Supplementary Exercise 10.18 of IPS7e: -------------------------------------- Linear regression model for Y = intra-arterial measurement of blood pressure x = oscillometric measurement of blood pressure for 81 prematurely born infants. Estimated regression line: yhat = 15 + 0.83*x. Standard error of the slope = 0.065. Test H0: beta1=0 against Ha: beta1>0 (positive correlation) t = (0.83-0)/0.065 = 12.77 DF = 81-2 = 79 P = P(t(79)>12.77) << 0.0005 We strongly reject the null hypothesis - there is clear evidence of a positive association between Y and X. Addition: Test H0: beta1=1 against Ha: beta1<>1, t = (0.83-1)/0.065 = -2.615 DF = 81-2 = 79 P = 2*P(t(79)>2.615) = 0.011 We reject the null hypothesis, there is some evidence that the slope differs from 1 (is less than 1). This means that there is a difference between the two scales used for measuring the blood pressure, the simple methods gives values systematically lower than the traditional one for high values of x (>88, see (*) below) whereas it gives higher values for small values of x (<88), due to the positive intercept. (*) technical addition: derivation of interval where x>y: The fitted line is y=15+0.83*x Solving the inequality y15/(1-0.83) or x>88 Second addition: The text of the exercise explains that the researchers computed prediction intervals and concluded the association between Y and x was not strong enough for clinical use. Our usual statistic for measuring strength of association is the correlation coefficient r, and we also use r^2 (R^2) as a measure of the predictive potential of a model. It is therefore of interest to estimate r. We can do this because the above test is the same as the test for H0: rho=0, and therefore 12.77 = t = r*sqrt((n-2)/1-r^2)), or t^2 = r^2 * (n-2) / (1-r^2), or t^2 = r^2 * (n-2 + t^2), or r^2 = t^2 / (n-2 + t^2) = 12.77^2 / (79 + 12.77^2) = 0.67, or r=0.82. It is not unusual for a correlation of 0.8 to be considered too weak for prediction, but that obviously also depends on the precision required by the prediction.