Anonymous asked in Education & ReferenceHomework Help · 2 months ago

I need help in Statistics? drug test success provides a "1-panel-thc" test for marijuana usage. Among 300 tested subjects, results from 27 subjects were wrong (either a false positive or a false negative) Using a 0.05 Significance Level to test the claim that 10% of results were wrong. 


2.The heights are measured for the simple random sample of supermodels Crawford,  Bundchen,Pestova, Christenson, Hume, Moss, Campbell, Schiffer, and Taylor. They have a mean height of 700 inches and a standard deviation of 1.5 Inches. Use a 0.01 Significance level to test the claim that the supermodels have heights with a mean that is greater than the mean height of 63.6 inches for the women in the general population.  


3. In a study of police gunsire reports during a recent year, it was found that among 540 shots fired by New York City Police, 182 hit their targets and among 283 shots fired by LA police , 77 hit their targets. We want to use a 0.05 Significance Level to test the claim that NYU police and LA police have the same proportion of hits. Test the claim by using a testing hypothesis. 


4. In a study of proctored and nonproctored tests in an online intermediate algebra course, researchers obtained data for tet results given below. Using a 0.01 Significance level to test the claim that students taking nonp exams get a higher mean than those taking proctored exams. Test the claim using a hypothesis test.  

Group 1(proctored) :n1=30, x̅1= 74.30, s1= 12.87  

Group 2 (nonp) :n2=32, x̅= 88.62, s2=22.09

1 Answer

  • Alan
    Lv 7
    2 months ago
    Favorite Answer

    1st several of these question are reverse of the 

    normal question given students. 

    That , they are normally want you to test if things 

    are different.  It is harder to prove that thi,ngs are the same. 

    However,  I used the normal method 

    to try to prove things are different, because if they are 

    different, then there definitely not the same.   

    I am not sure you can really show they are the same.  

    So possibly the correct answer, you cannot prove they are 

    the same.   So it is a possibility that your textbook, course, or 

    teacher is a troll asking you impossible quesitons.  

    This question 1. is a proportion question.  

    It is called a proportion when your probability is based 

    on what percentage or proportion of people do something. for a proportional the standard deviation = σ_p = sqrt[ P * ( 1 - P ) / n ]

    Unless  you actual know the size of the population  

    and then there is more complicated formula which 

    add a correction factor. 

    See this reference. 

    let's say the NULL hypothesis p = 0.10 

    the alternate hypothesis p <> 0.10 

    See this reference

    so test this hypothesis , 

    we assume p = 10%  

    and as a decimal = 0.10 

    so p = 0.10 (as a decimal) 

    1-p = 1-0.10 = 0.90 

    n = 300

    σ_p  = sqrt( 0.9*0.1/ 300)      

    σ_p =  0.017320508

    p_hat (test point) = 27/300 = 0.09  

    Z_test = (x-mean)/standard deviaton =

    Z_test =  (0.1-0.09)/  0.017320508

    Z_test = 0.577350269

    P(z<0.57) = 0.71566 

    P(z< 0.58) = 0.71904         

    I don't know how you show it is the same. 

    but you don't have enough evidence 

    to  Reject the NULL that they are the same 

    and cannot support that the claim they are different.  

    Critical_Z(for a 95 confid. int.)  = 1.96 

    since P(z< 1.96) = 0.975   

    and a 95 % confidence for a two-tail test 

    goes from 2.5 % to 97.5 % leaving 95 % in the middle.  

    since both 0.57 and 0.58 are less than 1.96  

    you cannot support that is different the 10%


    2. For this one, since there are only 9 data points. 

    You probably cannot use the Central Limit to say 

    it's normal. You will need to use a t-table.   

    population mean = 63.6   

    N = 9 , DF  = 8 

    T_test (for a sample avg)  =  (x_hat -mean)/(S/sqrt(N)) 

    where S = sample standard deviation 

    N = number of samples = 9 

    x_hat = 70    

    T_test = ( 70-63.5) /  (1.5/sqrt(9) )  =    13  

    so you want  x> average 

    you look for one tailed table value for 0.01 and

    DF = 8  

    T_Critical = 2.896

    read from a t-table 

    since T_Test > T_Critical

    13> 2.896    

    you can support the claim that they have a 

    higher average height than the standard 

    population and reject the NULL Hypothesis.


