FINAL EXAMINATION STUDY MATERIAL PART I UNIT V FINAL EXAMINATION STUDY MATERIAL PART I Copyright ©2011 Brooks/Cole, Cengage Learning

Inferential Statistics Involves Estimation Hypothesis Testing Purpose Make decisions about population characteristics or parameters on basis of sample statistics. Population? Copyright ©2011 Brooks/Cole, Cengage Learning

Reference Reading Material From Intro Stats 3RD Edition Chapter 18 Sampling Distribution For One Sample Proportion Sampling Distribution For One Sample Mean Chapter 22 Sampling Distribution For The Difference Between Two Proportions Chapter 24 Sampling Distribution For The Difference Between Two Means Copyright ©2011 Brooks/Cole, Cengage Learning

Reminders: Parameters, Statistics A statistic is a numerical value computed from a sample. Its value may differ for different samples. e.g. sample mean , sample standard deviation s, and sample proportion . A parameter is a numerical value associated with a population. Considered fixed and unchanging. e.g. population mean m, population standard deviation s, and population proportion p. Copyright ©2011 Brooks/Cole, Cengage Learning

Sampling Distribution For One Sample Proportion PROBLEM FORMULATION: SUPPOSE THAT p IS AN UNKNOWN PROPORTION OF ELEMENTS OF A CERTAIN TYPE S IN A POPULATION. ALTERNATIVELY, WE WISH TO ESTIMATE THE PROPORTION FALLING INTO A CATEGORY OF A CATEGORICAL VARIABLE. EXAMPLES PROPORTION OF LEFT - HANDED PEOPLE; PROPORTION OF HIGH SCHOOL STUDENTS WHO ARE FAILING A READING TEST; PROPORTION OF VOTERS WHO WILL VOTE FOR MR. X. Copyright ©2011 Brooks/Cole, Cengage Learning

Estimation of p TO ESTIMATE p, WE SELECT A SIMPLE RANDOM SAMPLE (SRS), OF SIZE SAY, n = 1000, AND COMPUTE THE SAMPLE PROPORTION. SUPPOSE THE NUMBER OF THE TYPE WE ARE INTERESTED IN, IN THIS SAMPLE OF n = 1000 IS x = 437. THEN THE SAMPLE PROPORTION IS COMPUTED USING THE FORMULA Copyright ©2011 Brooks/Cole, Cengage Learning

Estimation of p Copyright ©2011 Brooks/Cole, Cengage Learning

WHAT IS THE ERROR OF ESTIMATION? THAT IS, WHAT IS IS THERE A MODEL (PROBABILITY DISTRIBUTION MODEL) THAT CAN HELP US FIND THE BEST ESTIMATE OF THE TRUE PROPORTION OF p? LET’S START THE ANALYSIS BY FIRST ANSWERING THE SECOND QUESTION. Copyright ©2011 Brooks/Cole, Cengage Learning

THE APPROACH   Copyright ©2011 Brooks/Cole, Cengage Learning

Sample Distribution Table Sample of size n X = Number of type S Sample 1 Sample 2 . . . . . . Sample k Copyright ©2011 Brooks/Cole, Cengage Learning

Shape of Histogram of the p(hats) samples p(hats) p Copyright ©2011 Brooks/Cole, Cengage Learning

REMARKS ON OBSERVING THE HISTOGRAM THE HISTOGRAM ABOVE IS AN EXAMPLE OF WHAT WE WOULD GET IF WE COULD SEE ALL THE PROPORTIONS FROM ALL POSSIBLE SAMPLES. THAT DISTRIBUTION HAS A SPECIAL NAME. IT IS CALLED THE SAMPLING DISTRIBUTION OF THE PROPORTIONS. OBSERVE THAT THE HISTOGRAM IS UNIMODAL, ROUGHLY SYMMETRIC, AND IT’S CENTERED AT p. THE SAMPLING DISTRIBUTION FOLLOWS THE NORMAL MODEL. Copyright ©2011 Brooks/Cole, Cengage Learning

