Question

1. The tensile strength of a paper product is related to the amount of hardwood in the pulp. Ten samples are produced in the pilot plant, and the data obtained are shown in the following table.

Strength	Percent Hardwood	Strength	Percent Hardwood
160	10	181	20
171	15	188	25
175	15	193	25
182	20	195	28
184	20	200	30

1. Find the 95% confidence interval for the mean strength at 10% hardwood, and also the 95% confidence interval for a single observation at 10% hardwood (use software – in Statgraphics check Forecasts in Tables and Graphs).

Answer 1

X	Y	XY	X²	Y²
10	160	1600	100	25600
15	171	2565	225	29241
15	175	2625	225	30625
20	182	3640	400	33124
20	184	3680	400	33856
20	181	3620	400	32761
25	188	4700	625	35344
25	193	4825	625	37249
28	195	5460	784	38025
30	200	6000	900	40000
Ʃx =	Ʃy =	Ʃxy =	Ʃx² =	Ʃy² =
208	1829	38715	4684	335825

Sample size, n =	10
x̅ = Ʃx/n = 208/10 =	20.8
y̅ = Ʃy/n = 1829/10 =	182.9
SSxx = Ʃx² - (Ʃx)²/n = 4684 - (208)²/10 =	357.6
SSyy = Ʃy² - (Ʃy)²/n = 335825 - (1829)²/10 =	1300.9
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 38715 - (208)(1829)/10 =	671.8

Slope, b = SSxy/SSxx = 671.8/357.6 = 1.8786353

y-intercept, a = y̅ -b* x̅ = 182.9 - (1.87864)*20.8 = 143.82438

Regression equation :

ŷ = 143.8244 + (1.8786) x

Predicted value of y at x = 10

ŷ = 143.8244 + (1.8786) * 10 = 162.6107

Significance level, α = 0.05

Critical value, t_c = T.INV.2T(0.05, 8) = 2.306

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 1300.9 - (671.8)²/357.6 = 38.832774

Standard error, se = √(SSE/(n-2)) = √(38.83277/(10-2)) = 2.20320

95% confidence interval for the mean strength at 10% hardwood:

95% confidence interval for a single observation at 10% hardwood: