Question

am m on than one data point that value is the mean 3. For the data set shown below, complete parts (a) through (d) below x 3 4 5 7 80 y 5 7 6 13 15 (a) Find the estimates of Bo and B, Bombo (Round to three decimal places as needed.) P, by (Round to three decimal places as needed.) (b) Compute the standard error, the point estimate for a (Round to four decimal places as needed.) (c) Assuming the residuals are normally distributed, determine su (Round to three decimal places as needed) (d) Assuming the residuals are normally distributed test Ho B, = 0 versus H, B, *0 at the a= 0.05 level of significance. Use the P-value approach. The P alue for this test is (Round to three decimal places as needed) Make a statement regarding the null hypothesis and draw a conclusion for this test. Choose the correct answer below O A Reject There is not sufficient evidence at the 0.05 level of significance to conclude that a linear relation exists between x and OB. Do not reject Ho There is sufficient evidence at the r = 0.05 level of significance to conclude that a ligear relation exists between x and y O C. Do not reject There is not sufficient evidence at the = 0.05 level of significance to conclude that a linear relation exists between x and y OD. Reject Hy There is sufficient evidence at the c 005 level of significance to conclude that a linear relation exists between x and y

Answer 1

ate given &y 3 => Heze two variables Bo& Bi. a) => To find the estimates of Exigi= 284 Exis 27 exi2= 163 ni=5 na-5 syi= 46 sy;²= 504 T= 9.2 - B = {xiyi- nå7 - 284-5x5.48.9.2 Exi² ncseja 163- 5 (5.472 35.6 17.2 B = 2.07 Bi = 2.068767442 : y = Bot Bize. 5 = ßotsi so = - 2, 8 = 3.2 – ( 2 045 54 = 9.2 - 11-18 I stegs -1.98 (x-7)? 5-76 1.96 0-16 2-56 6.26 17.2 - b) BO = -98 & ß = 2.07 -1.98 Now to compute standaɛd cerot tee pt estimet 6 53 - Exi? -($32: 163 -5.4)2= 3.45 __ 6ý = yi? -(5)2 = Sou -(9.23?: 16.16 5 Pooled standard dein s= (ni61² +22 622 - 15x3.44 + 5x18-16 V nitn2-2 5+5-2 } = 12.25 s= 3.5

ezzot fo e boes se To compute standazd SE = 0.7 c) => Toe deteɛmine sbi sbi= v scyi-yi)2/2-2 4= 5.4 x'=x-5.4 = ( xi-5)2 9 x 5 -2.4 7 -1-4 x mrn it on - yx' -12 -9.8 -2.4 20.8 39 35-6 x12 5.76 1.96 0-16 2.56 6.76 17.2 % = âtô 1941 4.23 6.3 8.37 12-5| 14.58 (y-ŷ, z 977 0.7 -2.37 0.49 0.42 0-0 cynę? 0.5929 0.49 3.6164 0.2401 0.1766 7.1158 13 is 46 1-6 2.6 0 - b = - 2.0 7 . {x'y Ex12 35-6 17.2 á = P = doce que 11:58 الهی و y=-1.98+2-07 e. y=gatzats 7.12 1/25-2 sbi= 1-540562 4014728 0-371963 17.2 shi = 0.3714 sxx = {x? (Exi) ² - 163. { xi² an(50)2 = 163-565.4)? = 172 Sxx=

distrittd ate normally residual d Assuming the test Ho: BEO VIS HO: Bito T-statice = Bi Jelszeze son __38 2.45282 2.07 2007 0. 8439 12.25 17-2 I 688 I Ical = 2.452 2 critical rabe :- t (412, n-1) =t(0.05/2,4) 1.1-1 t-table = 2.77 cεitetia! if t-cal y z-tab at at level of we pray Reject Ho significance Decisions- tcal = 2.45 at tabe = 2.77 we may accept Ho at 5%. 2.o.s . nacº Peel - Ho: B = 0 . . Do not reject Ho. There is not sufficient infotme evidence at tel x=0.os level of significance to conclude that a linear Relation eseists beth Rx & Y o