CH-4-6 A Review of Probability Theory – Introduction to Statistics for Social Sciences : Book Guide

This Document Contains Chapters 4 to 6 Chapter 4 A Review of Probability Theory Learning Objectives: 1. Explain the concept of probability 2. Describe simple, mutually exclusive, and non-mutually exclusive probability problems using a Venn diagram. 3. Calculate basic probabilities of events from a random experiment. 4. Use the additive and multiplicative rules of probability. 5. Describe the difference between frequency distributions and probability distributions. Chapter Summary In this chapter, students were introduced to the most important concept underlying statistical analysis: probability theory. Students are exposed to the concepts of experiment, random outcomes and random experiment. Venn diagrams are used to illustrate the concept of probability and the terms and equations for calculating simple, mutually and non-mutually exclusive probability problems. A distinction is made between frequency and probability distributions Key Formulas The following represent the key formulas for this chapter. PowerPoint slides are provided for each chapter. In addition to these slides, a PDF file containing only the formulas are also provided. Simple Probability Mutually Exclusive Probability Non-Mutually Exclusive Probability Dependent Event Probability Independent Event Probability This Document Contains Chapters 4 to 6 Interactive Figures: The textbook contains interactive figures. You may wish to use these in a lecture. Students also have access to these. For this chapter, there are two interactive figures. 1. Figure 4.8 illustrates the Venn diagram. 2. Figure 4.11 illustrates the results of coin tossing. These interactive figures can be found on in the eBook and the Library under Chapter 4 Resources. Typical Lecture Material We have provided two sample lectures below. You may wish to add in additional discipline specific information to make these more relevant to your students. Lecture 1: Objective: Understand the concept of probability and the calculation of probabilities given simple, mutually exclusive and non-mutually exclusive events. Provide the following concept table to your students; and help them to fill in the definitions of each probability concept. The definitions that the students should come up with are in italics. Probability Concept Defintion Experiment A process that allows us to observe an outcome while controling or imposing certain restrictions during our observation. Random Experiment A repeatable experiment whose outcomes cannot be predictied with 100% certainty. Random Outcome/Event The outcome of interest that cannot be predicted with certainty. Sample Space All of the possible outcomes of a random experiment. Complement of an Event All other possible outcomes inside the sample space that are not the event we are interested in. Simple Probability The probability of a single event occuring, where no other possible events are considered. Mutually Exclusive Probability The probability of an outcome in relation to others, where only one outcome out of a set of possible outcomes can occur at the same time. Non-Mututally Exclusive Probability The probability of two (or more) outcomes occuring at the same time. Dependent Probability The probability of an event occuring given a prevous dependent event occurred. Independent Probability The probability of an event occuring given a previously independent event occured. Example 1: Read the following scenario to your students: Michelle was participating in a university research study regarding the effect of background music on studying. For the study, Michelle would be randomly assigned to one of three groups; Rock Music, Classical Music, or No Music. In total, 400 students were recruited for the research, of which 150 were randomly assigned to the No Music group. What is the probability that Michelle will be in the No Music group of the study? 