Question

A research firm collected data on a sample of n= 30 drivers to investigate the relationship between the age of a driver and the distance the driver can see. The data is given below Age Age 20 Distance 510 590 560 510 55 63 Distance 420 350 420 300 410 300 390 320 460 490 560 370 510 460 410 420 460 450 280 420 460 360 310 360 380 460

Answer 1

using our data, bet xi 1-1,2,;., 30 depresents age Y i=1,- «30 eepresents distance we have obtained, Exi = 1530 - x=51 Eli= 12700 - Y= 423.33 Exi² = 91782 Exili= 606350 Eyi ² = 5570000 .: Covariance = f Exili - 7 7 606350 - 51 (423.33) = -1378-163 sy - t Ex- 6' 91782 - (S12 - 458.4 sy - Evi" -(52 - 5570000 - (423:33) ?: 6458.378 . Corr coeff - Corasiance - - 1378.163 = -0.80097 St sy 458.464 58.378 30 30 30 =) As the decreases coeficient age increases, & model is distance at the driver good because of high can see conselation s Sxx i Syy = Sxx= - Sxx-{(xi-2² Bo ý - BX [1=30] {lyi-g ? - {yi - n j = 193666.7 {(x-x ² - {x² - n (à ² = 13752 - 0.8 19.3666-7 - 3.002 PPPPPP :. B. = -0.8V 13752 Bo = ý - 8 x = 423.33 + 3.002 * 51 57 6.432

. 2 1 the model is. Y= 576.4 à 340 B 576.4 3 il x=50 hou x*-so Ser = 13752 n = 30 N 8 - s7 6 4 + 3( So) = 42 4.4 (56) 2 we expect 42604 distance for so year old driver 5 9 95% Prediction interval of g 1 5o) in • ý (so ) & tga od 1th + (x _ z (2-1) 5 = 51 (+)² & bi-x} = Sxx= 13752 tra (0osle dogo (0.975) = 2004523 ŷ150) = 426.4 = I & (gi- yet when i = 57664- 3 ** Es 4 ar put all & & we get ye & ther, using computation & (yi-y? ² 69334.8 - Elgi- } ) 2 23900855 2.04. 2n => 1 2390185 ( 1+ 3o + 9352 i = 101.4009 951. prediction interval is. 496.4 I 101.4 - (325, 52708)

(m) 95% confidence interval for y/150lii 9 150 l it ta od t + (yn z ² (n-1) 5x. (n-1) sa A5 2390.85 = 18.23 & a5l. confidence interval is 426.4 & 18.23 = ( 108.17, 444.63)

(m) 95% confidence interval for y/150lii 9 150 l it ta od t + (yn z ² (n-1) 5x. (n-1) sa A5 2390.85 = 18.23 & a5l. confidence interval is 426.4 & 18.23 = ( 108.17, 444.63)