Question

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The following data represent the weight of a child riding a bike and the rolling distance achieved after going down a hill without pedaling.  

Weight (lbs.)	Rolling Distance (m.)
59	26
83	43
97	49
56	20
103	65
87	44
88	48
91	42
52	39
63	33
71	39
100	49
89	55
103	53
99	42
74	33

Find the 95% prediction interval for rolling distance when a child riding the bike weighs 106 lbs. (round to 4 decimal places

Answer 1

	Weight (X)	Distance (Y)	X * Y	X2	Ŷ	Sxx =Σ (Xi - X̅ )2	Syy = Σ( Yi - Y̅ )2	Sxy = Σ (Xi - X̅ ) * (Yi - Y̅)
	59	26	1534	3481	30.366	537.660	272.250	382.594
	83	43	3569	6889	42.925	0.660	0.250	0.406
	97	49	4753	9409	50.251	219.410	42.250	96.281
	56	20	1120	3136	28.796	685.785	506.250	589.219
	103	65	6695	10609	53.391	433.160	506.250	468.281
	87	44	3828	7569	45.018	23.160	2.250	7.219
	88	48	4224	7744	45.542	33.785	30.250	31.969
	91	42	3822	8281	47.112	77.660	0.250	-4.406
	52	39	2028	2704	26.703	911.285	12.250	105.656
	63	33	2079	3969	32.459	368.160	90.250	182.281
	71	39	2769	5041	36.646	125.160	12.250	39.156
	100	49	4900	10000	51.821	317.285	42.250	115.781
	89	55	4895	7921	46.065	46.410	156.250	85.156
	103	53	5459	10609	53.391	433.160	110.250	218.531
	99	42	4158	9801	51.298	282.660	0.250	-8.406
	74	33	2442	5476	38.216	67.035	90.250	77.781
Total	1315	680	58275	112639	680.000	4562.438	1874.000	2387.500

Equation of regression line is Ŷ = a + bX
b = ( n Σ(XY) - (ΣX* ΣY) ) / ( n Σ X2 - (ΣX)2 )
b = ( 16 * 58275 - 1315 * 680 ) / ( 16 * 112639 - ( 1315 )2)
b = 0.5233

a =( ΣY - ( b * ΣX ) ) / n
a =( 680 - ( 0.5233 * 1315 ) ) / 16
a = -0.5083
Equation of regression line becomes Ŷ = -0.5083 + 0.5233 X

X̅ = Σ (Xi / n ) = 1315/16 = 82.1875
Y̅ = Σ (Yi / n ) = 680/16 = 42.5

Estimated Error Variance (σ̂2) =
S2 = ( 1874 - 0.5233 * 2387.5 ) / 16 - 2
S2 = 44.6158
S = 6.6795

Predictive Confidence Interval of
Ŷ = -0.5083 + 0.5233 X
Ŷ = 54.9615

t(150.5333/2) = t(0.05/2) = 2.145
X̅ = (Xi / n ) = 1315/16 = 82.1875
= 54.9615

95% Predictive confidence interval is ( 39.3532 < < 70.5698 ).