Definition: Sampling Distribution For One Sample Proportion Statistics as Random Variables Each new sample taken  value of the sample statistic will change. The distribution of possible values of a statistic for repeated samples of the same size from a population is called the sampling distribution of the statistic. Many statistics of interest have sampling distributions that are approximately normal distributions Copyright ©2011 Brooks/Cole, Cengage Learning

Sampling Distributions What does the sampling distribution tell us? The sampling distribution allows us to make statements about where we think the corresponding population parameter is and how precise these statements are likely to be. Copyright ©2011 Brooks/Cole, Cengage Learning

Sampling Distributions Copyright ©2011 Brooks/Cole, Cengage Learning

Example – Gene For A Disease Suppose (unknown to us) 40% of a population carry the gene for a disease, (p = 0.40). We will take a random sample of 25 people from this population and count X = number with gene. Although we expect (on average) to find 10 people (40%) with the gene, we know the number will vary for different samples of n = 25. In this case, X is a binomial random variable with n = 25 and p = 0.4. Copyright ©2011 Brooks/Cole, Cengage Learning

Many Possible Samples Four possible random samples of 25 people: Note: Sample 1: X =12, proportion with gene =12/25 = 0.48 or 48%. Sample 2: X = 9, proportion with gene = 9/25 = 0.36 or 36%. Sample 3: X = 10, proportion with gene = 10/25 = 0.40 or 40%. Sample 4: X = 7, proportion with gene = 7/25 = 0.28 or 28%. Note: Each sample gave a different answer, which did not always match the population value of p. Although we cannot determine whether one sample will accurately reflect the population, statisticians have determined what to expect for most possible samples. Copyright ©2011 Brooks/Cole, Cengage Learning

Example – Gene For A Disease Copyright ©2011 Brooks/Cole, Cengage Learning

Example: Sampling Distribution For Smoking 18% of US adults smoke. How much would we expect the proportion of smokers in a sample of size 1000 to vary from sample to sample? A histogram was drawn to display the results of a simulation. The mean is 0.18 = the population proportion. The standard deviation was calculated as 0.0122. Normal: 68-95-99.7 rule works. 95% of all proportions are within 0.0244 of the mean. This is very close to the true value: 95.41 Copyright ©2011 Brooks/Cole, Cengage Learning

Example: Sampling Distribution For Smoking Copyright ©2011 Brooks/Cole, Cengage Learning

WHAT THEN IS THE APPROPRIATE PROBABILITY MODEL? ANSWER: IT IS AMAZING AND FORTUNATE THAT A NORMAL MODEL IS JUST THE RIGHT ONE FOR THE HISTOGRAMS OF SAMPLE PROPORTIONS. HOW GOOD IS THE NORMAL MODEL? IT IS GOOD IF THE FOLLOWING ASSUMPTIONS AND CONDITIONS HOLD. Copyright ©2011 Brooks/Cole, Cengage Learning

ASSUMPTIONS AND CONDITIONS INDEPENDENCE ASSUMPTION: THE SAMPLED VALUES MUST BE INDEPENDENT OF EACH OTHER. SAMPLE SIZE ASSUMPTION: THE SAMPLE SIZE, n, MUST BE LARGE ENOUGH REMARK: ASSUMPTIONS ARE HARD – OFTEN IMPOSSIBLE TO CHECK. THAT’S WHY WE ASSUME THEM. GLADLY, SOME CONDITIONS MAY PROVIDE INFORMATION ABOUT THE ASSUMPTIONS. Copyright ©2011 Brooks/Cole, Cengage Learning

