Question

The relationship between "strength" and "fineness" of cotton fibers was the subject of a study that produced the following data. (Give your answers correct to two decimal places.)

x, Strength	75	75	87	80	80	76	78	71	73	87
y, Fineness	4	4.3	4.1	4.4	4.9	4.6	4.7	4	4.4	4.7

(a) Find the 98% confidence interval for the mean measurement of fineness for fibers with a strength of 75.

Lower Limit -

Upper Limit -

(b) Find the 98% prediction interval for an individual measurement of fineness for fibers with a strength of 75.

Lower Limit -

Upper Limit -

Answer 1

X	Y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
75	4	10.240	0.168	1.312
75	4.3	10.240	0.012	0.352
87	4.1	77.440	0.096	-2.728
80	4.4	3.240	0.000	-0.018
80	4.9	3.240	0.240	0.882
76	4.6	4.840	0.036	-0.418
78	4.7	0.040	0.084	-0.058
71	4	51.840	0.168	2.952
73	4.4	27.040	0.000	0.052
87	4.7	77.440	0.084	2.552

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	782	44.1	265.600	0.9	4.88
mean	78.20	4.41	SSxx	SSyy	SSxy

sample size ,n    =10

here,   x̅   =78.20,ȳ =4.41

SSxx = Σ(x-x̅)² = 265.60

SSxy=Σ(x-x̅)(y-ȳ) =4.9

slope , ß1 = SSxy/SSxx =0.01837

intercept,ß0 = y̅-ß1* x̄ =2.97319

so, regression line isŶ    =2.97+0.02*x

SSE=(Sx*Sy - S²xy)/Sx = 0.80

std error ,Se = √(SSE/(n-2)) = 0.316

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a)

X Value=75

Confidence Level=98%

Sample Size , n=10

Degrees of Freedom,df=n-2 =8

critical t Value=tα/2 =2.896[excel function: =t.inv.2t(α/2,df) ]

Predicted Y (YHat), Ŷ    =2.97+0.02*75 = 4.35

margin of error,E=t*Std error=t* Se *√(1/n+(X-X̅)²/Sxx) = 0.3408

Confidence Lower Limit=Ŷ -E =4.01

Confidence Upper Limit=Ŷ +E =4.69

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c)

for prediction interval

margin of error,E=t*std error=t* Se *√(1 + 1/n+(X-X̅)²/Sxx) =    0.977

Prediction Interval Lower Limit=Ŷ -E =3.37

Prediction Interval Upper Limit=Ŷ +E =5.33