Construct a scatter plot usng excel for the gven data. Determne whether there s a postve lnear correlaton, negatve lnear correlaton, or no lnear correlaton. Complete the table and fnd the correlaton coeffcent r. The data for x and y s shown below. x -5-3 4 1-1 - 0 3-4 y 11-6 8-3 - 1 5-5 6 7 a) Scatter plot (.5 ponts) Scatter Plot 1 10 8 6 4 y 0-6 -4 - - 0 4 6-4 -6-8 x b) Type of correlaton (postve lnear correlaton, negatve lnear correlaton, or no lnear correlaton) ( ponts) Here the scatter plot does not show any strong pattern. Thus we may conclude that x and y are uncorrelated. c) Complete the table and fnd the correlaton coeffcent r rounded to 4 decmals. Part 1: Scatter plot (.5 ponts)
x y xy x y -5 11-55 5 11-3 -6 18 9 36 4 8 3 16 64 1-3 -3 1 9-1 - 1 4-1 - 4 1 0 5 0 0 5-5 -10 4 5 3 6 18 9 36-4 7-8 16 49-5 -8 85 370 Use the last row of the table to show the column totals. From the gven data we have n 10, x -5, y, x y -8, x 85 and y 370 We have, r n x ( x ) n y ( y ) 10*( 8) ( 5)* 10*85 ( 5) 10*370 () - 0.1044 Construct a scatter plot and nclude the regresson lne on the graph usng excel for the gven data. Determne whether there s a postve lnear correlaton, negatve lnear correlaton, or no lnear correlaton. Complete the table and fnd the correlaton coeffcent r. (References: example 1-4 pages 498-500; end of secton exercses 15 - pages 508-509) In an area of the Mdwest, records were kept on the relatonshp between the ranfall (n nches) and the yeld of wheat (bushels per acre). Ran fall, x 9.1 7.4 1 11.1 17.4 8.9 5.6 14. 14.6 Yeld, y 54.5 50. 6.8 63 86.4 53. 35.9 80 8.8 a) Scatter plot wth regresson lne (.5 ponts)
Scatter Dagram Yeld 100 90 80 70 60 50 40 30 0 10 0 y 4.3791x + 14.398 0 5 10 15 0 Ran fall b) Type of correlaton (postve lnear correlaton, negatve lnear correlaton, or no lnear correlaton) ( ponts) The scatter plot shows that as ran fall ncreases yeld also ncreases. Thus there exst a postve lnear correlaton between Ran fall and Yeld. c) Complete the table and fnd the correlaton coeffcent r rounded to 4 decmals. (5 ponts) x y xy x y 9.1 54.5 495.95 8.81 970.5 7.4 50. 371.48 54.76 50.04 1 6.8 753.6 144 3943.84 11.1 63 699.3 13.1 3969 17.4 86.4 1503.36 30.76 7464.96 8.9 53. 473.48 79.1 830.4 5.6 35.9 01.04 31.36 188.81 14. 80 1136 01.64 6400 14.6 8.8 108.88 13.16 6855.84 100.3 568.8 6843.09 13.91 384.98 Use the last row of the table to show the column totals. From the gven data we have
n 9, y 384.98 x 100.3, y 568.8, x y 6843.09, x 13.1 and We have, r n x ( x ) n y ( y ) 9*6843.09 100.3*568.8 9*13.1 (100.3) 9*384.98 (568.8) 0.9808 3. Usng the r calculated n problem test the sgnfcance of the correlaton coeffcent usng a 0.01 and the clam rho 0. Use the 7-steps hypothess test shown at the end of ths project. (References: example 7 page 505; end of secton exercses 3-8 pages 510-511) (5 ponts) (Note: Round the computed t to 3 decmals.) Answer: Let ρ be the populaton correlaton coeffcent. 1. H 0 : ρ 0 H a : ρ 0. a 0.01 3. We have, r n t r 0.9808 9 0.9808 13.313 4. Snce a 0.01, from Student s t table wth (n-) 7 degrees of freedom, the crtcal value s t 0 3.499
5. Rejecton regon: Reject Ho f t < -3.499 or t > 3.499 6. Decson: Here, t 13.313 > 3.499 So we reject the null hypothess Ho. 7. Interpretaton: Thus the correlaton between ran fall and yeld s statstcally sgnfcant at 1% level of sgnfcance. [If the alternatve hypothess s H a : ρ > 0, use the followng Let ρ be the populaton correlaton coeffcent. 1. H 0 : ρ 0 H a : ρ > 0. a 0.01 3. We have, r n t r 0.9808 9 0.9808 13.313 4. Snce a 0.01, from Student s t table wth (n-) 7 degrees of freedom, the crtcal value s t 0.998 5. Rejecton regon: Reject Ho f t >.998 6. Decson: Here, t 13.313 >.998
So we reject the null hypothess Ho. 7. Interpretaton: Thus there exst a postve correlaton between ran fall and yeld at 1% level of sgnfcance.] Secton 9.: Lnear Regresson (References: example 1-3 pages 514-516; end of secton exercses 13 - pages 518-50) 4. The data gven are the temperatures on randomly chosen days durng a summer class and the number of absences on those days. Temperature, x 7 85 91 90 88 98 75 100 80 Number of absences, y 3 7 10 10 8 15 4 15 5 a. Fnd the equaton of the regresson lne for the gven data. Round the lne values to the nearest two decmal places. (.5 ponts) Show your work. Let y denote the number of absences and x denote the temperature. Assume that x and y are lnearly related. Let y α + β x be the suggested lnear relatonshp. By the method of least squares the estmates of α and β are gven by, ˆ β n x x ( ) y ˆ x and ˆ β α n From the gven data we have n 9, x 779, y 77, x y 6995 and x 68163 We have,
ˆ β n x x ( ) 9* 6995 779*77 9*68163 (779) 0.4485 0.45 y ˆ x ˆ β α n 77 0.4485*779 9-30.677-30.7 Thus the ftted regresson equaton s Number of absences -30.7 + 0.45 Temperature b. Usng the equaton found n part a, predct the number of absences when the temperature s 95. Round to the nearest absence. (.5 ponts) If temperature 95, the predcted value of the number of absences s Number of absences -30.7 + 0.45*95-30.7 + 4.75 1.48 13