CONDITIONS RANDOMIZATION CONDITION: THE DATA VALUES MUST BE SAMPLED RANDOMLY. IF POSSIBLE, USE SIMPLE RANDOM SAMPLING DESIGN TO SAMPLE THE POPULATION OF INTEREST. 10% CONDITION: THE SAMPLE SIZE, n, MUST BE NO LARGER THAN 10% OF THE POPULATION OF INTEREST. SUCCESS/FAILURE CONDITION: THE SAMPLE SIZE HAS TO BE BIG ENOUGH SO THAT WE EXPECT AT LEAST 10 SUCCESSES AND AT LEAST 10 FAILLURES. THAT IS, Copyright ©2011 Brooks/Cole, Cengage Learning

Sampling Distribution for a Sample Proportion (The Central Limit Theorem) Let p = population proportion of interest or binomial probability of success. Let = sample proportion or proportion of successes. If numerous random samples or repetitions of the same size n are taken, the distribution of possible values of is approximately a normal curve distribution with Mean = p Standard deviation = s.d.( ) = This approximate distribution is sampling distribution of . Copyright ©2011 Brooks/Cole, Cengage Learning

Estimating the Population Proportion from a Single Sample Proportion In practice, we don’t know the true population proportion p, so we cannot compute the standard deviation of , s.d.( ) = . In practice, we only take one random sample, so we only have one sample proportion . Replacing p with in the standard deviation expression gives us an estimate that is called the standard error of . s.e.( ) = . . Copyright ©2011 Brooks/Cole, Cengage Learning

Standard Deviation Versus Standard Error of a Statistic • Standard deviation of a sampling distribution measures the variation among possible values of the sample statistic over all possible random samples. We include the name of the statistic being studied, e.g. the standard deviation of the mean. • Standard error describes the estimated value of the standard deviation of a statistic. We include the name of the statistic, e.g. the standard error of the mean. Copyright ©2011 Brooks/Cole, Cengage Learning

More Examples for which Rule Applies Election Polls: to estimate proportion who favor a candidate; units = all voters. Television Ratings: to estimate proportion of households watching TV program; units = all households with TV. Consumer Preferences: to estimate proportion of consumers who prefer new recipe compared with old; units = all consumers. Testing ESP: to estimate probability a person can successfully guess which of 5 symbols on a hidden card; repeatable situation = a guess. Copyright ©2011 Brooks/Cole, Cengage Learning

Example: Possible Sample Proportions Favoring a Candidate Suppose 40% all voters favor Candidate C. Pollsters take a sample of n = 2400 voters. Rule states the sample proportion who favor X will have approximately a normal distribution with mean = p = 0.4 and s.d.( ) = Histogram at right shows sample proportions resulting from simulating this situation 400 times. Copyright ©2011 Brooks/Cole, Cengage Learning

EXAMPLE ASSUME THAT 30% OF STUDENTS AT A UNIVERSITY WEAR CONTACT LENSES (A) WE RANDOMLY PICK 100 STUDENTS. LET REPRESENT THE PROPORTION OF STUDENTS IN THIS SAMPLE WHO WEAR CONTACTS. WHAT’S THE APPROPRIATE MODEL FOR THE DISTRIBUTION OF ? SPECIFY THE NAME OF THE DISTRIBUTION, THE MEAN, AND THE STANDARD DEVIATION. BE SURE TO VERIFY THAT THE CONDITIONS ARE MET. (B) WHAT’S THE APPROXIMATE PROBABILITY THAT MORE THAN ONE THIRD OF THIS SAMPLE WEAR CONTACTS? Copyright ©2011 Brooks/Cole, Cengage Learning

SOLUTION Copyright ©2011 Brooks/Cole, Cengage Learning

EXAMPLE INFORMATION ON A PACKET OF SEEDS CLAIMS THAT THE GERMINATION RATE IS 92%. WHAT’S THE PROBABILITY THAT MORE THAN 95% OF THE 160 SEEDS IN THE PACKET WILL GERMINATE? BE SURE TO DISCUSS YOUR ASSUMPTIONS AND CHECK THE CONDITIONS THAT SUPPORT YOUR MODEL. Copyright ©2011 Brooks/Cole, Cengage Learning

