Midterm Exam III Review Dr. Joseph Brennan Math 148, BU Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 1 / 25
Permutations and Combinations ORDER In order to count the number of possible ways to choose, without replacement, k objects from a collection of n distinct objects we must be specific as to we acknowledge order. A permutation is a choice where order matters. A combination is a choice where order does not matter. The only difference between a permutation and a combination is order. This leads to very similar counting formulas: ( ) n! n n! np k = = (n k)! k k! (n k)! Recall: An event E in the sample space S has probability P(E) = number of outcomes in E number of outcomes in S Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 2 / 25
iclicker A graduate class is comprised of 2 overachieving undergraduates, 7 first year graduates, 4 second year graduates, and 1 third year graduate. If class begins with random presentations by three students, what is the probability that an undergraduate and two first year graduates are chosen? (A) 0% (B) 6% (C) 11% (D) 19% (E) 27% ( 2 ) ( 1 7 ( 2) 14 ) 11% 3 Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 3 / 25
Law of Averages Law of Averages: If an experiment is independently repeated a large number of times, the percentage of occurrences of a specific event E will be the theoretical probability of the event occurring, but of by some amount - the chance error. Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 4 / 25
iclicker There are two hospitals: in the Hospital A 120 babies are born every day; in Hospital B 12 babies are born. On average, the ratio of baby boys to baby girls born every day in each hospital is 50/50. However, one day, in one of those hospitals, twice as many baby girls were born as baby boys. In which hospital was it more likely to happen? (A) Hospital A (B) Hospital B (C) Hospital C (D) Equally Likely Hospital B. The probability of a random deviation of a particular size (from the population mean), decreases with the increase in the sample size. Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 5 / 25
Random Variable Random Variable: An unknown subject to random change. Often a random variable will be an unknown numerical result of study. A random variable has a numerical sample space where each outcome has an assigned probability. There is not necessarily equal assigned probabilities. Any random variable X, discrete or continuous, can be described with A probability distribution. A mean and standard deviation. Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 6 / 25
Random Variables The distribution of a discrete random variable X is summarized in the distribution table: Value of X x 1 x 2 x 3... x k Probability p 1 p 2 p 3... p k The symbols x i represent the distinct possible values of X and p i is the probability associated to x i. p 1 + p 2 +... + p k = 1 (or 100%) Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 7 / 25
3 Coins Probability Histogram Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 8 / 25
Discrete Random Variable: µ Mean: The mean µ of a discrete random variable is found by multiplying each possible value by its probability and adding together all the products: µ = x 1 p 1 + x 2 p 2 +... + x k p k = k x i p i i=1 Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 9 / 25
Discrete Random Variable: σ Standard Deviation: The standard deviation σ of a discrete random variable is found with the aid of µ: σ = (x 1 µ) 2 p 1 + (x 2 µ) 2 p 2 +... (x k µ) 2 p k = k (x i µ) 2 p i i=1 When there are just two numbers, x 1 and x 2, in the distribution of X the distribution s standard deviation, σ, can be computed by using the following short-cut formula: where p i is the probability of x i. σ = x 1 x 2 p 1 p 2 Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 10 / 25
Box Models Box Model: A model framing a statistical question as drawing tickets (with or without replacement) from a box. The tickets are to be labeled with numerical values linked to a random variable. The expected value of a random variable is the average of the tickets occupying the box model. The standard deviation of a random variable is the standard deviation of the tickets. Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 11 / 25
iclicker Did you know there are 4, 8, 10, 12, and 20 sided die? Create a box model for rolling a 4-sided die. What is the standard deviation of the box? (A) 1.11 (B) 1.65 (C) 2 (D) 2.21 (E) 2.5 Solution: First find the mean: µ = 1 1 4 + 2 1 4 + 3 1 4 + 4 1 4 = 2.5 σ = 1.11 σ = (1 2.5) 2 1 4 + (2 2.5)2 1 4 + (3 2.5)2 1 4 + (4 2.5)2 1 4 Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 12 / 25
The Sum of n Independent Outcomes When the same experiment is repeated independently n times, the following is true for the sum of outcomes: The expected value of the sum of n independent outcomes of an experiment: nµ The standard error of the sum of n independent outcomes of an experiment: nσ The second part of the above rule is called the the Square Root Law. Note that the above rule is true for any sequence of independent random variables, discrete or continuous! Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 13 / 25
The Binomial Setting 1 There are a fixed number of n of repeated trials. 2 The trials are independent. In other words, the outcome of any particular trial is not influenced by previous outcomes. 3 The outcome of every trial falls into one of just two categories, which for convenience we call success and failure. 4 The probability of a success, call it p, is the same for each trial. 5 It is the total number of successes that is of interest, not their order of occurrence. NOTE: The Binomial Setting can be framed as a box model with only 1 s and 0 s where draws are performed with replacement. Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 14 / 25