1.a. (Simple Probability) Ask your students to represent this information in a Venn Diagram. Ask them to identify [P(A)], [P(Ac)] and the sample space [S]. Note: The completed Venn diagram should look like this: 1.b. Ask your students to calculate the simple probability of being randomly allocated to the No Music group. Note: The calculation is as follows P(No Music)= number of ways the even can happen total number of possible outcomes P(No Music) = 150 400 = 0.375 1.c. Ask your students to calculate the simple probability of being randomly allocated to a group other than the No Music group. Sample Space (n = 400) P(Music)=150 P(Ac) = 250 Sample Space (n = 400) P(D) = 100 Music Group Male P(A) = 90 No Music Female P(B) = 60 No Music Male P(C) = 150 Music Group Female P(Ac) = 1 – P(No Music) P(Ac) = 1 – 0.375 P(Ac) = 0.625 Example 2: (Mutually Exclusive Probability) Ask your students to keep in mind the information from the previous example. Shortly after joining the research study, Michelle learns that there will be fewer Males than Females randomly assigned to the No Music group. Provide your students with the following data table: No Music Group Music Group (consists of both rock and classical music groups) Total Male 60 100 160 Female 90 150 240 Total 150 250 400 Ask your students to represent this information in a Venn Diagram. Note: The completed Venn diagram should look like this: Ask your students to complete the following calculations: 1. The probability of being a female and being selected for the No Music Group. 2. The probability of being a male and being selected for the Music Group. Note: The calculation is as follows P(Female No Music)= 90 400 = 0.225 P(Male Music) = 100 400 = 0.25 2.b. (Non-Mutually Exclusive Probability) Now that Michelle knows the numbers of students that were allocated to each group, she is interested in knowing what the probability is that a student, participating in the research, is either Male or in the No Music group. Remind the students that they can be both male and in the No Music Group; so they need to consider the over-lap between these two ‘probability events’ of interest. Ask your students to 1. Identify what type of probability question this is. (Answer: non-mutually exclusive probability) 2. Calculate the probability of a student being male or in the No Music group. P(Event A or B) = P(A) + P(B) - P(A and B) P(Male or No Music) = P(Male) + P(No Music) - P(Male and No Music) P(Male or No Music) = 160 400 + 150 400 - ( 160 400 × 150 400) P(Male or No Music) = 0.40 + 0.375 - (0.40 × 0.375) P(Male or No Music) = 0.40 + 0.375 - 0. 15 = 0.625 Lecture 2: Objective: Understand the concept of dependent versus independent events and to discuss the difference between frequency and probability distributions. Remind your students about the distinction between dependent and independent events and that a distribution is the arrangement of any values based on the frequencies with which they occur. Example 1: (Dependent Events) Provide the following scenario to your students. Consider 10 golfers who are competing in a putting contest. Each contestant must get their ball in the hole from a distance of 10 feet. If they miss the putt on their first try, they may try one additional time. 4 out of 10 golfers made the shot on the first attempt. Of the 6 golfers who missed on the first attempt 3 made the shot on their second attempt, while the remaining 3 did not. a.) What are the possible outcomes for a single contestant? Answer: (1) get the ball in the hole on the first attempt (2) miss the hole on the first attempt but make it in on the second attempt (3) miss the hole on both the first and second attempts b.) What is the probability of missing the shot on the first attempt? Answer: P(missing the shot on the first attempt) = 6 ÷ 10 = 0.60 c.) What is the probability of missing the shot on both attempts? Answer: P(missing on both attempts) = P(missing the first attempt) × P(missing the second attempt given that they missed the first attempt) P(missing on both attempts) = (6 ÷ 10) × (3 ÷ 6) = 0.60 × 0.50 = 0.30 Example 2: (Independent Events) Provide the following scenario to your students. You are a first year social science student, where it is mandatory that you take a variety of introductory courses. Following the completion of your first year, you plan to apply to the School of Social Work program. Entrance into this program requires that students: (i) complete one year of undergraduate studies and obtain a minimum average mark of 80%, and (ii) write the program admissions exam. To strengthen your application, you aim to