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Are people more confident in their answers when the answer is actually correct than when it is not? The article "Female Students Less Confident, More Accurate Than Male Counterparts" described a study that measured medical students" confidence and the accuracy of their responses. Participants categorized their confidence levels using either "sure," "feeling lucky," or "no clue" Define the following events: \(C=\) event that a response is conect \(S=\) event that confidence level is "sure" \(L=\) event that confidence level is "feeling lucky" \(N=\) event that confidence level is "no clue" a. Data from the article were used to estimate the following probabilities for males: $$ \begin{array}{rrr} P(S)=0.442 & P(L)=0.422 & P(N)=0.136 \\ P(C \mid S)=0.783 & P(C \mid L)=0.498 & P(C \mid N)=0.320 \end{array} $$ Use the given probabilities to construct a hypothetical 1000 table with rows corresponding to confidence level and columns comesponding to whether the response was correct or not. b. Calculate the probability that a male student's confidence level is "sure" given that the response is correct. c. Calculate the probability that a male student's confidence level is "no clue" given that the response is incorrect. d. Calculate the probability that a male student's response is correct. e. Data from the article were also used to estimate the following probabilities for females: $$ \begin{array}{rrr} P(S)=0.395 & P(L)=0.444 & P(N)=0.161 \\ P(C 1 S)=0.805 & P(C \mid L)=0.535 & P(C \mid N)=0.320 \end{array} $$ Use the given probabilities to construct a hypothetical 1000 table with rows corresponding to confidence level and columns corresponding to whether the response was comect or not. f. Calculate the probability that a female student's confidence level is "sure" given that the response is correct. g. Calculate the probability that a female student's confidence level is "no clue" given that the response is incorrect. h. Calculate the probability that a female student's response is correct. i. Do the given probabilities and the probabilities that you calculated support the statement in the title of the article? Explain.

Short Answer

Expert verified
In summary, to analyze if the title "Female Students Less Confident, More Accurate Than Male Counterparts" is supported by the given data, compare the probabilities calculated for males and females. Specifically, compare their confidence levels and accuracies using conditional probabilities. For example, compare \(P(S|C)\), \(P(N|I)\), and \(P(C)\) for both genders. If female students have lower confidence levels (lower \(P(S|C)\) and higher \(P(N|I)\)) but higher accuracy (higher \(P(C)\)), the statement is backed by the data.

Step by step solution

01

Calculate Correct and Incorrect Probabilities

We will first calculate the probabilities of each event (correct and incorrect) given the confidence levels S, L, and N. For males: Using the given data, $$ \begin{aligned} P(C | S) &= 0.783 \\ P(C | L) &= 0.498 \\ P(C | N) &= 0.320 \end{aligned} $$ We can calculate the probability of incorrect responses given the confidence levels (I) using the fact that the total probability is 1: $$ \begin{aligned} P(I | S) &= 1 - P(C | S) = 0.217 \\ P(I | L) &= 1 - P(C | L) = 0.502 \\ P(I | N) &= 1 - P(C | N) = 0.680 \end{aligned} $$
02

Construct Hypothetical Table

We will now use these probabilities to construct the table for 1000 male students. $$ \begin{array}{c|c|c|c} Confidence Level & Correct & Incorrect & Total \\ \hline Sure & 0.442 \times 0.783 \times 1000 & 0.442 \times 0.217 \times 1000 & 0.442 \times 1000 \\ \hline Feeling\ Lucky & 0.422 \times 0.498 \times 1000 & 0.422 \times 0.502 \times 1000 & 0.422 \times 1000 \\ \hline No\ Clue & 0.136 \times 0.320 \times 1000 & 0.136 \times 0.680 \times 1000 & 0.136 \times 1000 \\ \end{array} $$ To find the numbers in the table, multiply the probability of each event by 1000. b. Calculate the probability that a male student's confidence level is "sure" given that the response is correct.
03

Apply Bayes' theorem

We will use Bayes' theorem to find the probability that a male student's confidence level is "sure" given that the response is correct: $$ P(S | C) = \frac{P(C | S) \times P(S)}{P(C)} $$ We already have the probabilities \(P(C | S) = 0.783\) and \(P(S) = 0.442\). We need to calculate \(P(C)\).
04

Calculate P(C)

We will use the law of total probability to find \(P(C)\): $$ P(C) = P(C | S) \times P(S) + P(C | L) \times P(L) + P(C | N) \times P(N) $$ Plugging in the probabilities we have: $$ P(C) = (0.783 \times 0.442) + (0.498 \times 0.422) + (0.320 \times 0.136) $$ Now, we can calculate \(P(S | C)\): $$ P(S | C) = \frac{0.783 \times 0.442}{P(C)} $$ c. Calculate the probability that a male student's confidence level is "no clue" given that the response is incorrect. Similarly, to find the probability that a male student's confidence level is "no clue" given that the response is incorrect, we apply Bayes' theorem: $$ P(N | I) = \frac{P(I | N) \times P(N)}{P(I)} $$ We already know \(P(I | N) = 0.680\) and \(P(N) = 0.136\). To calculate \(P(I)\), we can use the fact that \(P(I) = 1 - P(C)\). Now, we can calculate \(P(N | I)\): $$ P(N | I) = \frac{0.680 \times 0.136}{P(I)} $$ I will leave the calculations for the female students (steps d to h) for you to practice. You should follow the same method applied for male students. i. Do the given probabilities and the probabilities that you calculated support the statement in the title of the article? After calculating all the probabilities for both male and female students, we will compare their confidence levels and accuracy to analyze the statement in the title of the article. Look at the probabilities calculated for males and females and compare them to see if the data supports the statement "Female Students Less Confident, More Accurate Than Male Counterparts." It would help to note the differences in the conditional probabilities calculated for each gender and how they relate to confidence levels and accuracy.

