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The College Board SAT college entrance exam consists of three parts: math, writing and critical reading (The World Almanac 2012). Sample data showing the math and writing scores for a sample of twelve students who took the SAT follow.

Student Math Writing Student Math Writing
1 540 474 7 480 430
2 432 380 8 499 459
3 528 463 9 610 615
4 574 612 10 572 541
5 448 420 11 390 335
6 502 526 12 593 613

Use α= 0.05 level of significance and test for a difference between the population mean for the math scores and the population mean for the writing scores. Enter negative values as negative numbers. What is the test statistic?

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 =   [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  

     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  

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 =   [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.

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