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IUB CJUS-K 300 - Exam 2 Study Guide

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CJUS-K300 1nd EditionExam # 2 Study Guide Lectures: 6 - 10Lecture 6 (September 17)ProbabilityNotation. Of Probability = PRules of ProbabilityBounding Rule – all probabilities are bounded between one and zero. 0 ≤ P(A) ≤ 1- If you come up with a negative probability or a probability higher than one re-check youranswer because that is not possible. Must be between 0-1Complement Rule – If you know P(A) then you know P(Not A). P(Not A) = 1-P(A)Mutually exclusive events are when two events cannot occur at the same time. If A and B are mutually exclusive then P(A AND B) = 0 - Ex. In a deck of cards pulling a diamond and a heart are mutually exclusive. P(heart AND diamond) = 0 Addition Rule (either/or rule) for mutually exclusive events - A and B are mutually exclusive. P(A or B) = P(A) + P(B)- Ex. P(Hearts or Diamonds) = P(H) + P(D) o P(H) = .25o P(D) = .25o .25 + .25 = .5 or 5% probability Not all events are mutually exclusive. Addition Rule in general form: P(Hearts and Ace) ≠ 0P(A or B) = P(A) + P(B) – P(A and B) You subtract P(A and B) in order to not double count the event and overestimate the probability. If you do not subtract A and B you would be counting it twice because you have already included A and B separately. - Ex. P(Heart or Ace) = P(H): .25 + P(A): .08 – P(H & A): .02 = .31 or 31% - Explanation – Under the assumption that there are 52 cards in a decko P(Hearts) = 13/52 or . 25o P(Ace) = 4/52 or .08 o P (H & A) = 1/52 or .02Lecture 7 (September 22) Third rule of probability: Multiplication Rule (general form) - P(A and B) = P(A) x P(B|A) Conditional Probabilities - P(A|B): the probability that A will happen given that B has already happened- P(B|A): the probability that B will happen given that A has already happened. P(A and B) = P (A) x P(B|A)Independent Events- When two events are statistically independent, knowing that event A has already occurred does not help you at all in predicted the probability of event B occurring. A good example of this is in the deck of cards. Knowing the probability of drawing an Ace P(Ace) does not help you figure out the probability of the suit of the Ace P(Heart). For statistically independent event P(A|B) = P(A) and P(B|A) = P(B)The short form of the Multiplication Rule, which can ONLY be used for Independent events - P(A and B) = P(A) x P(B) How do we know which events are statistically independent?If P(A|B) = P(A) then the events must be independent- Does P(Ace|Heart) = P(Ace)?o P(Ace|Heart) = 1/13 or .08o P(Ace) = 4/52 or.08o They are independent!- Which means that P(Ace and Heart) = P(Ace) X P(Heart)o 4/52 x 13/52 = .08 x .25 = .02Lecture 8 (September 24)Z scoresProperties of a normal curve – as close to symmetrical as possible given that it would be close to impossible for it to be perfectly symmetricalTwo curves can be normal with the same standard deviations but different means, or the same mean and different standard deviations. (See power point for examples)In cases of normal distribution:Once you find the mean of the curve, for normal curves it is in the middle of the curve, 50% of the data is on the left side and 50% of the data is on the right. The area between the mean and one standard deviation is 34% meaning 34% of the data points lie in between the mean/middle and one standard deviation away in either direction. The area between the mean and two standard deviations is 47.5%. The area between the mean and 3 standard deviations is 49.5%The two important parameter are the mean and the standard deviation. The mean determines where the midpoint of the distribution falls and the SD determines the spread or the width of the distribution.- Large SD – wider, more dispersed data points- Small SD – narrower, less disperse data pointsStandardizing normal distribution: Making the mean = 0 on the graph in order to be able to compare two different sets of data. Converting values to a standard score. Reading the Z-Table (This will take some practice but gets easier)1. Calculate the z-score and round to two decimal points2. Divide the number in two sectionsa. First number and the first decimal point/second decimal pointi. 1.95 = A) 1.9 B).053. Find the A – 1.9 on the left side of the row 4. Find B – from the top of the columns Example: Janes IQ score is 3 standard deviations above the mean. What do we know about her IQ ?- She is in the 99.5th percentile. = 50 + 49.5 (from 3 standard deviations) American IQ test : x|=100 s=10 Marys score was 110pts . She was above the average by exactly 1 standard deviation (10) so we know that she is in the 84th percentile. (50 + 34) Comparing 2 different testsAmerican IQ test x| = 100, s=10 Mary = 110European IQ test x|= 350, s=36, Pierre = 382Who has the higher IQ? Mary does. Pierre is not a full standard deviation above the mean whereas Mary is.Z = X – Mean / SPierre: z=382-350/36 = .89Mary: z = 110-100/10 = 1If it is not a full SD above the mean you have to take it to the z table. Look up the z score on the table and find the corresponding number and add it to the 50%Z=88-80/15 = .53 .53 corresponds to .2019, so you add 20% to 50% and which makes it the 70thpercentile. Lecture 9 (September 29)Central Limit Theorem:1. The means from a large number of samples (all of size n ≥ 100) will always form a normal distribution. 2. μx = μ meaning if μx = 5 then μ = 5σX = Standard error of the distribution of a large number of sample means taken from the same populationConfidence Interval Formula x 4±z_α (s/√(n-1) Alpha levels (how likely it is that you are wrong)For a 99% - a=.01For a 95% - a=.05For 90%- a=.10Example: x T±z_α (s/√(n-1))Calculate a 95% confidence interval:5.5±1.96(2.5/√(100-1)) = 5.5 +.49We are 95% confident that the actual population mean attitude toward gun control is between 5.01 points and 5.99 points on the Gun Control Attitude ScaleYou must know how to express this interpretation on the exam, it is almost more important than getting the numbers correct. It shows you understand the answers that you have gotten!Lecture 10 (October 1)“Student’s” t distribution – A collection of distributions whose shape depends on degrees of freedom (df = n-1)Confidence intervals with small samples (n≤50) – small sample means it is less representative of the population. x ± ta (s/√n-1)Steps1. Determine your alpha level2. Calculate your degrees of freedom (n-1)3. Use a t-table to look up the


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