SOLUTION Copyright ©2011 Brooks/Cole, Cengage Learning

Sampling Distribution for Difference in Two Sample Proportions For the populations: p1 = population proportion for the first population. p2 = population proportion for the second population. Parameter: p1 – p2 = difference in popul proportions. For the samples: = sample proportion for sample from first popul. = sample proportion for sample from second popul. Statistic: = difference in sample proportions. Copyright ©2011 Brooks/Cole, Cengage Learning

Familiar Examples Estimating the difference between two populations with regard to the proportion falling into a category of a qualitative variable. Example research questions: 1. How much difference is there between the proportions that would quit smoking if taking the antidepressant buproprion (Zyban) versus if wearing a nicotine patch? 2. How much difference is there between men who snore and men who don’t snore with regard to the proportion who have heart disease? Population parameter: p1 – p2 = difference between the two population proportions. Sample estimate: = difference between the two sample proportions. Copyright ©2011 Brooks/Cole, Cengage Learning

Conditions Sampling distribution of difference in two independent sample proportions is approximately normal when: Condition 1: Sample proportions are available for two independent samples, randomly selected from the two populations of interest. Condition 2: All of the quantities n1p1, n1(1 – p1), n2p2, and n1(1 – p2) are at least 10. These quantities represent the expected numbers of successes and failures in each sample. Copyright ©2011 Brooks/Cole, Cengage Learning

Sampling Distribution for the Difference in Two Sample Proportions Mean = p1 – p2 Standard deviation = s.d.( ) = When we don’t know the populations proportions, we use the sample proportions, resulting in: Standard error = s.e.( ) = Copyright ©2011 Brooks/Cole, Cengage Learning

Example: Men, Women, Death Penalty Suppose 37% of women and 27% of men oppose death penalty, p1 = .37 and p2 = .27, for a difference p1 – p2 = .37 – .27 = .10 For independent random samples of 1017 women and 885 men, the sampling distribution of is approx normal with mean .10 and standard deviation: Note: 2008 survey gave observed difference of .36 – .285 = .075, which is not unusual. Copyright ©2011 Brooks/Cole, Cengage Learning

EXAMPLES FROM PRACTICE SHEET Copyright ©2011 Brooks/Cole, Cengage Learning

Sampling Distribution For One Sample Mean Suppose we want to estimate the mean weight loss for all who attend clinic for 10 weeks. Suppose (unknown to us) the distribution of weight loss is approximately N(8 pounds, 5 pounds). We will take a random sample of 25 people from this population and record for each X = weight loss. We know the value of the sample mean will vary for different samples of n = 25. What do we expect those means to be? Copyright ©2011 Brooks/Cole, Cengage Learning

Many Possible Samples Four possible random samples of 25 people: Note: Sample 1: Mean = 8.32 pounds, standard deviation = 4.74 pounds. Sample 2: Mean = 6.76 pounds, standard deviation = 4.73 pounds. Sample 3: Mean = 8.48 pounds, standard deviation = 5.27 pounds. Sample 4: Mean = 7.16 pounds, standard deviation = 5.93 pounds. Note: Each sample gave a different answer, which did not always match the population mean of 8 pounds. Although we cannot determine whether one sample mean will accurately reflect the population mean, statisticians have determined what to expect for most possible sample means. Copyright ©2011 Brooks/Cole, Cengage Learning

Familiar Examples Estimating the mean of a quantitative variable. Example research questions: What is the mean time that college students watch TV per day? What is the mean pulse rate of women? Population parameter: m = population mean for the variable Sample estimate: = sample mean for the variable Copyright ©2011 Brooks/Cole, Cengage Learning

