Question

This problem has several parts.

The table below shows the ages and bone densities for five women.

Age	40	42	47	51	50
Bone Density	340	335	319	319	310

Enter the slope and y-intercept of the regression line, using age as the independent variable. Round your answers to two decimal places.
	Slope:		Intercept:
Using your regression line (with rounded coefficients), predict the bone density of a 53-year-old woman to the nearest whole number.
	Estimate:
Did you just perform interpolation or extrapolation? A. interpolation B. extrapolation
Enter r for this data, each rounded to the nearest hundredth.
	Correlation:

Answer 1

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	230	1623	94	621.2	-226.00
mean	46.00	324.60	SSxx	SSyy	SSxy

sample size ,   n =   5          
here, x̅ = Σx / n=   46.00   ,     ȳ = Σy/n =   324.60  
                  
SSxx =    Σ(x-x̅)² =    94.0000          
SSxy=   Σ(x-x̅)(y-ȳ) =   -226.0          
                  
estimated slope , ß1 = SSxy/SSxx =   -226.0   /   94.000   =   -2.40
                  
intercept,   ß0 = y̅-ß1* x̄ =   435.20          
                  
so, regression line is   Ŷ =   435.20   +   -2.40   *x

......................

Predicted Y at X=   53   is                  
Ŷ =   435.19574   +   -2.404255   *   53   =   308
.................

A. interpolation

....................

correlation coefficient ,    r = Sxy/√(Sx.Sy) =   -0.94

.................

THANKS

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