Question

The following data are based on a random selection of these cars of different models and options. Let y be the dealer invoice (in thousands of dollars) for the given vehicle.

x	31.8	33.5	36.1	44.0	47.8
y	30.0	30.8	32.0	42.1	42.2

Σx = 193.2, Σy = 177.1, Σx² = 7657.54, Σy² = 6425.89, Σxy = 7010.56, and r ≈ 0.976.

(b) Use a 5% level of significance to test the claim that ρ > 0. (Use 2 decimal places.)

t
critical t

(c) Verify that S_e ≈ 1.5546, a ≈ 1.7787, and b ≈ 0.8706.

lower limit
upper limit

(d) Find the predicted dealer invoice when the list price is x = 45 (thousand dollars). (Use 2 decimal places.)


(e) Find a 95% confidence interval for y when x = 45 (thousand dollars). (Use 2 decimal place.)

(f) Use a 5% level of significance to test the claim that β > 0. (Use 2 decimal places.)

t
critical t

(g) Find a 95% confidence interval for β and interpret its meaning. (Use 3 decimal places.)

lower limit
upper limit

Answer 1

we have given as independent & n=s observation y as dependent variable. x=(31.8, 33.5, 36-1, 44.0, 47.8) 4 y=[ 30.0, 30.8, 32-0, 42'1, 42.2) we have given: 177.1, E9; Ez? = 7657.54 Exi=193.2, €4,? = 6425.89 Exy = 7010.56, 112 0.976

use to test level of significance a 5% the claim that P >0 + To Test SH 8=0 VS H: foo Test statistic : t = 87 In-z 2 I-8 1. t = (0.976 53 3) 1.6904815 6.21777052 1- 0.976?

7.762674 112 7.76 df degree of Freedom 3 n-2= critical ta ta,df to.os, 3 24 - 3 12:35 (from table) Decision here ;- we reject Ho if to ta, df to ta, df we reject Ho evidence to conclusion : There is sufficient fo ée conclude that result is significant.

с verify that se ² 1.5546 9 % 1.77874 br 0.8706. > = intercept ý = b ă f a Ć r b = slope. = hass aass zesum = ssay Ery- CEZ *EY; 4 2 کره . z Exi ² -Ceais n

hess %3D +0\0. 56 ( 1१3.2 * (+1) S 161.416 Ssxx hs-tsot - ( 193.252 5 192.292 2 :. = าม 0.8106 16+.46 | 92.292 _ (o. 8406 * ११. 2 9 - Eyi - (E FR:) = 140 5 5 या 1.+18+

ssuy -b ssay y = qt br 1.7787 + 0.8 106 * X Se = SSE where SSE = n-2 Ey;?- (843? = where Sluy n = 6425.89 (177.1)? S = 153.008 SSEE 7.249897 153.008 - 10.8706 * 167.416)

a.se = 2-416632 7.249897 3 1:55455 1:5546 211 x=45 find predicted dealer invoice when 40.95723379 LE) 11 1.7787 + (0.8706 +45) ell 40.96 g when s=45 (e) confidence interval for * se * =( 9 t to ,df tre-2) ? df essor ہدای

here d=1-0.95 0.05 :. t toos, 3 ti = (from table) 3.1824 con 40.96 ã Exi 193.2 38.64 5 r stan 192.292 se 1.5546

confidence interval) [ 40.96 t (3.1824 +1.5546* th +145-38.64) 192.292 = ( 40.96 † (3.1824+ 1.5546*0.6406)] (37.7907, 44.1292] [ 40.96 $ 3.1692) ali (37.80, 44.13 Tevel of significance to test use 5 %. 6 the claim that ß 70

vs To Test Hoi ß=0 H, : B0 Test statistics ta b-p seb. where seb se 11 1.5546 = 0:1121 Jssxx 192.292 t - 0.8706 0:1121 = 7-7662 Š +++ dFE D-R = 1 37 3

Critical value to, df = to-05, 3 all 2.35 (from table) Decision .!! we reject Ho if t statistics > toidf > ta,df we here t statistics reject Ho conclusion : there is sufficient evidence pro te result is significant Ho conclude that for ß. 9 957. confidence interval * ses * ß = (b £ tala,df

ta , df toos, 3 = f 3.1824 (from table) B C0.8706 1 (3.1824 * 0.1121)] = [0.8706 | 0.3567) C [0.5139, 1.2274] - < (0.514 1,223 *) Interpretation :- slope is lies within 0.514 to hence it is 1.227 interval does not include zero significant