obtain an 85% average for your first year courses. You know that you are doing really well in most of your classes with the exception of biochemistry and anatomy. You find out from the professors of these courses that the probability of getting an 85% or more in these courses is 0.25 for biochemistry and 0.55 for anatomy. What is the probability of getting an 85% or more in both biochemistry and anatomy? P (A and B) = P(A) × P(B) P (A and B) = 0.25 × 0.55 = 0.1375 Solutions to End-of-Chapter Problems Problem 4-1 a. P(heart) = ¼ (LO2) b. P(face or spade) = 11/26 (LO2) c. P(2of spades) = 1/52 (LO2) d. P(queen) = 1/13 (LO2) Problem 4-2 a. P(blue) = 1/10 (LO3) b. P(yellow or red) = 11/30 (LO3) c. P(white or yellow or blue) = 13/15 (LO3) d. P(not red) = 13/15 (LO3) Problem 4-3 P(contract disease) = 0.40 (LO4) Solutions to Interactive Exercises Question 4-1 In a random sample of 120 people, 51 had type O blood, 52 had type A blood, 10 had type B and the rest type AB. One person is selected at random, what is the probability that: (i) the person has blood type AB? (ii) the person has blood type either type A OR type B? (iii) the person does not have type B blood ? (iv) the person does not have either type O or type A? Feedback: (i) 7/120 (ii) (52+10)/120 (iii) 110/120 (iv) 17/120 Question 4-2 Consider three events X, Y and Z. A) P(X) = 0.25 P(Y) = 0.35 P(Z) = 0.70 B) P(X and Y) =0.15 P(Y and Z) =0.09 C) P(Y/Z) = 0.45 (i) Are Y and Z independent events? Why? (ii) Find P(X or Y) (iii) P(X/Y) (iv) Are you X and Y mutually exclusive events? Why? Feedback: (i) P(Y) is not equal to P(Y/Z), hence not independent. In other words, the event Z does influence event Y (ii) P(X orY) = P(X) + P(Y)-P(X and Y) = 0.25+0.35-0.15=0.45 (iii) P(X/Y) = P(X and Y) / P(Y) = 0.15/0.35= 0.43. (iv) P( X and Y) is not equal to zero, hence not mutually exclusive. Exercise 4-3 Two dice rolled. Find the probability of getting (i) a sum of 7 , 8 or 9 (ii) a sum greater than 10 or less than 3 Feedback: (i) 15/36 (ii) 4/36 Solutions to SPSS Exercises Question 4-1 The following cross tabulation was produced using the data file hourly wage data. sav contained within the SPSS library on your computer. The information is relative to their selection after graduating from nursing school and what they decide for a career path relative to their age. Age Range * Nurse Type Cross tabulation Count Nurse Type Hospital Office Total Age Range 18-30 106 43 149 31-45 391 175 566 46-65 190 95 285 Total 687 313 1000 Consider the following notation when answering the following questions: H=hospital, O=office, LOW=low age group 18 to 30, MIDDLE=middle age group 31 to 45, and HIGH=high age group 46 to 65. Answer the following questions relative to individuals chosen at random from this group. a) Recreate this cross tab using the data file in SPSS. LO: 3 Page: 108 b) What is P(H), P(MIDDLE), P(LOW)? Ans: P(H) = 0.687 , P(MIDDLE) = 0.566 , P(LOW) = 0.149? LO: 3 Page: 108 c) What is P(H or O)? are the events mutually exclusive? Ans: P(H or O) = 1 and the events are mutually exclusive LO: 3 Page: 112-117 d) What is P(H or high), P(H and High)? Ans: P(H or high) = 0.782 , P(H and High) = 0.190 LO: 4 Page: 112-117 e) What is P(Middle given O), P(O given MIDDLE)? Ans: P(Middle given O) = 0.559 , P(O given MIDDLE) = 0.309 LO: 4 Page: 112-117 f) Are the events MIDDLE and O independent? Would be so if P(MIDDLE and O) = P(MIDDLE)P(O). Ans: P(MIDDLE and O) = 0.175 , P(MIDDLE)P(O) = (0.566)(0.313) = 0.177 so dependent. OR P(MIDDLE given O) = P(MIDDLE) P(MIDDLE given O) = 0.559 , P(MIDDLE) = 0.566 so dependent again. LO: 4 Page: 112-117 Equations from Chapter 4 Scott R. Colwell and Edward M. Carter c 2012 Equation 4.1: Simple Probability P(A) = Number of ways the event can happen Total number of possible outcomes Where: P(A) = the probability of event A Equation 4.2: Simple Probability P(A) = s n Where: P(A) = the probability of event A s = the number of ways the event can happen n = the total number of possible outcomes Equation 4.3: The Sample Space P(S) = P(A) + P(Ac) Where: P(S) = the sample space P(A) = the probability of event A P(Ac) = the probability of the complement of A Equation 4.7: Mutually Exclusive Probability P(A or B) = P(A) + P(B) Where: P(A or B) = the probability of event A or event