X 1.75 2.30 1.91 4.77 2.35 .44 1.43 2.25 .44 2.59 1.42 1.86 1.71 1.13 3.13 1.40

ID: 3229270 • Letter: X

Question

1.75
2.30
1.91
4.77
2.35
.44
1.43
2.25
.44
2.59
1.42
1.86
1.71
1.13
3.13
1.40
.66
1.50
.88

1.02
1.37
1.49
2.28
2.01
.44
.61
1.48
.42
1.93
.91
1.45
1.17
.76
2.13
1.03
.10
1.14
.95

Background: A placebo is an inert substance that contains nothing nutritionally or biochem-

ically related to a disease. Although placebos should have no effect on any health variable,

placebos often work well as a treatment for many maladies. We almost never know why or un-

der which circumstances a placebo helps patients. But bene.ts for placebo have been observed

so often that no one doubts that they occur.

One study reported data from 19 different experiments that compared an antidepressant drug

for treating severely depressed patients against a placebo, a pill the same size and color as the actual drug, but otherwise inert. In each study half the participants received an FDA certified antidepressant drug and half received the placebo. The dependent variable was improvement on the Hamilton Depression Rating Scale.

Y = Mean improvement for patients who received placebo

X = Mean improvement for patients who received active drug

For example, in Study 1 y = 1.02 means that patients who received the placebo improved on

average 1.02 points on the Hamilton, compared to x = 1.75 for those who received the drug. In

Study 1, the placebo produced worse results. The "individual" in these data is actually a group

of individuals in a study.

A completely unexpected finding in this investigation was that the magnitude of the placebo

effect was related to the magnitude of the effect of the drug. In particular the greater the e.ect

of the active drug, the greater the placebo effect. The investigators did not know why this

occurred. Nonetheless we can use the effect of the active drug to predict the placebo effect with

considerable accuracy.

1.For this question use the total sample of N = 19 cases. Suppose a new study is conducted

in which the e.ect of the antidepressant drug is x = 2. (i) Calculate the predicted bene.t

for the placebo e.ect at x = 2. (ii) The class notes present an approximate method for the

standard error of prediction. Report the standard error of prediction for x = 2. (iii) Report

the 99% prediction interval for x = 2.

Explanation / Answer

Solution

X = mean improvement for patients who received the drug.

Y = mean improvement for patients who received the placebo

Back-up Theory

Regression model Y = + X + , where is the error term, which is assumed to be Normally distributed with mean 0 and variance 2. Then,

Let (xi, yi) be a pair of sample observation on (X, Y), i= 1, 2, …., n, where n = sample size.

Then, Mean X = Xbar = (1/n)sum of xiover I = 1, 2, …., n; ……………….(1)

Variance of X, V(X) = (1/n)Sxx where Sxx

= sum of (xi – Xbar)2 over i = 1, 2, …., n ………………………………..(2)

Standard Deviation of X = SDX = sq.rt of V(X). ……………………………(3)

Similarly, Mean Y = Ybar =(1/n)sum of yiover i= 1, 2, …., n;…………….(4)

Variance of Y, V(Y) = (1/n)Syy where Syy

= sum of (yi – Ybar)2 over i = 1, 2, …., n ………………………………………………(5)

Standard Deviation of Y = SDY = sq.rt of V(Y). ………………………….……………….(6)

Covariance of X and Y, Cov(X, Y)

= (1/n)Sxy where Sxy = sum of {(xi – Xbar)(yi – Ybar)} over i = 1, 2, …., n………(7)

Now, to work out solution,

Part (i)

Estimated Regression of Y on X is given by: Ycap = a + bX, where

b = Cov(X, Y)/V(X) = Sxy/Sxx and a = Ybar – b.Xbar..…………………….(8)

Calculations

xbar

1.78526316

ybar

1.19421053

Sxx

18.8164737

Syy

6.62366316

Sxy

10.0223789

0.53263853

0.24331058

So, estimated Y, Ycap = 0.243 + 0.533X and hence the predicted benefit

for the placebo effect at x = 2 is 0.243 + (2 x 0.533) = 1.309 = 1.31 ANSWER

Part(ii)

The standard error of prediction = s[(1/n) + {(xi – Xbar)2/Sxx}] where s = standard error of the residual.

which is given by s2 = (Syy – b2Sxx)/(n - 2) = 0.075609 or s = 0.275 and hence

Standard error of prediction for x = 2 is 0.062 ANSWER

Part (iii)

100(1 - )% Confidence Interval (CI) for ycap at x = x0 is (a + bx0) ± tn – 2,/2xSE(Ycap)

So,99% prediction interval for x = 2 is: (1.122, 1.496) ANSWER