The table below gives the age and bone density for five randomly selected women.

Question

The table below gives the age and bone density for five randomly selected women. Using this data, consider the equation of the regression line, yˆ=b0+b1x y ^ = b 0 + b 1 x , for predicting a woman's bone density based on her age. Keep in mind, the correlation coefficient may or may not be statistically significant for the data given. Remember, in practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant.

Step 1 of 6: Find the estimated slope. Round your answer to three decimal places.

Step 2 of 6: Find the estimated y-intercept. Round your answer to three decimal places.

Step 3 of 6: Determine if the statement "Not all points predicted by the linear model fall on the same line" is true or false.

Step 4 of 6: Find the estimated value of y when x=53x=53. Round your answer to three decimal places.

Step 5 of 6: Substitute the values you found in steps 1 and 2 into the equation for the regression line to find the estimated linear model. According to this model, if the value of the independent variable is increased by one unit, then find the change in the dependent variable yˆy^.

Step 6 of 6: Find the value of the coefficient of determination. Round your answer to three decimal places.

Age 40 45 53 57 63 Bone Density 350 348 341 327 324

Explanation / Answer

calculation procedure for regression
mean of X = X / n = 51.6
mean of Y = Y / n = 338
(Xi - Mean)^2 = 339.2
(Yi - Mean)^2 = 570
(Xi-Mean)*(Yi-Mean) = -420
b1 = (Xi-Mean)*(Yi-Mean) / (Xi - Mean)^2
= -420 / 339.2
= -1.238
bo = Y / n - b1 * X / n
bo = 338 - -1.238*51.6 = 401.892
value of regression equation is, Y = bo + b1 X
Y'=401.892-1.238* X

1.
estimated slope = -1.238
2.
the estimated y-intercept = 401.892
3.
true,
Not all points predicted by the linear model fall on the same line
4.
estimated value of y, when x=53
Y' = 401.892 -(1.238*53) =336.278

5.
if the value of the independent variable is increased by one unit then y value
Y'=401.892-1.238* X
now x value = 53+1 =54
Y' = 401.892-1.238*54 =335.04

6.

calculation procedure for correlation

sum of (x) = x = 258

sum of (y) = y = 1690

sum of (x^2)= x^2 = 13652

sum of (y^2)= y^2 = 571790

sum of (x*y)= x*y = 86784

to caluclate value of r( x,y) = covariance ( x,y ) / sd (x) * sd (y)

covariance ( x,y ) = [ x*y - N *(x/N) * (y/N) ]/n-1

= 86784 - [ 5 * (258/5) * (1690/5) ]/5- 1

= -84

and now to calculate r( x,y) = -84/ (SQRT(1/5*86784-(1/5*258)^2) ) * ( SQRT(1/5*86784-(1/5*1690)^2)

=-84 / (8.237*10.677)

=-0.955

value of correlation is =-0.955

coeffcient of determination = r^2 = 0.912

properties of correlation

1. If r = 1 Corrlation is called Perfect Positive Corrlelation

2. If r = -1 Correlation is called Perfect Negative Correlation

3. If r = 0 Correlation is called Zero Correlation

& with above we conclude that correlation ( r ) is = -0.9551< 0, perfect nagative correlation

Line of Regression Y on X i.e Y = bo + b1 X X Y (Xi - Mean)^2 (Yi - Mean)^2 (Xi-Mean)*(Yi-Mean) 40 350 134.56 144 -139.2 45 348 43.56 100 -66 53 341 1.96 9 4.2 57 327 29.16 121 -59.4 63 324 129.96 196 -159.6

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