B P(A) = the probability of event A P(B) = the probability of event B Equation 4.9: Non-Mutually Exclusive Probability P(A or B) = P(A) + P(B) - P(A and B) Where: P(A or B) = the probability of event A or event B P(A) = the probability of event A P(B) = the probability of event B P(A and B) = the probability of event A and event B Equation 4.10: Dependent Events P(A and B) = P(A)P(B given A) Where: P(A and B) = the probability of event A and event B P(A) = the probability of event A P(B given A) = the probability of event B given event A Equation 4.12: Independent Events P(A and B) = P(A)P(B) Where: P(A and B) = the probability of event A and event B P(A) = the probability of event A P(B) = the probability of event B Chapter 5 The Normal Curve Learning Objectives: 1. Identify why a probability distribution is a theoretical distribution. 2. Explain the normal curve and its importance 3. Describe the areas under the normal curve. 4. Transform raw scores to standardized scores (z-scores) and percentiles. 5. Find the area(s) under the normal curve using Appendix A: Area under the normal curve. 6. Draw inferences based on z-scores. 7. Transform percentiles to z-scores. 8. Transform percentiles to z-scores to raw scores. Chapter Summary In this chapter, students are shown the link between probability theory, probability distributions and the normal curve. Students see, for the first time, the normal curve. A number or normal curves are presented in relation to simple sample data in order to illustrate the importance of probability theory, the link between the sample and the population; and related probability terms. The chapter ends by demonstrating the important skill set of transforming raw scores (data) to standardized scores and vice versa. Key Formulas The following represent the key formulas for this chapter. PowerPoint slides are provided for each chapter. In addition to these slides, a PDF file containing only the formulas are also provided. Convert Raw to Standardized Score Convert Standardized to Raw Score Interactive Figures: The textbook contains interactive figures. You may wish to use these in a lecture. Students also have access to these. For this chapter, there is one interactive figure. 1. Figure 5.11 illustrates the areas under the normal curve These interactive figures can be found on in the eBook and the Library under Chapter 5 Resources. Typical Lecture Material We have provided two sample lectures below. You may wish to add in additional discipline specific information to make these more relevant to your students. Lecture 1: Objective: Understand the importance of the standard normal curve; and how to apply the standard normal curve to a known population mean and standard deviation. Review the following concept table with your students; and help them to fill in the definitions of each probability concept. The definitions that the students should come up with are in italics. Probability Concept Defintion Population The total number of indiiduals, objects or items that you are interested in, Sample A subset of the population. Normal Curve A theoretical probability distribution that has a symmetrical bell shape. Sample Mean The average score in a sample distribution Sample Variance A measure of the extent to which the data varies from its mean. Sample Standard Deviation The average amount, measured in standard units in which the data scores vary (positively or negatively) from its sample mean. Population Mean The theoretical average of population on a measure of interest. Population Standard Deviation The theortical average amount, measured in standard units, in which the data scores theoretically vary positively or negatively) from the theoretical population mean. Population Variance A theoretical measure of the extent to which the sample data varies from the theoretical population mean. Area under the Normal Curve Represents 100% of the cases or possible outcomes of interest. Raw score It is the observed score on our variable of interst. Standardized Score A raw score that has been converted into the number of standard deviations that raw score is from the theoretical population mean. Example 1: Reiterate to your students the focus of the chapter was to communicate the importance of understanding and committing the properties of the standard normal curve to memory. 