion when the true probability of getting heads/win is 0.47?

c) How would you conduct the test if the number of times the policeman played was only 4? Explain without running your test again but sketch how would you calculate the probabilities. Choose a hypothetical estimate if you need to.

Answer 1

a)

Ho :   p =    0.5                  
H1 :   p <   0.5       (Left tail test)          
                          
Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   40                  
Sample Size,   n =    100                  
                          
Sample Proportion ,    p̂ = x/n =    0.4000                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0500                  
Z Test Statistic = ( p̂-p)/SE = (   0.4000   -   0.5   ) /   0.0500   =   -2.0000
                          

p-Value   =   0.022750132   [excel function =NORMSDIST(z)]              
Decision:   p-value<α , reject null hypothesis

                      
There is enough evidence that proportion is less than 0.5 and not fair game

b)

true proportion,   p=   0.47                      
                              
hypothesis proportion,   po=    0.400                      
significance level,   α =    0.05                      
sample size,   n =   100                      
                              
std error of sampling distribution,   σpo = √(po*(1-po)/n) = √ (   0.400   *   0.600   /   100   ) =   0.0490
std error of true proportion,   σp = √(p(1-p)/n) = √ (   0.47   *   0.53   /   100   ) =   0.0499
Zα/2   = ±   1.960   (two tailed test)                  
We will fail to reject the null (commit a Type II error) if we get a Z statistic between                       -1.960   and   1.960
these Z-critical value corresponds to some X critical values ( X critical), such that                              
-1.960   ≤(p^ - po)/σpo≤   1.960                      
-1.960   *σpo + po≤ p^ ≤   1.960   *σpo + po                  
0.3040   ≤ p^ ≤   0.4960                      
                              
now, type II error is ,ß =        P(0.304< p^ < 0.496)       =P( (0.304-p) /σp < Z < (0.496-p)/σp )              
       =P( (0.304-0.47)/0.0499) < Z < (0.496-0.47)/0.0499 )                      
so, P(   -3.326   < Z <   0.521   ) = P ( Z ≤   0.521   ) - P ( Z ≤   -3.326   )
       =   0.699   -   0.000   =   0.6985  

c)

How would you conduct the test if the number of times the policeman played was only 4?

Here n would be very small and hence we would calculate probability by large sample.

thanks

revert back for doubt

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