The Binomial Distribution Let X denote the number of successes under the binomial setting. Then X is a random variable which may take values 0, 1, 2, 3,..., n. In particular, X = 0 means no successes in n trails. Only failures were observed. X = n means the outcomes of all n trails are successes. X = 5 means 5 successes in n trials. It turns out that X has a special discrete distribution which is called the binomial distribution. The probabilities of values of X are computed as ( ) n P(X = k) = p k (1 p) n k, k = 0, 1, 2,..., n. (1) k So the binomial distribution is a probability distribution of a random variable X which has 2 parameters: p (probability of success) and n (the number of trials). Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 15 / 25
Binomial Mean and Standard Deviation Let X be a binomial random variable with parameters n (number of trials) and p (probability of success in each trial). Then the mean and standard deviation of X are µ = np, σ = np(1 p). Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 16 / 25
Binomial Distribution and Normal Curves NORMAL APPROXIMATION for BINOMIAL COUNTS Let X be a random variable which has a binomial distribution with parameters n and p. When n is large, the distribution of X is approximately normal. X is approximately normal with mean np and standard deviation np(1 p). As a rule, we will use this approximation for values of n and p that satisfy np 10 and n(1 p) 10. Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 17 / 25
The Central Limit Theorem (CLT) The Central Limit Theorem: When drawing at random with replacement from a box, the probability histogram for the sum will approximately follow the normal curve, even if the contents of the box do not. The larger the number of draws, the better the normal approximation. The sample size n should be at least 30 (n 30) before the normal approximation can be used. For symmetric population distributions the distribution of x is usually normal-like even at n = 10 or more. For very skewed populations distributions larger values of n may be needed to overcome the skewness. Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 18 / 25
Parameters & Statistics Parameter: A numerical fact about a population. Statistic: A numerical fact about a sample. An investigator knows a statistic and wants to know a parameter. Probability Methods: Sampling techniques which implements an objective chance process to choose subjects from the population, leaving no discretion to the interviewer. It is possible to compute the chance that any particular individual in the population will get into the sample. Simple Random Sampling: A sampling technique where selection of individuals is equally likely and drawing for the sample is performed without replacement. Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 19 / 25
Bias Sampling Bias: A bias in which a sample is collected in such a way that some members of the intended population are less likely to be included than others. The bias can lead to an over/underrepresentation of the corresponding parameter in the population. Almost every sample in practice is biased because it is practically impossible to ensure a perfectly random sample. Non-response Bias: A bias that results when respondents differ in meaningful ways from nonrespondents. Respondents and nonrespondents can differ in ways beyond their willingness to answer a questionnaire. Quota Sampling: A sampling method in which interviewers are assigned a fixed quota of subjects to interview. Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 20 / 25
Variable Type Given n draws with replacement from a box with Mean µ (average for quantitative and percent for qualitative). Standard Deviation σ. Sum Average Number Percent Expected Value: n µ µ n µ µ Standard Error: n σ σ/ n n σ σ/ n Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 21 / 25
The Correction Factor When drawing without replacement, to get the exact SE you must multiply by the correction factor: number of objects in box number of draws number of objects in box 1 When the number of tickets in the box is large relative to the number of draws, the correction factor is nearly one. Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 22 / 25
Normal Curve for SE for Averages and Percentages Suppose 1,000 draws are made with replacement from a box whose average ticket value is 200. The standard error for averages is found to be 10. There is about a 68% chance for the average of the 1, 000 draws to be in the range 190 to 210. Suppose 1,000 draws are made with replacement from a 0 1 box whose percent of 1 s was 15%. The standard error for percent is found to be 0.5%. There is about a 68% chance for the percentage of successful draws of the 1, 000 draws to be in the range 14.5% to 15.5%. Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 23 / 25
iclicker Historically, Five Star Pizza has recorded Thursday night drivers as having an average delivery time of 28 minutes with a standard deviation of 5 minutes. If the manager were to take a sample of 100 orders at random on Thursday nights spread over a year, what is the chance that the average delivery time is greater than 30 minutes? (A) 0% (D) 10% (B) 2% (E) 40% (C) 5% Solution: The box has µ = 22 and σ = 5. EV = 28 SE = 5 100 = 0.5 We can answer this question using the normal curve: ( ) 30 28 P(X > 30) = 1 P Z < = 1 P(Z < 4) 0 0.5 Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 24 / 25
iclicker Five Star Pizza has not been keeping track of their Monday night delivery time and the manager has been receiving complaints. The manager takes a simple random sample of delivery 40 deliveries, and finds a sample mean of 36 minutes and a sample standard deviation of 12 minutes. What is the 90% confidence interval for the average Monday night delivery time? (A) [35,37] (C) [32,40] (E) [28,44] (B) [34,38] (D) [30,42] Solution: Begin by finding the z-score for confidence: The margin of error is found (90/2 + 50) = 95 with z-score: 1.66 m = z C SE = 1.66 12 40 4 [ x m, x + m] = [32, 40] Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 25 / 25