The relationship between \"strength\" and \"fineness\" of cotton fibers was the

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.)

(a) Draw a scatter diagram. (Do this on paper. Your instructor may ask you to turn in this work.)

(b) Find the 98% confidence interval for the mean measurement of fineness for fibers with a strength of 71.

(c) Find the 90% prediction interval for an individual measurement of fineness for fibers with a strength of 71.

x, Strength 69 70 77 77 70 73 84 76 87 69 y, Fineness 4.6 4.3 3.9 3.7 4.9 4.2 4.5 4 3.9 4.5

Explanation / Answer

a.) This can be easily done using 2D cartesian cordinate system and plotting each data point.

b.)

X = 752 , Y = 42.5 , X*Y = 3185 , X2 = 56910

a = (Y*X2X*XY) / (nX2(X)2) = (42.5*56910752*3185) / (10*56910(752)2)

= 6.55

b = (n*X*YX*Y) / (n*X2(X)2 = (10*3185752*42.5) / (10*56910(752)2)

= -0.031

Equation of regression line

y = a + b*x

y = 6.55 - 0.031x

Now, we can find corresponding value of fitness for strength 71.

y = 6.55 - 0.031*71

= 4.349

Mean value of y = 4.25

SD = 0.3778

SE = 0.3778 / 10

= 0.1194

Z value for 98% confidence level = 2.33

Confidence interval = [mean - Z*SE, mean + Z*SE]

= [4.25 - 2.33*0.119, 4.25 + 2.33*0.119]

= [3.97, 4.52]

c.) Similarly as previous one. Here Z value for 90% level = 1.64

Prediciton interval = [mean - (Z*SD*( 1 + 1/n)), mean + (Z*SD*( 1 + 1/n))]

= [4.25 - (1.64*0.3778*( 1 + 1/10)), 4.25 - (1.64*0.3778*( 1 + 1/10))]

= [3.6, 4.89]

X Y X*Y X2 69 4.6 317.4 4761 69 4.5 310.5 4761 70 4.3 301 4900 70 4.9 343 4900 73 4.2 306.6 5329 76 4 304 5776 77 3.9 300.3 5929 77 3.7 284.9 5929 84 4.5 378 7056 87 3.9 339.3 7569 Sum = 752 Sum = 42.5 Sum = 3185 Sum = 56710

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