Question #280495

A study was conducted at Virginia Tech to de- termine if certain static arm-strength measures have an influence on the "dynamie lift" characteristics of an individual. Twenty-five individuals were subjected to strength tests and then were askod to perform a weight- lifting test in which weight was dynamically lifted over- head. The data are given here.


Dynamie Lift, y Arm Strength


17.3 19.3 19.5 19.7


Individual Z17 71.7 48,3 23 88.3 75.0 22.9 91.7 23.1 100.0 73.3 65.0 26.4 26.8 27.6 28.1 75.0 10 88.3 68.3 96.7 76.7 78.3 11 28.2 12 13 28.7 29.0 14 29.6 15 29.9 29,9 30.3 31.3 36.0 39.5 60.0 16 71.7 17 85.0 18 85.0 19 88.3 20 100.0 21 40.4 100.0 22 44.3 100.0 23 44.6 91.7 24 50.4 55.9 100.0 25 71.7


(a) Estimate o and di for the lincar regression curve HYje = Bo + Bz.


(b) Find a point estimate of pyY ja0


(c) Plot the residuals versus the r's (arm strength). Comment

1
Expert's answer
2021-12-17T10:43:50-0500

a)

Arm Strength, XX


17.3,19.3,19.5,19.7,22.9,23.1,26.4,17.3,19.3,19.5,19.7,22.9,23.1,26.4,

26.8,27.6,28.1,28.2,28.7,29.0,29.6,26.8,27.6,28.1,28.2,28.7,29.0,29.6,

29.9,29.9,30.3,31.3,36.0,39.5,40.4,29.9,29.9,30.3,31.3,36.0,39.5,40.4,

44.3,44.6,50.4,55.944.3,44.6,50.4,55.9

Dynamic Lift, YY


71.7,48.3,88.3,75.0,91.7,100.0,73.3,71.7,48.3,88.3,75.0,91.7,100.0,73.3,

65.0,75.0,88.3,68.3,96.7,76.7,78.3,65.0,75.0,88.3,68.3,96.7,76.7,78.3,

60.0,71.7,85.0,85.0,88.3,100.0,100.0,60.0,71.7,85.0,85.0,88.3,100.0,100.0,

100.0,91.7,100.0,71.7100.0,91.7,100.0,71.7


Xˉ=1niXi=778.725=31.148\bar{X}=\dfrac{1}{n}\sum _iX_i=\dfrac{778.7}{25}=31.148

Yˉ=1niYi=205025=82\bar{Y}=\dfrac{1}{n}\sum _iY_i=\dfrac{2050}{25}=82

SSXX=iXi21n(iXi)2SS_{XX}=\sum_iX_i^2-\dfrac{1}{n}(\sum _iX_i)^2

=26591.63(778.7)225=2336.6824=26591.63-\dfrac{(778.7)^2}{25}=2336.6824

SSYY=iYi21n(iYi)2SS_{YY}=\sum_iY_i^2-\dfrac{1}{n}(\sum _iY_i)^2

=172891.46(2050)225=4791.46=172891.46-\dfrac{(2050)^2}{25}=4791.46

SSXY=iXiYi1n(iXi)(iYi)SS_{XY}=\sum_iX_iY_i-\dfrac{1}{n}(\sum _iX_i)(\sum _iY_i)

=65164.04778.7(2050)25=1310.64=65164.04-\dfrac{778.7(2050)}{25}=1310.64

β1=SSXYSSXX=1310.642336.6824=0.5609\beta_1=\dfrac{SS_{XY}}{SS_{XX}}=\dfrac{1310.64}{2336.6824}=0.5609

β0=Yˉβ1Xˉ=821310.642336.6824(31.148)\beta_0=\bar{Y}-\beta_1\cdot\bar{X}=82-\dfrac{1310.64}{2336.6824}(31.148)

=64.5292=64.5292

The regression equation is:


μYX=64.5292+0.5609X\mu_{Y|X}=64.5292+0.5609X

b)


μY30=64.5292+0.5609(30)=81.3562\mu_{Y|30}=64.5292+0.5609(30)=81.3562

c)


r=SSXYSSXXSSYYr=\dfrac{SS_{XY}}{\sqrt{SS_{XX}}\sqrt{SS_{YY}}}

=1310.642336.68244791.46=0.3917=\dfrac{1310.64}{\sqrt{2336.6824}\sqrt{4791.46}}=0.3917

0.2<r<0.40.2<r<0.4

Positive weak correlation.


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