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 76 69 71 76 83 72 78 74 80 82 y, Fineness 4.3 4.5 4.7 4.0 3.9 4.0 5.0 4.7 4.1 4.5 (a) Draw a scatter diagram. (Do this on paper. Your instructor may ask you to turn in this work.) (b) Find the 95% confidence interval for the mean measurement of fineness for fibers with a strength of 74. Lower Limit Upper Limit (c) Find the 98% prediction interval for an individual measurement of fineness for fibers with a strength of 74. Lower Limit Upper Limit

Answer 1

	x	y	x^2	y^2	xy
1	76	4.3	5776	18.49	326.8
2	69	4.5	4761	20.25	310.5
3	71	4.7	5041	22.09	333.7
4	76	4	5776	16	304
5	83	3.9	6889	15.21	323.7
6	72	4	5184	16	288
7	78	5	6084	25	390
8	74	4.7	5476	22.09	347.8
9	80	4.1	6400	16.81	328
10	82	4.5	6724	20.25	369
Total	761	43.7	58111	192.19	3321.5
Mean	76.1	4.37
Sxx	198.9
Syy	1.221
Sxy	-4.07

Using the formula above

=-4.07/198.9

= -0.02

= 4.37 - (-0.02) * 76.1

= 5.927

Therefore the regression equation is

= 5927 - 0.02x

Error variance =

=0.1422

Now we have all the required values. To find the prediction and confidence interval we find the predicted (mean y and y). This is found by sub values of 'x' in the regression equation

Sxx =198.9

=0.1422

			variance		df (n -2)	c..v.	Lower limit	Upper limit
Mean	74	4.413	0.0174	0.05 (1-0.95)	8	2.7515	4.0503	4.7757
Predicted	74	4.413	0.1596	0.02  (1-0.98)	8	3.3554	3.0725	5.7534

The critical value is found using t-dist tables with df and /2