Euparkeria is a rare disease. The symptoms are not clear at first, but grow worse as time progresses. There is a test which detects Euparkeria early, and allows treatment to start before the symptoms become serious. Note that the symbol ‘Pos’ indicates a positive result on the test. The symbol E indicates that the patient has Euparkeria. The symbol indicates that the patient does not have Euparkeria. Note: Ec is “E complement”               Pr (Pos| E) = .920                      Pr (E) = .0456               Pr (Pos| Ec) = .028                  Note: Pr (E) + Pr (Ec) =1               What is Pr (Pos Ո E)? 

Euparkeria is a rare disease. The symptoms are not clear at first, but grow worse as time progresses. There is a test which detects Euparkeria early, and allows treatment to start before the symptoms become serious. Note that the symbol ‘Pos’ indicates a positive result on the test. The symbol E indicates that the patient has Euparkeria. The symbol indicates that the patient does not have Euparkeria. Note: Ec is “E complement”               Pr (Pos| E) = .920                      Pr (E) = .0456               Pr (Pos| Ec) = .028                  Note: Pr (E) + Pr (Ec) =1               What is Pr (Pos)? 

The following data consists of WAT (word aptitude test) scores for a group of 25 young women students enrolled at a local college.   146.4 148.8 150.2 151.1 152.5 147.4 149.2 150.3 151.3 152.9 147.8 149.5 150.5 151.6 153.4 148.3 148.6 149.7 149.9 150.8 150.9 151.8 152.2 153.9 154.8                               Put the data in your calculator to answer the following questions. What is the IQR for this data?

An instructional resources company claims to have a method to train students which will increase their scores on any standardized test. Concerned that their admission process would be adversely affected, a large university decides to test this claim. The registrar selects, using a method which is independent and random, a group of n =200 high school students. This group is randomly divided in two, n1=100 students are sent to be trained to increase their test scores and n2=100 students are shown an inspirational movie to act as a control group.             After training both groups are administered an old PDQ. test, the following results are obtained Sample mean 1=257 and sample mean 2=249. To interpret the results the following assumptions are made:               a. The distribution of the scores before training was normal N (250, 50)            b.      Any changes due to training represent a change in the location of the mean                      rather the shape of the distribution, hence σ1 = σ2 = 50   Test Ho: µ1 = µ2 vs. Ha: µ1 > µ2  at the α =.05 level of significance.   The following triple (z-score, p-value, Your Conclusion) is produced when the data is properly input in to your calculator.

A group of students has been trained to increase their score on the math segment of the S.A.T.  For the general population, S.A.T.-Math scores are distributed N (505,105). It is reasonable to assume that the shape of the distribution is unchanged by training effects, but is shifted, hopefully to the right. It follows that we assume that σ=105. For a random sample, of size n = 100 the sample mean  S.A.T.-math score is 529.   Compute a 99% confidence interval for µ, the mean score of the population of trained students.

A group of students has been trained to increase their score on the math segment of the S.A.T.  For the general population, S.A.T.-Math scores are distributed N (505,105). It is reasonable to assume that the shape of the distribution is unchanged by training effects, but is shifted, hopefully to the right. It follows that we assume that σ=105. For a random sample, of size n = 100 the sample mean S.A.T.-math score is 529.   Find n in order to estimate µ with a margin of error M=5, and with 99% confidence (σ =105). Remember to round up to the next highest whole number.

In 2007 the scores on the College Aptitude Test (C. A. T.) were distributed normally with mean 500 and standard deviation 100, briefly N (500, 100)  Find the Z score for a CAT  score of 450

The following data consists of WAT (word aptitude test) scores for a group of 25 young women students enrolled at a local college.   146.4 148.8 150.2 151.1 152.5 147.4 149.2 150.3 151.3 152.9 147.8 149.5 150.5 151.6 153.4 148.3 148.6 149.7 149.9 150.8 150.9 151.8 152.2 153.9 154.8                               Put the data in your calculator to answer the following questions. Compute the 5- number summary.

A rural southern county has a revenue shortfall. A property tax initiative is placed on the ballot. A large independent simple random sample of voters yields the following information:          Let event A be the voter favors the proposed tax,          and event B be that the voter is a property owner. Then Pr (A|B) = .394, Pr (A| Bc) =.943 and Pr (B) = .764, Pr (Bc) = .236   NOTE: Pr (B) + Pr (Bc) = 1 by definition. Given this data and the following formula answer the following question, what is P(A)?   Pr (A) = Pr(A|B) Pr(B) + Pr(A| Bc)Pr(Bc) 

Which of the following is not a justification for the popularity of intercollegiate athletics?