Answer to Question #280495 in Statistics and Probability for Mayyy

Question

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

Answer 1

a)

Arm Strength, $X$

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

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

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

$$44.3,44.6,50.4,55.9$$

Dynamic Lift, $Y$

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

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

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

$$100.0,91.7,100.0,71.7$$

$$\bar{X}=\dfrac{1}{n}\sum _iX_i=\dfrac{778.7}{25}=31.148$$

$$\bar{Y}=\dfrac{1}{n}\sum _iY_i=\dfrac{2050}{25}=82$$

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

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

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

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

$$SS_{XY}=\sum_iX_iY_i-\dfrac{1}{n}(\sum _iX_i)(\sum _iY_i)$$

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

$$\beta_1=\dfrac{SS_{XY}}{SS_{XX}}=\dfrac{1310.64}{2336.6824}=0.5609$$

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

$$=64.5292$$

The regression equation is:

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

b)

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

c)

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

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

$$0.2<r<0.4$$

Positive weak correlation.