1.a. Ask your students to draw a standard normal curve with corresponding area measurements and cumulative percentages. Note: It should look like the following. Then pose various questions to the students such as: a) What percentage of the population fall within 1 standard deviation from the population mean? Answer: 68.26% b) What percentage of the population fall within 2 standard deviations from the population mean? Answer: 95.44% Example 1.b. Read the following scenario to your students: 0.13% 2.15% 34.13% 34.13% 13.59% 2.15% 13.59% 0.13% -3σ -2σ -1σ µ +1σ +2σ +3σ 0.13% 2.28% 15.87% 50.0% 84.13% 97.72% 99.87% You are interested in examining anxiety levels among undergraduate students who are required to complete a mandatory introduction to Social Statistics class. The anxiety scale is out of 100; with a population mean (µ) of 75 and a standard deviation (σ) of 5. Ask your students to create a normal probability curve theoretically representing your data. Then pose various questions to the students such as: a) What percentage of the population has an anxiety score ranging between 70 and 90? b) If the entire population of students taking this fall statistics course is 2000, what number if students can we expect to have anxiety scores ranging between 70 and 90? c.) A student with an anxiety score of 62 wants to know how they compare with the remaining population of students. What might you tell them? Lecture 2: Objective: To understand standardized scores and their purpose; to learn how to convert a raw score to a standardized score and vice versa. Example 1: Remind students of the previous lecture’s example: You are interested in exploring anxiety levels among undergraduate students who are required to complete a mandatory introduction to Social Statistics course. The anxiety scale is out of 100; with a population mean (µ) of 75 and a standard deviation (σ) of 5. You have come across two students in class that you have noted are extremely hard working. They both study diligently and meet with you on a regular basis to prep for exams; however, they perform very differently on assignments. You wonder to yourself how many standard deviations each of them is from the population mean with respect to their ‘stats course’ anxiety; as maybe it is their level of anxiety which might be explaining the differences in performance. Mary has an anxiety score of 76. Mark has an anxiety score of 87. Pose the following questions to your students: 1. How many standard deviations are Mary and Mark’s anxiety scores from the mean of the student population? 2. What percentage of students score higher than lower than Mark and Mary on the anxiety scale? 3. If we know that Mary’s friend Scott is -1.25 standard deviations from the mean, what is his raw anxiety score? 4. What about Mark’s friend, Colleen? Colleen is 1.33 standard deviations from the mean. What is her raw anxiety score? 5. In what percentile does Colleen fall under the standard normal curve? 6. What is the probability of a randomly selected student having an anxiety score greater than 63? 7. What is the probability of a randomly selected student having an anxiety score less than 50? Solutions to End-of-Chapter Problems Problem 5-1 a. P(0 <= z = 2.01) = 0.0222 (LO2/LO3) (LO5) c. P(-1.09 <= z <= 1.11) = 0.7286 (LO2/LO3) (LO5) d. P(z <= -3.05) = 0.0011 (LO2/LO3) (LO5) Problem 5-2 a. P(daily census totals less than 8500) = 0.05750 or 5.75% (LO5/LO6) b. P(daily census totals greater than 8500) = 1 - 0.05750 = 0.9425 or 94.25% (LO5/LO6) c. P(daily census totals between 6000 and 12000) = 0.7726 or 77.26% (LO6) Problem 5-3 a. Minimum score = 588 (LO7) b. Minimum score = 452 (LO7) c. Score to not go to summer school = 183 (LO7) Solutions to Interactive Exercises Question 5-1 The average heating cost of a house in a city has an approximately normal distribution with a mean of $150/month and a standard deviation of $20.00/month. What is the probability that a randomly selected household pays a) more than $190.00 / month? b) less than $130.00/month c) between $130 and $190 /month Feedback: a) 0.0228 b) 0.1587 c) 0.1359 Question 5-2 The weight of Kalahari male lions, is normally distributed with a mean of $194 kg and SD of 3.5 kg. a) What is the probability that a randomly selected cat weighs (i) less than 194 kg? (ii) between 191 kg and 197 kg? b) Is it unusual that a randomly selected cat weigh more than 204 kg? Explain your reasoning. c) What should the weight of the cat be so that only 10% of all the cats weigh greater than this one? Feedback: a) (i) = 0.50 (ii) = 0.5961 b) Probability = 0.0026. This is a small probability. c) 209.7575 Question 5-3 The test scores in a mathematical reasoning test are normally distributed with a mean of 75 points. 