Respuesta :
Answer:
Yes, there is a difference between the population mean for the math scores and the population mean for the writing scores.
Test Statistics = Â [tex]\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }[/tex] follows [tex]t_n_- _1[/tex] .
Step-by-step explanation:
We are provided with the sample data showing the math and writing scores for a sample of twelve students who took the SAT ;
Let A = Math Scores ,B = Writing Scores  and D = difference between both
So, [tex]\mu_A[/tex] = Population mean for the math scores
    [tex]\mu_B[/tex] = Population mean for the writing scores
 Let [tex]\mu_D[/tex] = Difference between the population mean for the math scores and the population mean for the writing scores.
       Null Hypothesis, [tex]H_0[/tex] : [tex]\mu_A = \mu_B[/tex]   or  [tex]\mu_D[/tex] = 0
   Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu_A \neq \mu_B[/tex]    or  [tex]\mu_D \neq[/tex] 0
Hence, Test Statistics used here will be;
      [tex]\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }[/tex] follows [tex]t_n_- _1[/tex]   where, Dbar = Bbar - Abar
                                [tex]s_D[/tex] = [tex]\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}[/tex]
                                n = 12
Student     Math scores (A)      Writing scores (B)     D = B - A
   1            540               474                  -66
   2            432              380                   -52 Â
   3            528              463                   -65
   4            574              612                    38
   5            448              420                   -28
   6            502              526                   24
   7            480              430                   -50
   8            499              459                  -40
   9            610               615                    5
   10            572              541                    -31
   11            390              335                   -55
   12            593              613                    20 Â
Now Dbar = Bbar - Abar = 489 - 514 = -25
 Bbar = [tex]\frac{\sum B_i}{n}[/tex] = [tex]\frac{474+380+463+612+420+526+430+459+615+541+335+613}{12}[/tex]  = 489
 Abar =  [tex]\frac{\sum A_i}{n}[/tex] = [tex]\frac{540+432+528+574+448+502+480+499+610+572+390+593}{12}[/tex] = 514
 ∑[tex]D_i^{2}[/tex] = 22600   and  [tex]s_D[/tex] = [tex]\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}[/tex] = [tex]\sqrt{\frac{22600 - 12*(-25)^{2} }{12-1} }[/tex] = 37.05
So, Test statistics = Â [tex]\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }[/tex] follows [tex]t_n_- _1[/tex]
              = [tex]\frac{-25 - 0}{\frac{37.05}{\sqrt{12} } }[/tex] follows [tex]t_1_1[/tex]  = -2.34
Now at 5% level of significance our t table is giving critical values of -2.201 and 2.201 for two tail test. Since our test statistics doesn't fall between these two values as it is less than -2.201 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region .
Therefore, we conclude that there is a difference between the population mean for the math scores and the population mean for the writing scores.
