Question

by the 1. A government agency is responsible for awarding funds to projects according to the potential scientific research projects. A company examines the relation between the amount of funding (in $10 000) requested by the proposals (x) and the length of time between submission of the project proposal and project approval (Y). For the most recent 7 proposals the following data are obtained: 14 20 5 10 5000 5001000 Time (in months) Amount (in $10 000) a) Estimated regression line f , for the model Y-A, + β, x + e was found to be } = 5.898 + 0.0149 x. b) For the model in part (a) test the hypothesis that e) For the above data. Let z -In x. Find the estimated regression line Y for the model d) Which of the two models [model in part (a) or in part (c)] explains better variation in Y? Why?

Answer 1

a)

X	Y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
1	3	56169.00	41.327	1523.57
5	4	54289.00	29.469	1264.86
10	6	51984.00	11.755	781.71
50	8	35344.00	2.041	268.57
100	11	19044.00	2.469	-216.86
500	14	68644.00	20.898	1197.71
1000	20	580644.00	111.755	8055.43

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	1666	66	866118.000	219.7	12875
mean	238.00	9.43	SSxx	SSyy	SSxy

estimated slope , ß1 = SSxy/SSxx =   12875.0   /   866118.000   =   0.0149
                  
intercept,   ß0 = y̅-ß1* x̄ =   5.898
                  
so, regression line is   Ŷ =   5.898 +   0.0149   *x

b)

SSE=   (SSxx * SSyy - SS²xy)/SSxx =    28.325
      
std error ,Se =    √(SSE/(n-2)) =    2.38013

Ho:   ß1=   0          
H1:   ß1╪   0          
n=   7              
alpha =   0.02              
estimated std error of slope =Se(ß1) = Se/√Sxx =    2.38013   /√   866118.0   =   0.003
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    0.0149   /   0.0026   =   5.812
                  
Degree of freedom ,df = n-2=   5              
p-value =    0.0021              
decison :    p-value<α , reject Ho  

reject Ho and conclude  that slope is significantly different from zero

c)

here, X = ln(x)
                  

X	Y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
0	3	13.32	41.327	23.47
1.6094379	4	4.16	29.469	11.08
2.3025851	6	1.82	11.755	4.62
3.912023	8	0.07	2.041	-0.37
4.6051702	11	0.91	2.469	1.50
6.2146081	14	6.58	20.898	11.72
6.9077553	20	10.61	111.755	34.44

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	25.55157957	66	37.473	219.7	86.4510441
mean	3.65	9.43	SSxx	SSyy	SSxy

estimated slope , ß1 = SSxy/SSxx =   86.5   /   37.473   =   2.3070
                  
intercept,   ß0 = y̅-ß1* x̄ =   1.0074          
                  
so, regression line is   Ŷ =   1.0074   +   2.3070   *x

4)

from X and Y data points,

R² =    (Sxy)²/(Sx.Sy) =    0.8711 or 87.11%

and from z and y data points

R² =    (Sxy)²/(Sx.Sy) =    0.9077 pr 90.77%

so, model in part c) explains better variation in Y