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Question:Consider the clinical trial of depression drugs in Example2.1.4. Suppose that a patient is selected at random from the 150 patients in that study and we recordY, an indicator of the treatment group for that patient, andX, an indicator of whether or not the patient relapsed. Table 3.3contains the joint p.f. ofXandY.

Response(X)

Treatment Group(Y)

Impramine(1)

Lithium(2)

Combination(3)

Placebo(4)

Relapse(0)

0.120

0.087

0.146

0.160

No relapse(1)

0.147

0.166

0.107

0.067

a. Calculate the probability that a patient selected at random from this study used Lithium (either alone or in combination with Imipramine) and did not relapse.

b. Calculate the probability that the patient had a relapse(without regard to the treatment group).

Short Answer

Expert verified
  1. The probability that a patient selected at random from this study used Lithium (either alone or in combination with Imipramine) and did not relapse is 0.273.
  2. The probability that the patient had a relapse (without regard to the treatment group) is 0.513.

Step by step solution

01

Given information

Referring to Example 2.1.4, a patient was selected randomly from 150 patients.

Y= An indicator of the treatment group for that patient.

X= An indicator of whether or not the patient relapsed.

The table shows the proportions of clinical depression studies that provide the proportions of responses regarding the treatments.

02

Calculate the probability

a.

Let us consider the joint probability function of the above table f(x,y),where x=0,1 and y=1,2,3,4.

The probability of a randomly selected patient from the study used Lithium alone or in combination with Imipramine and did not relapse is\(\left\{ {Y \in \left\{ {2,3} \right\}} \right\} \cap \left\{ {X = 1} \right\}\).

So, the probability is,

\(\begin{array}{c}f\left( {1,2} \right) + f\left( {1,3} \right) = 0.1666 + 0.107\\ = 0.273\end{array}\)

03

Calculate the probability

b.

The probability that the patient had relapsed without regard to treatment groups is

\(\begin{array}{c}\Pr \left( {X = 0} \right) = f\left( {0,1} \right) + f\left( {0,2} \right) + f\left( {0,3} \right) + f\left( {0,4} \right)\\ = 0.120 + 0.087 + 0.146 + 0.160\\ = 0.513\end{array}\)

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