Example: Mean Hours of Sleep for College Students Survey of n = 190 college students. “How many hours of sleep did you get last night?” Sample mean = 7.1 hours. If we repeatedly took samples of 190 and each time computed the sample mean, the histogram of the resulting sample mean values would look like the histogram at the right: Copyright ©2011 Brooks/Cole, Cengage Learning

The Normal Curve Approximation Rule for Sample Means (The Central Limit Theorem) Let m = mean for population of interest. Let s = standard deviation for population of interest. Let = sample mean. If numerous random samples of the same size n are taken, the distribution of possible values of is approximately a normal curve distribution with Mean = m Standard deviation = s.d.( ) = This approximate distribution is sampling distribution of . Copyright ©2011 Brooks/Cole, Cengage Learning

Standard Error of the Mean In practice, the population standard deviation s is rarely known, so we cannot compute the standard deviation of , s.d.( ) = . In practice, we only take one random sample, so we only have the sample mean and the sample standard deviation s. Replacing s with s in the standard deviation expression gives us an estimate that is called the standard error of . s.e.( ) = . For a sample of n = 25 weight losses, the standard deviation is s = 4.74 pounds. So the standard error of the mean is 0.948 pounds. Copyright ©2011 Brooks/Cole, Cengage Learning

ASSUMPTIONS AND CONDITIONS INDEPENDENCE ASSUMPTION: THE SAMPLED VALUES MUST BE INDEPENDENT OF EACH OTHER SAMPLE SIZE ASSUMPTION: THE SAMPLE SIZE MUST BE SUFFICIENTLY LARGE. REMARK: WE CANNOT CHECK THESE DIRECTLY, BUT WE CAN THINK ABOUT WHETHER THE INDEPENDENCE ASSUMPTION IS PLAUSIBLE. Copyright ©2011 Brooks/Cole, Cengage Learning

CONDITIONS RANDOMIZATION CONDITION: THE DATA VALUES MUST BE SAMPLED RANDOMLY, OR THE CONCEPT OF A SAMPLING DISTRIBUTION MAKES NO SENSE. IF POSSIBLE, USE SIMPLE RANDOM SAMPLING DESIGN TO ABTAIN THE SAMPLE. 10% CONDITION: WHEN THE SAMPLE IS DRAWN WITHOUT REPLACEMENT (AS IS USUALLY THE CASE), THE SAMPLE SIZE, n, SHOULD BE NO MORE THAN 10% OF THE POPULATION. LARGE ENOUGH SAMPLE CONDITION: IF THE POPULATION IS UNIMODAL AND SYMMETRIC, EVEN A FAIRLY SMALL SAMPLE IS OKAY. IF THE POPULATION IS STRONGLY SKEWED, IT CAN TAKE A PRETTY LARGE SAMPLE TO ALLOW USE OF A NORMAL MODEL TO DESCRIBE THE DISTRIBUTION OF SAMPLE MEANS Copyright ©2011 Brooks/Cole, Cengage Learning

The Normal Curve Approximation Rule for Sample Means Normal Approximation Rule can be applied in two situations: Situation 1: The population of measurements of interest is bell-shaped and a random sample of any size is measured. Situation 2: The population of measurements of interest is not bell-shaped but a large random sample is measured. Note: Difficult to get a Random Sample? Researchers usually willing to use Rule as long as they have a representative sample with no obvious sources of confounding or bias. Copyright ©2011 Brooks/Cole, Cengage Learning

Examples for which Rule Applies Average Weight Loss: to estimate average weight loss; weight assumed bell-shaped; population = all current and potential clients. Average Age At Death: to estimate average age at which left-handed adults (over 50) die; ages at death not bell-shaped so need n  30; population = all left-handed people who live to be at least 50. Average Student Income: to estimate mean monthly income of students at university who work; incomes not bell-shaped and outliers likely, so need large random sample of students; population = all students at university who work. Copyright ©2011 Brooks/Cole, Cengage Learning

