In an experiment designed to describe the dose-response curve for vitamin K, ind

Question

In an experiment designed to describe the dose-response curve for vitamin K, individual rats were depleted of their vitamin K reserves and then fed dried liver for 4 days at different dosage levels. The response variable (for each rat) was measured as the concentration of a clotting agent needed to clot a sample of its blood in 3 minutes. More precisely, the researcher recorded the natural log of concentration (Y = concentration) and the natural log of dose (X = dose) for 12 rats. The least-squares estimate of the regression line was mu{ln Y | ln X) = 6.760 - 1.785(ln X) (a) A 95% confidence interval for the slope of the regression line is (-1.902, -1.668). Use this interval to calculate a 95% confidence interval for the effect of increasing dosage by 50% on the median response (on the original scale). (b) A 95% prediction interval for In Y at value of X = 10 is (2.343, 2.957). Calculate and interpret the 95% prediction interval for concentration when dose is 10 on the original scale.

Explanation / Answer

a) a 95 % confidence interval for the slope of the regression line (e-1.902,e-1.668)=(0.149270,0.188624) on the origibal scale.

b) a 95% prediction interval for ln Y at value X=10 is (2.343,2.957)

So,  95% prediction interval for Y at value X=10 is (e2.343,e2.957)=(10.4124,19.2402)

So, when X=10, the predicted value on the original scale will be between (10.4124,19.2402) with probability 0.95

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