Are people more confident in their answers when the answer is actually correct than when it isnot? Astudy was done on medical students' confidence and the accuracy of their responses.Participantscategorized their confidence levels using either "sure", "feeling lucky" or "no clue".Define the following eventsC= event that a response is correctS= event that confidence level is "sure"L = event that confidence level is "feeling lucky"N = event that confidence level is "no clue"P(S) = 0.442 P(L) = 0.422 P(N) = 0.136P(C|S) = 0.783 P(C | L) = 0.498 P(C| N) = 0.320Not Ctotal34696442L210212442N4492136total60040010001. Calculate the probability that a student's confidence level is "sure" given that the response iscorrect.2. Calculate the probability that a student's confidence level is "no clue" given that the responseisincorrect.3. Calculate the probability that the student's response is correct.4. Calculate the probability that the student's response is correct or his confidence level is"feelinglucky"5. Calculate the probability that the student's response is incorrect and his confidence level is"sure".

Name: Are people more confident in their answers when the answer is actuall
Uploaded: 2024-03-20T07:06:41.000Z
Channel: Bartleby
Description: Are people more confident in their answers when the answer is actually correct than when it isnot? Astudy was done on medical students' confidence and the accuracy of their responses.Participantscategorized their confidence levels using either "sure

NOTE: As per the guidelines, we are supposed to solve only first three sub parts only. a). Consider,…

Math

Probability

P(S) = 0.442 P(L) = 0.422 P(N) = 0.136 P(C | S) = 0.783 P(C | L) = 0.498 P(C | N) = 0.320 %3D Not C total 346 96 442 210 212 442 N 44 92 136 total 600 400 1000 1. Calculate the probability that a student's confidence level is "sure" given that the response is correct. 2. Calculate the probability that a student's confidence level is "no clue" given that the response is

P(S) = 0.442 P(L) = 0.422 P(N) = 0.136 P(C | S) = 0.783 P(C | L) = 0.498 P(C | N) = 0.320 %3D Not C total 346 96 442 210 212 442 N 44 92 136 total 600 400 1000 1. Calculate the probability that a student's confidence level is "sure" given that the response is correct. 2. Calculate the probability that a student's confidence level is "no clue" given that the response is

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

See similar textbooks

Similar questions

A College Alcohol Study interviewed an SRS of 1000 college students about their drinking habits. Suppose that half of all college students "drink to get drunk" at least once in a while. That is sample proportion, p-hat 0.5. %3D 9. Estimate a 95% confidence interval for the actual percentage among all students. Round the number to nearest hundredth. O47% to 53% O 50% to 55% 43% to 52% O 44% to 56%
A survey of randomly selected university students found that 91 of the 101 first-year students and 96 of the 136 second-year students surveyed had purchased used textbooks in the past year. (a) Construct a 95% confidence interval for the difference the proportions of first-year and second-year university students who purchased used textbooks. (Round your answers to 4 decimal places, if needed.) Answer: (b) If a 98% confidence interval for the difference in proportion of first-year and second-year university students who purchased used textbooks is (0.0809, 0.3093). Give an interpretation of this confidence interval. O We are 98% confident that the first-year students at this university bought between 8.09% and 30.93% more used textbooks than second-year students. O We know that 98% of the first-year students bought between 8.09% and 30.93% higher than the proportion of second-year students who bought used textbooks. O We are 98% confident that, at this university, the proportion of…
A newsgroup is interested in constructing a 99% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 536 randomly selected Americans surveyed, 428 were in favor of the initiative. Round answers to 4 decimal places where possible. a. With 99% confidence the proportion of all Americans who favor the new Green initiative is between and
Beth wants to determine a 90 percent confidence interval for the true proportion p of high school students in the area who attend their home basketball games. Out ofn randomly selected students she finds that that exactly half attend their home basketball games. About how large would n have to be to get a margin of error less than 0.04 for p?
Suppose you are told that a 95% confidence interval for the average price of a gallon of regular gasoline in your state is from $3.37 to $3.60. Use the fact that the confidence interval for the mean is in the form x − E to x + E to compute the sample mean and the maximal margin of error E. (Round your answers to two decimal places.) x = $ E = $
Are the statements true or false?
During the 2009 holiday season, 436 randomly sampled American adults were surveyed regarding their average spending on Black Friday. A 95% confidence interval based on this sample is ($80.31,$89.11). (c) Determine the sample mean x and the sample standard deviation s of the sample. (d) Which would be the confidence level corresponding to an interval ($81.27,$88.15) for the data found in (c)? (e) If the values of the sample mean and sample standard deviation found in (c) do not change, find the sample size N for which a 99% confidence interval would have a margin of error smaller than 5.
Suppose you are told that a 95% confidence interval for the average price of a gallon of regular gasoline in your state is from $3.37 to $4.18. Use the fact that the confidence interval for the mean is in the form x − E to x + E to compute the sample mean and the maximal margin of error E. (Round your answers to two decimal places.) x=$ E=$
When finding the upper bound, UU, of a confidence interval given the lower bound, LL, and the margin of error, EE, we use the formula U=L+2EU=L+2E. Find the upper bound for the proportion of babies that are born preterm if the lower bound is L=0.364L=0.364 and the margin of error is E=0.176E=0.176. Round your answer to 3 decimal places.U=
The paper "The Effects of Adolescent Volunteer Activities on the Perception of Local Society and Community Spirit Mediated by Self-Conception"+ describes a survey of a large representative sample of middle school children in South Korea. One question in the survey asked how much time per year the children spent in volunteer activities. The sample mean was 14.76 hours and the sample standard deviation was 16.54 hours. USE SALT (a) Based on the reported sample mean and sample standard deviation, explain why it is not reasonable to think that the distribution of volunteer times for the population of South Korean middle school students is approximately normal. (Round your answer to nearest percent.) If the distribution of volunteer times is approximately normal, for the sample standard deviation of s = 16.54 hours and the sample mean of x = 14.76 hours, approximately % of volunteer times would be negative. Therefore, it ---Select--- reasonable to think that the distribution of volunteer…
The following table shows 1000 Engineer school applicants classified according to scores made on a college entrance examination and the quality of the high school from which they graduated, as rated by a group of educators: Quality of High Schools Superior Scores Average Low 300 100 High 260 340 A 95% confidence interval for the population proportion of applicants that a low score on the examination and graduated from an average high school is: (0.081 , 0.118) None of these (0.271,0.328) O (0.310, 0.369)
Call me: A sociologist wants to construct a 99% confidence interval for the proportion of children aged 8 - 10 living in New York who own a smartphone.(a) A survey by the National Consumers League estimated the nationwide proportion to be 0.40 . Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.02 ?