Question

Problem 1.21) Airfreight breakage. A substance used in biological and medical research is shipped by air freight to users in cartons of 1,000 ampules. The data below, involving 10 shipments, were collected on the number of times the carton was transferred from one aircraft to another over the shipment route (X) and the number of ampules found to be broken upon arrival (Y). Assume the first-order regression model (1.1) is appropriate.

i: 1,2,3,4,5,6,7,8,9,10

xi: 1,0,2,0,3,1,0,1,2,0

yi: 16,9,17,12,22,13,8,15,19,11

Problem 2.25) Refer to Airfreight breakage Problem 1.21.

a. Set up the ANOVA table. Which elements are additive?

b. Conduct an F test to decide whether or not there is a linear association between the number of times a carton is transferred and the number of broken ampules; control sigma risk at .05. State the alternatives, decision rules, and conclusion.

c. Obtain the t^* statistic for the test in part (b) and demonstrate numerically its equivalence to the F^* statistic obtained in part b.

d. Calculate R^2 and r. What proportion of the variation in Y is accounted for by introducing X into the regression model?

Please only solve problem 2.25

Answer 1

x	y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
1	16	0.0000	3.2400	0.0000
0	9	1.0000	27.0400	5.2000
2	17	1.0000	7.8400	2.8000
0	12	1.0000	4.8400	2.2000
3	22	4.0000	60.8400	15.6000
1	13	0.0000	1.4400	0.0000
0	8	1.00	38.44000	6.2000
1	15	0.00	0.64000	0.000
2	19	1.00	23.04	4.80
0	11	1.00	10.24	3.20

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	10.00	142.00	10.00	177.60	40.00
mean	1.00	14.20	SSxx	SSyy	SSxy

sample size ,   n =   10      
here, x̅ = Σx / n=   1.000   ,     ȳ = Σy/n =   14.200
              
SSxx =    Σ(x-x̅)² =    10.0000      
SSxy=   Σ(x-x̅)(y-ȳ) =   40.0      

SSE=   (SSxx * SSyy - SS²xy)/SSxx =    17.6000
SST = SSyy=177.600

SSR = SST-SSE=160.000

a)

Anova table
variation	SS	df	MS	F-stat	p-value
regression	160.000	1	160.000	72.727	0.0000
error,	17.600	8	2.200
total	177.600	9

b)

Ho: regression model is not useful

H1: regression model is useful

F critical value(0.05,1,8) = 5.318

decision rule: reject Ho, if F >5.318

F stat = 72.727

since, F stat >5.318, Reject ho

hence, it is concluded that regression model is useful.

c)

estimated slope , ß1 = SSxy/SSxx =   40.0   /   10.000   =   4

std error ,Se =    √(SSE/(n-2)) =    1.4832
estimated std error of slope =Se(ß1) = Se/√Sxx =    1.483   /√   10.00   =   0.4690
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    4.0000   /   0.4690   =   8.5280

t² = 8.528² = 72.727

So, F stat = (t -stat)²

d)

correlation coefficient ,    r = Sxy/√(Sx.Sy) =   0.949
      
R² =    (Sxy)²/(Sx.Sy) =    0.9009
0.9009 proportion of the variation in Y is accounted for by introducing X into the regression model