5.48% of the students scored more than 80 points. What is the standard deviation of the scores? Feedback: Standard deviation = 3.125 Solutions to SPSS Exercises Exercise 5-1 Answer the following questions to the best of your ability using SPSS. You should be able to find the associated probabilities using the transform function. a) For the standard normal curve find the specific areas given below and compare your results using the statistical table provided in the text. a. To the left of z=-1.32 – 0.0934 b. To the right of z=1.52 – 0.0643 c. Between z=0.32 and z=1.92 – 0.3471 d. Between z=-2.42 and z=-1.77 – 0.0306 LO: 2, 3, 5 Page: 136-140, 143 b) Let z be a standard normal variable. Find the given probabilities using SPSS. a. P(-1.20 z  2.64) – 0.8808 b. P(z  -1.20) – 0.8849 c. P(-0.73  z  3.12) – 0.7664 LO: 2, 3, 5 Page: 136-140, 143 c) Let x have a normal distribution with mean () = 10 and standard deviation () = 2. Use SPSS to find the probability that a value selected at random from this distribution will be between 11 and 14. P(11  x  14) = .2857 LO: 4 Page: 140 Equations from Chapter 5 Scott R. Colwell and Edward M. Carter c 2012 Equation 5.2: Standardized Score z = x −µ σ Where: z = is the standardized score x = is any score along the x-axis µ = is the mean value of all of the scores σ = is the standard deviation Equation 5.13: Solving for x x = zσ + µ Where: z = is the standardized score x = is any score along the x-axis µ = is the mean value of all of the scores σ = is the standard deviation Chapter 6 Getting From the Sample to the Population: Sample Distribution versus Sampling Distributions Learning Objectives: 1. Define statistical inference. 2. Explain the difference between a sample statistic and a population parameter. 3. Explain the difference between a random and non-random sample. 4. State and explain four different ranacdom sampling methods. 5. Explain sampling error. 6. Explain the term “sampling distribution of means.” 7. Briefly explain why the three statements of the central limit theorem are true. 8. Explain what is meant by the mean of the sampling distribution means and standard error of the mean. 9. Describe why we can use the sampling distribution to make inference about a population based on a random sample for that population. Chapter Summary In this chapter, students broaden their knowledge about the normal curve and probability theory. The chapter begins with drawing the distinctions between a sample statistic and a population parameter; and then moves onto describing four different types of sampling. Students are then exposed to the three arguments of the central limit theorem, its associated formulas, and the its implications for making inferences from the sample to the population of interest. Key Formulas The following represent the key formulas for this chapter. PowerPoint slides are provided for each chapter. In addition to these slides, a PDF file containing only the formulas are also provided. Mean of the Sampling Distribution Means Standard Error of the Mean Interactive Figures: The textbook contains interactive figures. You may wish to use these in a lecture. Students also have access to these. For this chapter, there are three interactive figures. 1. Figure 6.3 provides an example of a sampling distribution of means. 2. Figure 6.8 illustrates the standard error and sampling distributions. 