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Chapter 19: Problem 747

Suppose we wish to test whether a new drug aids the recovery of patients suffering from a mental illness when it is known that some patients recover spontaneously without any treatment at all. Starting with \(\mathrm{m}+\mathrm{n}\) patients, we create two groups. One group of \(\mathrm{m}\) patients gets no treatment and thus acts as a control group for the treatment group, which contains n patients. At the end of the experiment, the numbers of patients who recovered are \(j\) and \(k\). If in fact that drug has no effect, then it is as if there were \(j+k\) among the \(n+m\) who were bound to recover anyway. Find the probability that \(\mathrm{k}\) recover in the treatment group only by virtue of drawing a random sample of n from \(\mathrm{n}+\mathrm{m}\).

Short Answer

Expert verified
The probability of k patients recovering in the treatment group randomly when the drug has no effect can be calculated using the following equation: \[P = \frac{\binom{m}{j} \binom{n}{k}}{\binom{m + n}{j + k}}\]

Step by step solution

01

Identify the total possible outcomes

As there are a total of m + n patients and j + k recoveries, we need to find the total possible ways to allocate the j + k recoveries among the m + n patients, which can be done using combinations. In this case, we will calculate the combinations as: \[\binom{m + n}{j + k}\]
02

Identify the favorable outcomes

We want to find the probability that k recoveries occur in the treatment group due to random chance. Thus, we need to find the number of ways in which j recoveries can be distributed among m patients in the control group and k recoveries can be distributed among n patients in the treatment group. This can be found using the product of combinations as follows: \[\binom{m}{j} \binom{n}{k}\]
03

Calculate the probability

Finally, we can calculate the probability of k recoveries happening in the treatment group by random chance using the following equation: \[P = \frac{\textrm{Favorable outcomes}}{\textrm{Total possible outcomes}}\] Plugging in values calculated in the previous steps: \[P = \frac{\binom{m}{j} \binom{n}{k}}{\binom{m + n}{j + k}}\] This gives us the probability of k patients recovering in the treatment group randomly when the drug has no effect.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Control Group
A control group is a fundamental part of any scientific study that aims to demonstrate the efficacy of a treatment. It consists of subjects who do not receive the experimental treatment, allowing for a comparison between treated and untreated subjects.
  • This group remains untreated to ensure that any changes observed in the experimental group can be attributed to the treatment itself, rather than external factors or spontaneous recovery.
  • For example, in our exercise, the control group consists of \( m \) patients who do not receive the drug. This establishes a baseline for natural recovery rates without the influence of the treatment.
Control groups help eliminate biases and provide reliable data for hypothesis testing, ensuring that the results of the experiment are due to the treatment and not a placebo effect.
Treatment Group
The treatment group is the set of participants who receive the experimental variable, which, in this case, is the new drug. By observing this group, researchers assess the effectiveness of the treatment compared to the control group.
  • This group consists of \( n \) patients who are administered the drug to see if there is a significant difference in recovery compared to those who did not receive the treatment.
  • Determining the number of recoveries in this group, \( k \), helps in evaluating the actual impact of the drug compared to those recovering spontaneously.
The distinction between control and treatment groups is essential for identifying the genuine efficacy of the treatment and for understanding its potential benefits or harms.
Combinatorics
Combinatorics is a branch of mathematics that involves counting, arranging, and structuring various elements within a set. It's crucial in calculating the ways in which elements can be grouped, arranged, or selected.
In the context of our exercise, combinatorics allows us to calculate the different ways recoveries can be distributed among patients. We use combinations to determine:
  • The total ways to allocate \( j+k \) recoveries among \( m+n \) patients, calculated as \( \binom{m+n}{j+k} \).
  • The different ways \( j \) recoveries could be assigned to \( m \) control group patients, given by \( \binom{m}{j} \).
  • The ways \( k \) recoveries could be assigned to \( n \) treatment group patients, given by \( \binom{n}{k} \).
Combinatorics provides the mathematical framework for determining probabilities by counting and arranging these possibilities.
Probability
Probability measures how likely an event is to occur within a given set of possible outcomes. In hypothesis testing, probability helps in assessing whether the observed data can be attributed to random chance instead of the tested treatment.
To solve our exercise problem, we calculate:
  • The probability \( P \) that \( k \) recoveries occur in the treatment group due to random chance, represented by the ratio \( \frac{\binom{m}{j}\binom{n}{k}}{\binom{m+n}{j+k}} \).
  • This fraction shows us how likely it is to observe \( k \) recoveries among the treated patients just by randomly drawing them from all patients expected to recover, without any drug effect.
Understanding probability in this context aids in making informed conclusions about the effectiveness of the drug and the role of chance in recovery rates.

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Most popular questions from this chapter

What is the relationship between the \(\mathrm{X}^{2}\) statistic and sample size given that everything else remains constant?