Example: Hypothetical Mean Weight Loss Suppose the distribution of weight loss is approximately N(8 pounds, 5 pounds) and we will take a random sample of n = 25 clients. Rule states the sample mean weight loss will have a normal distribution with mean = m = 8 pounds and s.d.( ) = pound Histogram at right shows sample means resulting from simulating this situation 400 times. Empirical Rule: It is almost certain that the sample mean will be between 5 and 11 pounds. Copyright ©2011 Brooks/Cole, Cengage Learning

EXAMPLES FROM PRACTICE SHEET Copyright ©2011 Brooks/Cole, Cengage Learning

Effect Of Increasing the Size of the Sample Suppose we take n = 100 people instead of just 25. The standard deviation of the mean would be s.d.( ) = pounds. For samples of n = 25, sample means are likely to range between 8 ± 3 pounds => 5 to 11 pounds. For samples of n = 100, sample means are likely to range only between 8 ± 1.5 pounds => 6.5 to 9.5 pounds. Larger samples tend to result in more accurate estimates of population values than smaller samples. Copyright ©2011 Brooks/Cole, Cengage Learning

Sampling Distribution for Difference in Two Sample Means Let m1 = population mean for first population. Let m2 = population mean for second population. Parameter: m1 – m2 = difference in population means. Let = sample mean for sample from first population. Let = sample mean for sample from second population. Statistic: = difference in sample means. Let s1 = population standard deviation for first population. Let s2 = population standard deviation for second population. Let s1 = sample std deviation for sample from first population. Let s2 = sample std deviation for sample from second population. Copyright ©2011 Brooks/Cole, Cengage Learning

Familiar Examples Estimating the difference between two populations with regard to the mean of a quantitative variable. Example research questions: 1. How much difference is there in average weight loss for those who diet compared to those who exercise to lose weight? 2. How much difference is there between the mean foot lengths of men and women? Population parameter: m1 – m2 = difference between the two population means. Sample estimate: = difference between the two sample means. Copyright ©2011 Brooks/Cole, Cengage Learning

Conditions for Sampling Distribution of to be Approx Normal An important condition in this situation is that the two samples must be independent. How? Take separate random samples from each of two populations such as men and women. Take a random sample from a population and divide the sample into two groups based on a categorical variable such as smoker and nonsmoker. Randomly assign participants in a randomized experiment to two treatment groups such as exercise or diet. Copyright ©2011 Brooks/Cole, Cengage Learning

Conditions for Sampling Distribution of to be Approx Normal In addition to independent samples, one of the following two situations must hold: Situation 1: The populations of measurements are both bell-shaped and random samples of any size are measured. Situation 2: Large random samples are measured from each population. Arbitrary definition of large is both samples are at least 30, but extreme outliers or extreme skewness in either sample may require even larger samples. Copyright ©2011 Brooks/Cole, Cengage Learning

Standard Error of the Mean Difference Standard deviation of : s.d.( ) = . Standard error of : s.e.( ) = . The standard error is used to estimate the standard deviation. Copyright ©2011 Brooks/Cole, Cengage Learning

EXAMPLES FROM PRACTICE SHEET Copyright ©2011 Brooks/Cole, Cengage Learning

The Central Limit Theorem (CLT) The Central Limit Theorem states that if n is sufficiently large, the sample means of random samples from a population with mean m and finite standard deviation s are approximately normally distributed with mean m and standard deviation . Technical Note: The mean and standard deviation given in the CLT hold for any sample size; it is only the “approximately normal” shape that requires n to be sufficiently large. Copyright ©2011 Brooks/Cole, Cengage Learning

Sampling Distribution for Any Statistic Every statistic has a sampling distribution, but the appropriate distribution may not always be normal, or even approximately bell-shaped. Construct an approximate sampling distribution for a statistic by actually taking repeated samples of the same size from a population and constructing a relative frequency histogram for the values of the statistic over the many samples. Copyright ©2011 Brooks/Cole, Cengage Learning