3. Figure 6.9 illustrates a skewed population versus normal sample distribution These interactive figures can be found on in the eBook and the Library under Chapter 6 Resources. Typical Lecture Material We have provided two sample lectures below. You may wish to add in additional discipline specific information to make these more relevant to your students. Lecture 1: Objective: Understand the concept of statistical inference, the distinction between sample statistics and population parameters; and learn four different types of sampling. Review the following concept table with your students; and help them to fill in the definitions of each statistical concept. The definitions that the students should come up with are in italics. Statistical Concept Defintion Statistical Inference Process of making statements about the broader population based on the use of sample data from that population. Sample Statistic A numeric value derived from the sample data. Population Parameter A numeric value in the population. Generalizable Refers to statiscal demonstration that are results from our sample can be applied to the population of interest (i.e. the population from which we extracted/selected our sample). Probability Sampling The method of creating a sample from a population using random selection. Non-Probability Sampling The method of creating a sample from a ppulation that does not utilize random selection. Sampling Error The difference between our sample mean and the population mean. Point Estimate A statistic (e.g. sample mean) that is used to estimate a population parameter. Example 1: Ask your students to match the corresponding statistical notation with its appropriate identification Example 2: Read the following scenario to your students. Population Standard Deviation Sample Mean Population Mean Population Variance Sample Standard Deviation Sample Variance 𝜎 𝑥 𝜎2 𝑠2 s 𝜇 You are interested in exploring Canadian high school students’ levels of motivation to attend University. The total population of high school students in Canada is 30; and your goal is to obtain a random sample of 10 students in order to complete your study of post-secondary motivation. The good thing about your research area is that every high school in Canada completes an ‘Assessment of Post-Secondary Motivation’ with each graduating student; which means you have access to the entire population’s data on levels of post-secondary motivation. Unfortunately you only have enough research funding to actually analyze 10 students data (your sample). The following table outlines each population member’s unique identification number, their motivation score (out of 20, with higher numbers meaning more motivation to attend post- secondary school), whether they reside in an Urban or Rural setting and whether or not they are male or female. Population ID # Motivation Score Geography U = Urban R = Rural Gender Male = 0 Female = 1 1 18 R 0 2 17 R 1 3 12 U 1 4 10 U 0 5 3 U 0 6 4 U 0 7 16 R 0 8 19 R 1 9 20 R 1 10 1 U 0 11 0 R 1 12 5 U 1 13 7 R 0 14 9 U 1 15 13 U 0 16 14 U 1 17 15 R 0 18 18 U 1 19 11 R 1 20 8 R 0 21 19 R 1 22 18 U 0 23 15 U 1 24 12 U 0 25 16 U 0 26 14 U 0 27 13 R 1 28 10 U 0 29 11 R 0 30 18 U 0 Population Mean (µ) = 12.2 Males = 17 (56.67% of population) Rural = 13 (43.3% of population) Population Standard Deviation = 5.598 Pose the following questions to your students: 1. Which members of the population belong to your sample using a simple random sampling approach? 2. Which members of the population belong to your sample using a systematic sampling approach? What is the sample’s mean motivation score? 3. Which members of the population belong to your sample using a stratified random sampling approach? 4. Which members of the population belong to your sample using a cluster sampling approach? Lecture 2: Objective: To conceptually understand the central limit theorem, the theorem’s three tenants and the computation of the mean of the sampling distribution means and the standard error of the mean. Remind your student’s of the post-secondary motivation assessment example from the previous lecture; and post on the blackboard or overhead projector the corresponding data table. Pose the following questions to your students: 1. What is the standard error of the mean? 2. If you used a random numbers table to generate 10 simple random samples (n =10) of the post-secondary motivation data, what would be the mean of the sampling distribution means? 