/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 82 A medical research team wishes t... [FREE SOLUTION] | 91Ó°ÊÓ

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A medical research team wishes to evaluate two different treatments for a disease. Subjects are selected two at a time, and then one of the pair is assigned to each of the two treatments. The treatments are applied, and each is either a success (S) or a failure (F). The researchers keep track of the total number of successes for each treatment. They plan to continue the chance experiment until the number of successes for one treatment exceeds the number of successes for the other treatment by 2 . For example, they might observe the results in the table for Exercise \(6.82\) given below. The chance experiment would stop after the sixth pair, because Treatment 1 has 2 more successes than Treatment \(2 .\) The researchers would conclude that Treatment 1 is preferable to Treatment \(2 .\) Suppose that Treatment 1 has a success rate of \(.7\) (that is, \(P(\) success \()=.7\) for Treatment 1\()\) and that Treatment 2 has a success rate of \(.4 .\) Use simulation to estimate the probabilities in Parts (a) and (b). (Hint: Use a pair of random digits to simulate one pair of subjects. Let the first digit represent Treatment 1 and use \(1-7\) as an indication of a success and 8,9 , and 0 to indicate a failure. Let the second digit represent Treatment 2, with 1-4 representing a success. For example, if the two digits selected to represent a pair were 8 and 3 , you would record failure for Treatment 1 and success for Treatment 2\. Continue to select pairs, keeping track of the total number of successes for each treatment. Stop the trial as soon as the number of successes for one treatment exceeds that for the other by \(2 .\) This would complete one trial. Now repeat this whole process until you have results for at least 20 trials [more is better]. Finally, use the simulation results to estimate the desired probabilities.) a. Estimate the probability that more than five pairs must be treated before a conclusion can be reached. (Hint: \(P(\) more than 5\()=1-P(5\) or fewer \() .\) ) b. Estimate the probability that the researchers will incorrectly conclude that Treatment 2 is the better treatment.

Short Answer

Expert verified
The two probabilities required in the exercise depend on random digits chosen for each trial. Therefore, actual values cannot be provided without the digit inputs for the simulation. In general, for any such exercise, the more trials conducted, the more accurate the expected probabilities.

Step by step solution

01

Understanding the Simulation

Each digit 1-7 represents a success in Treatment 1 with a probability of 0.7 and each digit 1-4 represents a success in Treatment 2 with a probability of 0.4. Thus, establish a table to record the results with three columns: Pair, Treatment 1, and Treatment 2. This is where results for each simulation trial will be recorded.
02

Run the Simulation

Randomly select two digits to simulate one pair of subjects in each round, where the first digit represents Treatment 1 and the second represents Treatment 2. Record 'S' for success and 'F' for failure corresponding to each Treatment in the table accordingly. Repeat this process until the number of successes for one treatment exceeds that of the other by 2. This completes one trial. Repeat this whole process for at least 20 trials. The more trials conducted, the more accurate the result, as per the Law of Large Numbers.
03

Estimate Probability for Part (a)

After all the trials, count in how many trials more than five pairs had to be treated before a conclusion could be reached. The probability is estimated as \( P(\) more than 5 pairs\() = \frac{\text{No. of trials with more than 5 pairs}}{\text{Total no. of trials}} \). Alternatively, this could be calculated as 1 minus the probability of 5 or fewer pairs.
04

Estimate Probability for Part (b)

Similarly, count how many times the researchers incorrectly conclude that Treatment 2 is the better treatment. The probability estimate would be \( P(\) Treatment 2 better\() = \frac{\text{No. of trials where Treatment 2 is better}}{\text{Total no. of trials}} \).

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Key Concepts

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

Probability Estimation
Probability estimation is a fundamental concept in statistics where we determine the likelihood of a particular outcome occurring. In the context of our medical research scenario, it involves estimating the chances that one of two treatments will be deemed superior based on their success rates. This process requires a rigorous approach wherein we set up a simulation to mimic the likely outcomes of a real-world experiment.

