Question

An experiment was conducted to observe the effect of an increase in temperature on the potency of an antibiotic. Three 1-ounce portions of the antibiotic were stored for equal lengths of time at each of the following Fahrenheit temperatures: 40, 55, 70, and 90. The potency readings observed at the end of the experimental period were

Potency reading, y: 49 38 27 24 38 33 19 28 16 18 23

Temperature, x: 40 40 40 55 55 55 70 70 70 90 90

a) Find the least-squares line appropriate for these data.

b) Calculate the 95% confidence intervals for B₀ and B₁

Answer 1

a)

Formula sheet

			Sxx	Syy	Sxy
	(x)	(y)	(x-xbar)^2	(y-ybar)^2	(x-xbar)(y-ybar)
	40	49	=(G3-$G$15)^2	=(H3-$H$15)^2	=(G3-$G$15)*(H3-$H$15)
	40	38	=(G4-$G$15)^2	=(H4-$H$15)^2	=(G4-$G$15)*(H4-$H$15)
	40	27	=(G5-$G$15)^2	=(H5-$H$15)^2	=(G5-$G$15)*(H5-$H$15)
	55	24	=(G6-$G$15)^2	=(H6-$H$15)^2	=(G6-$G$15)*(H6-$H$15)
	55	38	=(G7-$G$15)^2	=(H7-$H$15)^2	=(G7-$G$15)*(H7-$H$15)
	55	33	=(G8-$G$15)^2	=(H8-$H$15)^2	=(G8-$G$15)*(H8-$H$15)
	70	19	=(G9-$G$15)^2	=(H9-$H$15)^2	=(G9-$G$15)*(H9-$H$15)
	70	28	=(G10-$G$15)^2	=(H10-$H$15)^2	=(G10-$G$15)*(H10-$H$15)
	70	16	=(G11-$G$15)^2	=(H11-$H$15)^2	=(G11-$G$15)*(H11-$H$15)
	90	18	=(G12-$G$15)^2	=(H12-$H$15)^2	=(G12-$G$15)*(H12-$H$15)
	90	23	=(G13-$G$15)^2	=(H13-$H$15)^2	=(G13-$G$15)*(H13-$H$15)
Total	=SUM(G3:G13)	=SUM(H3:H13)	=SUM(I3:I13)	=SUM(J3:J13)	=SUM(K3:K13)
mean	=G14/5	=H14/5
				b1	=K14/I14	Sxy/Sxx
				b0	=H15-K16*G15	ybar-b1*xbar

			Sxx	Syy	Sxy
	(x)	(y)	(x-xbar)^2	(y-ybar)^2	(x-xbar)(y-ybar)
	40	49	9025	184.96	1292
	40	38	9025	605.16	2337
	40	27	9025	1267.36	3382
	55	24	6400	1489.96	3088
	55	38	6400	605.16	1968
	55	33	6400	876.16	2368
	70	19	4225	1900.96	2834
	70	28	4225	1197.16	2249
	70	16	4225	2171.56	3029
	90	18	2025	1989.16	2007
	90	23	2025	1568.16	1782
Total	675	313	63000	13855.76	26336
mean	135	62.6
				b1	0.418031746	Sxy/Sxx
				b0	6.165714286	ybar-b1*xbar

The  regression line is

y = 6.1657 + 0.4180 x

b)

Since, n=11 and standard deviation is unknown, hence we use t statistic to calculate confidence interval

lower limit for b0	0.8905611
Upper limit for b0	11.440867
lower limit for b1	0.4467487
upper limit for b1	0.3893148

Formula sheet

lower limit for b0	=K17-TINV(0.025,9)*SQRT(STDEV(H3:H13)*((1/11)+(G15^2/I14)))
Upper limit for b0	=K17+TINV(0.025,9)*SQRT(STDEV(H3:H13)*((1/11)+(G15^2/I14)))
lower limit for b1	=K16-T.INV(0.025,9)*SQRT(STDEV(H3:H13)/I14)
upper limit for b1	=K16+T.INV(0.025,9)*SQRT(STDEV(H3:H13)/I14)

0.8906 < b0 < 11.4409

0.3893 < b1 < 0.4467