    Not sure of this answer, but I followed 

    Khan Academy examples disregarding my 1st instincts. 

    so 1st find p and  σ_p1  (here this value are for 

    just samples of the full data, so 

    call LA test group 1 

    call NYC test group 2 

    so find p1 and σ_p1

    p1 = 182/540 =   91/270  

    p1 = approx. 0.337037037  

    in the standalone case 

    σ_p1=  sqrt ( (91/270)(179/270)/ 540) 

    σ_p1=   approx.  0.020341668

    p2 = 77/283

    p2=   approx. 0.272084806

    σ_p2=sqrt(  (77/283)*(206/283)/ 283)   

    σ_p2= 0.02994799    

    to show the they are the same, 

    you would like to check if the difference is zero. 

    D= p2-p1=  0.337037037 -  0.272084806

    D =  0.064952231    

    σ_diff^2  =   (σ_p1)^2 + (σ_p2)^2

    σ_diff =  sqrt(  (σ_p1)^2 + (σ_p2)^2 )   

    My first thought would be to use 

    individual σ_p1 and σ_p2

    but video from Khan academy says 

    since you are assuming they are equal  

    then you should use the combined fraction  

    p1 +p2   

    p_combined = (77+182)/ (283+540)  = 

    p_combined =0.314702309

    so Khan Academy say   in video below

    since you are assuming they are equal

    used p_combined to 

    the variance for the both o_p1^2 and o_p2^2

    so if you follow there advice 


    o_p1 = o_p2= sqrt(  0.314702309 *(1-0.314702309 )/829) 

    o_p1,2=  0.016187869

    (σ_p1)^2=  0.000262047

    σ_diff^2 = (σ_p1)^2 + (σ_p2)^2

    σ_diff^2=  0.000262047 + 0.000262047

    σ_diff= sqrt( 0.000524094) 

    σ_diff = 0.022893104

    (while if you used the original values , you get  

    0.033 approx.)   

    so using the Khan Academy assumption 

    z_test = p_diff/ σ_diff =  

    z_test = 0.064952231/  0.022893104 = 

    z_test =  2.837196376

    since this based on the Khan academy assumption

    to pool the proportion since you are assume 

    they are the same  

    since 2.837 > 1.96 (Z_critical for two-sided 95% check)  

    you can reject the NULL and support that they 

    are different and are not the same   

    However, if you decided to just use the original 

    individual σ_p1 and σ_p2 , 

    then you get the opposite answer  

    since  1.946 < 1.96    

    then you cannot support they are different. 


    Here, I would use Welch t-test equations

    Group 1(proctored) :n1=30, x̅1= 74.30, s1= 12.87

    Group 2 (nonp) :n2=32, x̅2= 88.62, s2=22.09

    1st,  you are never going to show  that they are  

    higher , because only test data shows that they are lower. 

    That is, unless you made a typo.   

    nonp is 88.62  which is greater than 74.30 so you have no hope

    in showing proctor is higher and should have 

    went home. 

    so if you agree with wikipedia that a student t-test 

    with the equations  should be used. 

    s_delta_minus =sqrt(s_1^2/n_1+s_2^2/n_2) 

    s_delta_minus = sqrt ( 12.87^2/30 + 22.09^2/32 ) =  

    s_delta_minus=   4.557437122

    Wikipedia says use this

     Welch–Satterthwaite equation to determine the DF

     D.F. = (s_1^2/n_1 + s_2^2/n_2)^2/ 

     ((s_1^2/n_1)^2/(n_1-1) + (s_2^2/n_2)^2/(n_2-1).

    DF = ( 12.87^2/30 + 22.09^2/32)^2 / 

     ( ((12.87^2)/30)^2/ (29) + ( 22.09^2/32)^2/ 31 ))

    DF = 50.44342102

     DF = 50 rounded 

    so we want a one-sided , since the question is . 

    a greater than 

    DF = 50 , and   alpha =0.01 one sided 

    t_crtical (DF =50, alpha =0.01) = 2.405

    Using this calculator : 


    t_test = (x1_bar- x2_bar)/ s_minus_delta

    t_test  =    (74.30 - 88.62) /  4.557437122

    t_test = -3.142116856   

    since  -3.14 is not greater than 2.405    

    You cannot support the claim  that the proctored 

    test score are higher.   

    If you really wanted to, you could support the claim 

    that they are lower since   -3.14<  -2.405  on the low end. 

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