Question

*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 the a risk at .05. State the alternatives, decision rule, 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 R2 and r. What proportion of the variation in Y is accounted for by introducing -X into the regression model?

i	X	Y
1	1	16
2	0	9
3	2	17
4	0	12
5	3	22
6	1	13
7	0	8
8	1	15
9	2	19
10	0	11
SUM	10	142
AVG	1	14.2

Answer 1

ANSWER:

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