Question

an instructer for a test preparation course claims that the course will improve the test scores of students...

Test: Practice Final Exam Time Limit: 03:00 00 Submit Test This Question: 3 pts 2 of 34 This Test: 100 pts possible 0 An instructor for a fost preparation course claims that the course will improve the test scores of students. The table shows the crtical reading scores for 14 students the first two times they took the test. Before taking the test for the second time, the students took the instructor's course to try to improve the critical reading test scores Ala 0.01, is there enough evidence to support the instructor's cam? Complete parts (a) through 00 Student 1 2 3 4 5 6 8 Score on first test 2000442038 220 503 543 Scene on second test 20141 388 10 11 12 13 14 15 12333 414381 Identify the rejection regions. Choose the correct answer below. O A T-2650 2650 OB. t>3.012 Oct-3012 ont>3.012 OD T-2650 (c) Coloured and - Tanaman Click to select your answers) Save for Later 0 e 2:37 PM 7/29/2020 de 3 $ 4 % 5 C 9 3 6 7 8 Q и E R T Y U o P

Answer 1

2 را gi 397 -75 solution we want to test preparation cousse claims that the course will improve the test score of students. U=14 no. Mxilly annoub=0 1. Mx My as nor Molo First second test test di=H-Y Ques (B) ms Hi 401 -4 16 taitical left 3uu 419 5625 tailed at 320 306 - 66 4356 (neat=144= 34 Gou 635 -31 961 =13df 496 449 47 2209 ?d=:01 530 564 - 26 676 329 300 29 t = -2,65 503 633 - 130 16900 13,101 543 601 -50 3364 405 539 - 54 2916 so Rejection Region 412 442 -30 go 0 t<-21650 333 350 -25 625 414 447 -33 loog ang 381 431 -50 2500 Edi Edi²=42970 = -506 и aus ① đ = Edi -506=-36.1429 Rounded value up to 3-decimal = đ= -36,143 = 5h [ Edi?- noty - Vitan ( 42978-14x (36-14333 S d = 43.5798 sd @ Standard - t-statistics (t) = d ntur,& sd/su t-stat -36, 143 XJ TU 43.5798 -3.1032 t-Stat Rounded It = -31103

e Here we see that t-stat is less than t-Critical €2.650) so we Rejectno ( Mullhypothesis) option - O Reject the null hypothesis ② At 14. significance level there is enough evidence to Reject the null hypothesish conclude that secand test scose Emplores the process score-