Question

A chemical reaction is run 12 times, and the temperature x;(n °C) and the yield y, (In percent of a theoretical maximum) Is recorded each time. The following summary statistics are recorded: I=65.0, y = 29.03, ( 2)' = 6032.0, (4-3) = 835.42, (1, -1) (4. - u) = 1988.5 Let Bo represent the hypothetical yield at a temperature of O'C. and let B represent the increase in yleld caused by an Increase in temperature of 1°C. Assume that assumptions 1 through 4 for errors in linear models hold. Find a 95% confidence interval for the mean yleld at a temperature of 40°C. Round the answers to three decimal places. The 95% confidence interval is (16.731 24.895)

please provide me correct answer

Answer 1

n=	12
SSx=Σx2-(Σx)2/n=	6032
SSy=Σy2-(Σy)2/n=	835.42
SP=Σxy-(ΣxΣy)/n=	1988.5
SST=Syy=	835.42
SSE =Syy-(Sxy)2/Sxx=	179.8941
SSR =(Sxy)2/Sxx =	655.5259
b1= SSxy/Sx =	0.3297

predicted value at X=40 is:0.33*40+7.60219827586207=

20.789

a)

standard error of CI=s*√(1/n+(x0-x̅)2/Sxx)=		45.2088
for 95 % CI value of t=		2.2280	(from excel:tinv(0.05,10)
margin of error E=t*std error=		100.7252
lower confidence bound=xo-E=		-79.937
Upper confidence bound=xo+E=		121.514

95% confidence interval =( -79.937 , 121.514)

b)

standard error of PI=s*√(1+1/n+(x0-x̅)2/Sxx)=		113.9145
for 95 % CI value of t=		2.2280	(from excel:tinv(0.05,10)
margin of error E=t*std error=		253.8016
lower confidence bound=xo-E=		-233.013
Upper confidence bound=xo+E=		274.590

95% prediction interval =(-233.013 , 274.590)