To estimate the probabilities in the medical study, we use a random digit method. Each digit from 1-7 is mapped to success for Treatment 1, reflecting its success rate of 0.7. Similarly, digits 1-4 represent success for Treatment 2, given its success rate of 0.4. This mapping is key because it allows the simulation to closely replicate the true probability of each treatment's success in an actual trial. By running the simulation multiple times—ideally more than 20 trials—we gather enough data to calculate a reliable estimation of the probability for each treatment's outcome.
Experimental Design
Experimental design refers to planning how to conduct an experiment and gather data in a way that leads to reliable and interpretable results. In the medical study example, the design involves selecting subject pairs and administering the two treatments, recording successes and failures until one treatment's success exceeds the other's by 2.

To translate this into a simulation, we use a structured approach, random digit selection to represent treatment success or failure. The use of this method allows an accurate reflection of the real-world randomness and variability that would be present in a true medical trial. Keeping track of outcomes in a systematically designed table is crucial as it ensures clarity and ease of analysis once the simulation is complete. An effectively designed experiment, simulated or otherwise, is the backbone of quality data collection and consequential analysis, leading to sound conclusions.
Success Rate Analysis
Success rate analysis involves examining the outcomes of treatments or trials to determine which one is more effective. In the context of our simulation, success rate analysis is conducted by comparing the accumulated successes of the two treatments after each simulated trial. By tracking these results, researchers can evaluate which treatment consistently performs better.

The analysis comes down to numerically summarizing the trials, which involves counting the number of times each treatment outcome occurs under simulated conditions. Assessing the success rate gives us an objective measure to compare treatments, which in this case, could lead to identifying the preferred treatment for the disease. Moreover, the analysis includes identifying scenarios where the researchers may incorrectly conclude the effectiveness of a treatment, which is an essential aspect of error evaluation in clinical research.

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Most popular questions from this chapter

Five hundred first-year students at a state university were classified according to both high school GPA and whether they were on academic probation at the end of their first semester. The data are $$ \begin{array}{lcccc} && {\text { High School GPA }} \\ & 2.5 \text { to } & 3.0 \text { to } & 3.5 \text { and } & \\ \text { Probation } & <3.0 & <3.5 & \text { Above } & \text { Total } \\ \hline \text { Yes } & 50 & 55 & 30 & 135 \\ \text { No } & 45 & 135 & 185 & 365 \\ \text { Total } & 95 & 190 & 215 & 500 \\ \hline \end{array} $$ a. Construct a table of the estimated probabilities for each GPA-probation combination. b. Use the table constructed in Part (a) to approximate the probability that a randomly selected first-year student at this university will be on academic probation at the end of the first semester. c. What is the estimated probability that a randomly selected first-year student at this university had a high school GPA of \(3.5\) or above? d. Are the two outcomes selected student has a bigh school GPA of \(3.5\) or above and selected student is on academic probation at the end of the first semester independent outcomes? How can you tell? e. Estimate the proportion of first-year students with high school GPAs between \(2.5\) and \(3.0\) who are on academic probation at the end of the first semester. f. Estimate the proportion of those first-year students with high school GPAs \(3.5\) and above who are on academic probation at the end of the first semester.

Four students must work together on a group project. They decide that each will take responsibility for a particular part of the project, as follows: Because of the way the tasks have been divided, one student must finish before the next student can begin work. To ensure that the project is completed on time, a schedule is established, with a deadline for each team member. If any one of the team members is late, the timely completion of the project is jeopardized. Assume the following probabilities: 1\. The probability that Maria completes her part on time is \(.8\). 2\. If Maria completes her part on time, the probability that Alex completes on time is \(.9\), but if Maria is late, the probability that Alex completes on time is only \(.6 .\) 3\. If Alex completes his part on time, the probability that Juan completes on time is \(.8\), but if Alex is late, the probability that Juan completes on time is only \(.5 .\) 4\. If Juan completes his part on time, the probability that Jacob completes on time is \(.9\), but if Juan is late, the probability that Jacob completes on time is only .7. Use simulation (with at least 20 trials) to estimate the probability that the project is completed on time. Think carefully about this one. For example, you might use a random digit to represent each part of the project (four in all). For the first digit (Maria's part), \(1-8\) could represent on time and 9 and 0 could represent late. Depending on what happened with Maria (late or on time), you would then look at the digit representing Alex's part. If Maria was on time, \(1-9\) would represent on time for Alex, but if Maria was late, only \(1-6\) would represent on time. The parts for Juan and Jacob could be handled similarly.

