an article in a medical journal presents data relating ulna

an article in a medical journal presents data relating ulna length(x) to height(y). There were fourteen data points with ulna lengths ranging from 22.5cm to 28.2 and corresponding heights ranging from 158cm to 174cm. Regression analysis resulted in the following summary statistics:the sum of Xi=346.1; the same of Yi=2310;Sxx=36.463571;Sxy=137.60;Syy=626.00 (a)Determine the equation of the regression line (b) Calculate r^2 (c) Calculate a 95% confidence interval for the slope of the regression line (d)Calculate a 95% prediction interval for the height of an individual with a 25.0cm ulna

Solution

Given n = 14       X_i = 346.1     Y_i = 2310   SS_xx = 36.463571   SS_xy = 137.60    SS_yy =626

X = X_i / n = 346.1/ 14 = 24.7214

Y = Y_i / n = 2310/14 = 165

Y= a+ bx

b=SS_xy/SS_xx = 137.60/36.463571 = 3.7736

a= Y   -   b X = 165 - (3.7736)(24.7214) = 71.7113

r² = b(SS_xy)/SS_yy   = (3.7736)(137.60)/626)   = 0.8295

95% confident interval for the slop of the line

Standard deviation of error is

S_e = ((SS_yy) – b(SS_xy))/n-2

       (626 – 3.7736(137.60))/14-2

     8.8961

=2.9826

S_b = S_e/SS_xx = 2.9826/36.463571 = 0.4939

The table value for t_tab at 5% level of significance is 1.782 for right tail test at 12 degree of freedom

The 95% confident interval for slop is b±t(S_b)

3.7736 ±   (1.782)(0.4939)

[2.8935 , 4.6537]