A preparation of insulin is being studied to determine the effects on reducing t

Question

A preparation of insulin is being studied to determine the effects on reducing the blood-sugar level in rats. 25 rats of identical breed are randomly injected with different dosages of insulin (X) and reduction rate in their blood-sugar levels (Y) are recorded. The results are summarized as: X = 150 X2 = 1000 Y = 50 Y2 = 693 XY = 450 And the incomplete Analysis of Variance table is given as:

M.S. Source D.F. 225 225 Regression Error Total 24 593 From the above information, the 99% predicted interval for the reduction rate in blood-sugar level while injected 10 units of insulin dosage is O (12.9787, 23.0213) o (5.7003, 30.2997) o (2.9789, 13.0213) o (4.2997 20.2997) o None of above

Explanation / Answer

MS=SS/DF so DF(regression)=MS(regression)/SS(regression)=225/225=1

error DF=Total DF- regression DF=24-1=23

error SS= Total SS - Regression SS=593-225=368

error MS=error SS/ error DF=368/23=16

the regression anlaysis has been done using Y=a+bX=-7+1.5x

for X=10, Y=-7+10*1.5=8

a=(sum(y)*sum(x2)-sum(x)*sum(xy))/(n*sum(x2)-(sum(x))2)=(50*1000-150*450)/(25*1000-150*150)=-7

b=(n*sum(xy)-sum(x)*sum(y))/ (n*sum(x2)-(sum(x))2)=(25*450-150*50)/(25*1000-150*150)=1.5

(1-alpha)*100% prediction interval of yp=yp ±t(alpha/2,error df)*sqrt(MSE)*sqrt(1+1/n+(xp-xbar)/Sxx)

99% prediction interval of yp=8 ±t(0.01/2,23)*sqrt(16)*sqrt(1+1/25+(10-6)/225)=

=8 ±2.81*sqrt(16)*sqrt(1+1/25+(10-6)/225)=8 ±11.56=(-3.56,19.56)

right choice is none of above

source DF SS MS F F-critical(0.01) Regression 1 225 225 14 8 Error 23 368 16 Total 24 593

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.