A theater complex is currently showing four \(\mathrm{R}\) rated movies, three \(\mathrm{PG}-13\) movies, two \(\mathrm{PG}\) movies, and one G movie. The following table gives the number of people at the first showing of each movie on a certain Saturday: $$ \begin{array}{ccc} \text { Theater } & \text { Rating } & \begin{array}{c} \text { Number of } \\ \text { Viewers } \end{array} \\ \hline 1 & \mathrm{R} & 600 \\ 2 & \mathrm{PG}-13 & 420 \\ 3 & \mathrm{PG}-13 & 323 \\ 4 & \mathrm{R} & 196 \\ 5 & \mathrm{G} & 254 \\ 6 & \mathrm{PG} & 179 \\ 7 & \mathrm{PG}-13 & 114 \\ 8 & \mathrm{R} & 205 \\ 9 & \mathrm{R} & 139 \\ 10 & \mathrm{PG} & 87 \\ \hline \end{array} $$ Suppose that a single one of these viewers is randomly selected. a. What is the probability that the selected individual saw a PG movie? b. What is the probability that the selected individual saw a \(\mathrm{PG}\) or a \(\mathrm{PG}-13\) movie? c. What is the probability that the selected individual did not see an \(\mathrm{R}\) movie?

Consider the following information about travelers on vacation: \(40 \%\) check work e-mail, \(30 \%\) use a cell phone to stay connected to work, \(25 \%\) bring a laptop with them on vacation, \(23 \%\) both check work e-mail and use a cell phone to stay connected, and \(51 \%\) neither check work e-mail nor use a cell phone to stay connected nor bring a laptop. In addition \(88 \%\) of those who bring a laptop also check work e-mail and \(70 \%\) of those who use a cell phone to stay connected also bring a laptop. With \(E=\) event that a traveler on vacation checks work e-mail, \(C=\) event that a traveler on vacation uses a cell phone to stay connected, and \(L=\) event that a traveler on vacation brought a laptop, use the given information to determine the following probabilities. A Venn diagram may help. a. \(P(E)\) b. \(P(C)\) c. \(P(L)\) d. \(P(E\) and \(C)\) e. \(P\left(E^{C}\right.\) and \(C^{C}\) and \(\left.L^{C}\right)\) f. \(P(E\) and \(\operatorname{Cor} L)\) g. \(\quad P(E \mid L)\) h. \(P(L \mid C)\) i. \(P(E\) and \(C\) and \(L)\) j. \(P(E\) and \(L)\) k. \(P(C\) and \(L)\) 1\. \(P(C \mid E\) and \(L)\)

The article "Birth Beats Long Odds for Leap Year Mom, Baby" (San Luis Obispo Tribune, March 2 , 1996) reported that a leap year baby (someone born on February 29) became a leap year mom when she gave birth to a baby on February 29,1996 . The article stated that a hospital spokesperson said that only about 1 in \(2.1\) million births is a leap year baby born to a leap year mom (a probability of approximately .00000047). a. In computing the given probability, the hospital spokesperson used the fact that a leap day occurs only once in 1461 days. Write a few sentences explaining how the hospital spokesperson computed the stated probability. b. To compute the stated probability, the hospital spokesperson had to assume that the birth was equally likely to occur on any of the 1461 days in a four- year period. Do you think that this is a reasonable assumption? Explain. c. Based on your answer to Part (b), do you think that the probability given by the hospital spokesperson is too small, about right, or too large? Explain.

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