3. If you take another 5 simple random samples (n =10) from the population, what does your mean of the sampling distributions change to? Is it closer to the mean of the population? Why or why not? Solutions to End-of-Chapter Problems Problem 6-1 a. Sample Number… …Includes …has a sample mean (x¯ ) 1 2 2 2 2 2 4 3 3 2 6 4 4 2 8 5 5 4 2 3 6 4 4 4 7 4 6 5 8 4 8 6 9 6 2 4 10 6 4 5 11 6 6 6 12 6 8 7 13 8 2 5 14 8 4 6 15 8 6 7 16 8 8 8 Therefore the mean of the sampling distribution means (𝜇𝑥̅)is 5. (LO6/LO7/LO8) b. The standard error of the mean (𝜎𝑥̅)is 1.581. (LO6/LO7/LO8) Avg. of pairs Squared deviations of the average pairs from overall mean 5 2,2 2 9 2,4 3 4 2,6 4 1 2,8 5 0 4,2 3 4 4,4 4 1 4,6 5 0 4,8 6 1 6,2 4 1 6,4 5 0 6,6 6 1 6,8 7 4 8,2 5 0 8,4 6 1 8,6 7 4 8,8 8 9 Mean of averages = 5 Average of squared deviations = 2.5 The mean of the average of the pairs is 5, which is the mean of their original numbers 2,4,6,8. The variance of the average of the pairs is 2.5 so the population variance of the 16 numbers in column 2 is 2.5. The third column is just the intermediate step in calculating the variance of the 16 numbers in column 2. c. Since 𝜎 = 𝜎𝑥̅×√𝑛 then 1.581 × √2 = 1.581 × 1.414 = 2.236 which is the same value as σ. Problem 6-2 a. Standard deviation = 3.75. (LO6/LO7/LO8) b. Standard deviation = 3. (LO6/LO7/LO8) c. If the 25 are sampled without replacement, then the samples would be considered dependent. As such, our formula for standard deviation would be incorrect. (LO6/LO7/LO8) Problem 6-3 a. Standard error = 0.50 (LO6/LO7/LO8) b. Standard error = 0.40 (LO6/LO7/LO8) c. Standard error = 0.33 (LO6/LO7/LO8) d. Standard error = 0.29 (LO6/LO7/LO8) Solutions to Interactive Exercises Question 6-1 The working hours of resident doctors are 80 hours a week on an average with standard deviation of 12 hours. If a sample of 16 people is selected and their average working hours is computed: a) Describe the sampling distribution. b) If you want to the probability that the average working time of 16 doctors is less than 75 hours, what assumption would you make about the population? c) What is the probability that that the average working time of 16 doctors is less than 75 hours? Feedback: a) Mean = 80 Standard deviation = 12/4 = 3 Shape = There is no information about the shape of the population distribution. Also, the sample size is too small to apply Central limit theorem. b) The population is Normal. c) 0.9525 Exercise 6-2 The mean age of the participants of a city marathon indicated that the population mean is 31.4 years with standard deviation of 7.2 years. If a random sample of 36 were selected, within which two values does the middle 95% exist? Feedback: (29.048, 33.752) Exercise 6-3 Surgical wait time is defined as the time between when a surgery is ordered and when it is performed. In a particular province the wait time for knee replacement surgery is approximately normally distributed with a mean of 254 days and a standard deviation of 18 days. In less than how many days is 95% of the operations are done? Feedback: 324 days Solutions to SPSS Exercises Question 6-1 Assume that scores on a test are normally distributed in the population with a mean of 100 and a standard deviation of 15. You have 6 people with scores of 60, 75, 95, 110, 135, and 140. You want to tell them what proportion of people in the population have scores less than or equal to theirs. a) Input the data and save the data file. Call the variable - scores. b) Compute the standardized scores for the data values using transform. Compute a variable named – z-scores – using this function. Ans: -2.67 -1.67 -.33 .67 2.33 2.67 c) Compute the cumulative probabilities for the standardized scores – call this variable cumprobs. Ans: .0038 .0478 .3694 .7475 .9902 .9962 d) Check to see that your computed results to 4 decimal places is the same as those of the table that you could compute manually if you did not have SPSS on hand. Ans: They should be identical. LO: 6, 7, 8 Page: 170-179 Equations from Chapter 6 Scott R. Colwell and Edward M. Carter c 2012 Equation 6.1: Mean of the Sampling Distribution Means µx¯ = Px¯ n Where: µx¯ = mean of the sampling distribution means Px¯ = the sum of the sampling means n = the number of samples Equation 6.3: Standard Error of the Mean σx¯ = √σ n Where: σx¯ = standard error of the mean σ = population standard deviation n = the sample size Solution Manual for Introduction to Statistics for Social Sciences Scott R. Colwell, Edward M. Carter 9780071319126