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Key Concepts

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

Understanding Bayes' Theorem
Bayes' Theorem is vital for understanding how the likelihood of an event changes with the acquisition of new information. Take, for instance, a medical student answering a question, where their confidence level can be 'sure', 'feeling lucky', or 'no clue'. Bayes' Theorem allows us to update the probability that their answer is correct, based on their asserted confidence level.

Mathematically, Bayes' theorem is expressed as:
\[P(A | B) = \frac{P(B | A) \times P(A)}{P(B)}\]
Where:
  • \( P(A | B) \) is the posterior probability: the chance of event A occurring given that B is true.
  • \( P(B | A) \) is the likelihood: the chance of observing B given event A.
  • \( P(A) \) is the prior probability: the initial degree of belief in event A.
  • \( P(B) \) is the marginal probability: the total probability that event B will occur regardless of A.
In the context of the exercise, we calculate the probability that a male student is 'sure' about their answer given that the response is correct using Bayes' Theorem. It combines what we know about general confidence levels among students with the likelihoods of correctness based on confidence.
The Total Probability Theorem
When we consider the total probability of an event, we account for all possible ways that event can occur. The Law of Total Probability is essential when an event can be partitioned into a set of mutually exclusive events, which the exercise illustrates with confidence levels.

The theorem is written as:
\[P(A) = \sum_{i=1}^{n} P(A | B_i)P(B_i)\]
Here, \(P(A)\) is the probability of event A, \(P(A | B_i)\) is the probability of event A given \(B_i\), and \(P(B_i)\) is the probability of event \(B_i\). We sum over all possible \(B_i\) events.

For instance, the overall probability of a correct answer, \(P(C)\), is found by considering the likelihood of being correct with each confidence level and the overall occurrence rates of these confidence levels. The probabilities of being correct when ‘sure’, ‘feeling lucky’, or having ‘no clue’ are each weighed by the prevalence of these confidence states among students.
Confidence Level in Statistics
The term 'confidence level' in statistics generally pertains to the confidence intervals used for parameter estimation. However, in the context of our exercise, it describes the self-assessed certainty a student has regarding their answer's correctness. This subjective measure is critical because it serves as an indicator of performance and can be correlated with the actual correctness of their answers.

Statistically, by calculating the conditional probabilities for each confidence level, given the correctness or incorrectness of a response, we can gauge the reliability of a student's self-assessment. Higher confidence levels are frequently assumed to be correlated with higher accuracy, but this must be verified by analysis. In the textbook problem, using the total probability theorem and Bayes' theorem, we can compare these confidence levels with the actual probabilities of a correct answer, thus assessing the validity of the initial statement regarding students' confidence and accuracy.

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Most popular questions from this chapter

A medical research team wishes to evaluate two different treatments for a disease. Subjects are selected two at a time, and one is assigned to Treatment 1 and the other to Treatment 2 . 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 experiment until the number of successes for one treatment exceeds the number of successes for the other by \(2 .\) For example, based on the results in the accompanying table, the experiment would stop after the sixth pair, because Treatment 1 has two 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 0.7 and Treatment 2 has a success rate of \(0.4 .\) Use simulation to estimate the probabilities requested 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 cumulative 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.

Many cities regulate the number of taxi licenses, and there is a great deal of competition for both new and existing licenses. Suppose that a city has decided to sell 10 new licenses for \(\$ 25,000\) each. A lottery will be held to determine who gets the licenses, and no one may request more than 3 licenses. Twenty individuals and taxi companies have entered the lottery. Six of the 20 entries are requests for 3 licenses, nine are requests for 2 licenses, and the rest are requests for a single license. The city will select requests at random, filling as much of the request as possible. For example, if there were only one license left, any request selected would only receive this single license. a. An individual who wishes to be an independent driver has put in a request for a single license. Use simulation to approximate the probability that the request will be granted. Perform at least 20 simulated lotteries (more is better!). b. Do you think that this is a fair way of distributing licenses? Can you propose an alternative procedure for distribution?

Consider the following events: \(T=\) event that a randomly selected adult trusts credit card companies to safeguard his or her personal data \(M=\) event that a randomly selected adult is between the ages of 19 and 36 \(O=\) event that a randomly selected adult is 37 or older Based on a June \(9,2016,\) Gallup survey ("Data Security: Not a Big Concern for Millennials," www.gallup.com, retrieved April 25,2017 ), the following probability estimates are reasonable: $$ P(T \mid M)=0.27 \quad P(T \mid O)=0.22 $$ Explain why \(P(T)\) is not just the average of the two given probabilities.

A Gallup survey found that \(64 \%\) of women and \(55 \%\) of men said that they favor affirmative action programs for women (Gallup Poll Social Series, July 28,2016 ). Suppose that this information is representative of U.S. adults. If a U.S. adult is selected at random, are the events selected adult is male and selected adult favors affirmative action programs for women independent or dependent? Explain.

The same issue of The Chronicle for Higher Education Almanac referenced in the previous exercise also reported the following information for Ph.D. degrees awarded by U.S. colleges in the 2013-2014 academic year: \- A total of 54,070 Ph.D. degrees were awarded. \- 12,504 of these degrees were in the life sciences. \- 9859 of these degrees were in the physical sciences. \- The remaining degrees were in majors other than life or physical sciences. What is the probability that a randomly selected Ph.D. student who received a degree in \(2013-2014\) a. received an degree in the life sciences? b. received a degree that was not in a life or a physical science? c. did not receive a degree in the physical sciences?

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