Suppose that a sales region has been divided into five territories, each of which was judged to have an equal sales potential. The actual sales volume for several sampled days is indicated below. \begin{tabular}{|c|c|c|c|c|c|} \hline & \multicolumn{5}{|c|} { T e r r i t o r y } \\ \hline & A & B & C & D & E \\ \hline Actual Sales & 110 & 130 & 70 & 90 & 100 \\ \hline \end{tabular} What are the expected sales for each area? How many degrees of freedom does the chi-square statistic have? Compute \(\mathrm{X}^{2}\) and use the table of \(\mathrm{X}^{2}\) values to test the significance of the difference between the observed and expected levels of sales.

To find out whether there is a significant association between the sensitivity of the skin to sunlight and the color of a person's eyes, we could take a random sample of, say, 100 people, and test them all with a standard dose of ultra violet rays, noting both their reaction to this test and their eye color. Both of these features are noted for each person in the trial, so the observations are matched. The results could then be tabulated in an association table like this \begin{tabular}{|l|l|l|l|l|l|} \hline \multicolumn{2}{|l|} {} & \multicolumn{3}{|c|} { Ultra-violet reaction } & \multirow{2}{*} {\(\mathrm{R}\)} \\ \cline { 3 - 5 } \multicolumn{2}{|l|} {} & \(++\) & \(+\) & \(-\) & \\ \hline \multirow{2}{*} { Eye Color } & Blue & 19 & 27 & 4 & 50 \\ \cline { 2 - 6 } & Grey or green & 7 & 8 & 5 & 20 \\ \cline { 2 - 6 } & Brown & 1 & 13 & 16 & 30 \\ \hline C & & 27 & 48 & 25 & \(100=\mathrm{N}\) \\ \hline \end{tabular} Is this relationship significant? What are some measures of association which describe the strength of association? How strongly associated are these variables?

When you use a \(\mathrm{X}^{2}\) test to reject a null hypothesis, you must always look back at the data to make sure that they support your alternative. You have taken a census of the insect population of your rose garden. On the basis of several large samples you conclude that the insect population is distributed as follows: \(\begin{array}{llll}\text { Ladybugs } & 20 \% & \text { Inch worms } 20 \% & \text { Weevils }\end{array}\) \(30 \%\) Aphids \(\quad 25 \%\) Brown spiders \(5 \%\) Now you treat your garden with an insecticide that is supposed to control the undesirable weevils, aphids, and brown spiders without affecting ladybugs or inch worms. To check the effect of the insecticide you collect 150 insects at random. Your sample is composed as follows: \(\begin{array}{llllll}\text { Ladybugs } & 25 & \text { Ladybugs } & 25 & \text { Inch worms } 45\end{array}\) \(\begin{array}{lllll}\text { Weevils } & 45 & \text { Aphids } & 25 & \text { Brown spiders } 10\end{array}\) Use the census to determine the predicted frequencies. Compute \(\mathrm{X}^{2}\) and answer these two questions: (a) Has the distribution of insects changed significantly (at the \(5 \%\) level 1 ? (b) Has the insecticide had the intended effect?

Let the result of a random experiment be classified by two attributes. Let these two attributes be eye color and hair color. One attribute of the outcome, eye color, can be divided into certain mutually exclusive and exhaustive events. Let these events be: \(\mathrm{A}_{1}:\) Blue eyes \(\mathrm{A}_{2}:\) Brown eyes \(\mathrm{A}_{3}:\) Grey eyes \(\mathrm{A}_{4}:\) Black eyes \(\quad \mathrm{A}_{5}:\) Green eyes The other attribute of the outcome can also be divided into certain mutually exclusive and exhaustive events. Let these events be: \(\mathrm{B}_{1}:\) Blond hair \(\mathrm{B}_{2}:\) Brown hair \(\quad \mathrm{B}_{3}:\) Black hair \(\mathrm{B}_{4}:\) Red hair. The experiment is performed by observing \(\mathrm{n}=500\) people and each of them are categorized according to eye color and hair color. Let \(\mathrm{A}_{\mathrm{i}} \cap \mathrm{B}_{\mathrm{j}}\) be the event that a person with eye color \(\mathrm{A}_{\mathrm{i}}\) and hair color \(\mathrm{B}_{\mathrm{i}}\) is observed, \(\mathrm{i}=1,2,3,4,5\) and \(j=1,2,3,4\). Let \(X_{i j}\) be the observed frequency of event \(\mathrm{A}_{\mathrm{i}} \cap \mathrm{B}_{\mathrm{j}}\). Test the hypothesis that \(\mathrm{A}_{\mathrm{i}}\) and \(\mathrm{B}_{\mathrm{j}}\) are independent attributes.
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