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Why do we need a synonym? Perhaps because for most people their first encounter(s) with Statistics were somewhat negative, daunting, impenetrable, arcane, etc. Why do we need Statistics? Because, as Norman and Streiner put it, “the world is full of variation, and sometimes it‘s hard to tell real differences from natural variation.” Statistics addresses the variability among people or within one person from one time to the next. The word “statistic” has a purely superficial resemblance to the word “sadistic”. But the word actually comes from the Latin word for the “state”, because the first data collection was for the purposes of the state – tax collection and military service. Birth and mortality rates appeared in England in the 17th century, about the same time that French mathematicians were laying the groundwork for probability by studying gambling problems. Applications to studies of heredity, agriculture and psychology were developed by the great English scientists, Galton, Pearson, and Fisher, who gave us many of techniques we use today: design of experiments, randomization, hypothesis testing, regression, and analysis of variance. With such a diversity of origin, it is not surprising that the word “statistics” means different things to different people. So is it possible to give a concise definition of the word “statistics”. Here are two. 1. Statistics involves the collection, summarization, presentation and interpretation of numerical facts, and the generalization of these facts. 2. Statistics is decision-making in the face of uncertainty. Although these definitions sound benign, some people have such a fear of this part of the research process that they never get beyond it. Here are four statements about Statistics. Statistical thinking will one day be as necessary for efficient citizenship as the ability to read and write. H.G. Wells “Data, data, data!” he cried impatiently. I can’t make bricks without clay.” Sherlock Holmes A statistical analysis, properly conducted, is a delicate dissection of uncertainties, a surgery of suppositions. M.J. Moroney Knowledge of statistical methods is not only essential for those who present statistical arguments; it is also needed by those on the receiving end. R.G.D. Allen And one more: “Statistics… the most important science in the whole world; for upon it depends the practical application of every other science and of every art; the one science essential to all political and social administration, all education, all organization based on experience, for it only gives results of our experience.” Question: Who was the author of this quote? Can you guess the “demographics” of the author: sex, living or dead, profession, etc. [ADD A LINK TO THE ANSWER] Answer: The author of the quote is Florence Nightingale, who gained fame for her nursing career, but who was a driving force in the collection and analysis of hospital medical statistics. It is fascinating that a nurse, who provided care one patient at a time, understood that you can’t know what works on an individual unless you have studied it in the aggregate. By the way, FN was elected a Fellow of the Statistical Society of London (now the Royal Statistical Society) in 1858, and became an Honorary Member of the American Statistical Association in 1874. Presentation of Material 1. Two Types of Data Origin of the word: The word ‘data’ is plural (the singular is ‘datum’); it comes from the Latin meaning ‘to give’; so in the current sense, data are the information given to us to analyze and interpret. Note that to be grammatically correct you should say, “data are” not “data is”. By saying (or writing) “data are” you demonstrate a much higher level of intelligence! We begin by thinking about types of data or variables. The simplest classification is a dichotomy. Data (or variables) are either categoric or measurement. The first principle of data analysis is to understand which type of data you have. But not only is this the first principle, it is undoubtedly the most important one. If you do not know what type of data you have you cannot choose an appropriate analysis! Categoric data: (also called discrete, or count data) – variables are categoric if observations can be put into distinct “bins”. In other words, there are a limited number of possible values that the variable can take. There are three subtypes of categoric data: Binary: the most basic categoric data, there are only two possible values; for example: Yes/No, Survive/Die, Accept/Reject, Male/Female, 0/1. Nominal: extension of binary to more than two categories, but the categories are unordered. Nominal means “named”. For example: marital status, eye colour, diagnosis type Ordinal: extension of binary to more than two categories, but the categories are ordered. Ordinal means “ordered”. For example, a 3-point scale of change – better, the same, worse; highest level of education; severity of illness. Typical 3-point, 5-point, or 7-point response scales of agreement, satisfaction, etc. are ordinal in nature. Measurement data (also called continuous, or interval) – Measurement variables are characterized by the involvement of some kind of measurement process such as a measuring instrument or questionnaire. Also, there are a large number of possible values with little repetition of each value. For example: age (in years), height and weight, systolic blood pressure, psychosocial instrument measures of depression or anxiety, visual analogue scale for pain or quality or life. Some variables can be expressed as more than one type of data. For example, age in years is a measurement variable, but can be turned into a categoric variable. It depends on the mechanism of measurement and the future use of the data. Some variables can be created from other variables. For example, psychosocial tests are usually “scored” by summing up responses to a number of individual items. While individual items are ordinal, the sum approximates a measurement variable. For example, the Beck Depression Inventory has 21 questions, each answered on a 4-point scale (0 to 3); the overall score is the sum of the responses to the 21 questions. In general, there is more information in measurement data than in categoric data. Recap: If you do not know what type of data you have you cannot choose an appropriate analysis!  Figure: Types of Data – the type of data often determines the type of analysis Exercise: Employees at the Belltown Smelter (affectionately known as the BS) must complete an employee questionnaire which is kept on file by the Human Resources Department. Following is a sample of the questions. For each, decide whether it is categoric or measurement, or possibly either. Date of birth Highest level of education Number of jobs in past 10 years Type of residence Number of children Before-taxes income in the last year before joining BS Anticipated after-taxes income after the first year with BS Job satisfaction score using a 10-item (each with 5-point scale) tool Alcohol consumption Absenteeism (# days of worked missed in a year) [LINK TO ANSWERS} Answers to Exercise on Types of Data: Date of birth: will be used to compute age, which is a measurement variable Highest level of education: likely to be categoric (e.g. less than high school, high school diploma, trade school, college degree, university degree). Note that “number of years of education” would be measurement, but would not very useful for analysis Number of jobs in past 10 years: probably categoric (e.g. 1, 2, 3 or more) Type of residence: categoric Number of children: likely to be treated as categoric (Yes or No re dependent children) Before-taxes income in the last year before joining BS: measurement Anticipated after-taxes income after the first year with BS: measurement *Job satisfaction score using a 10-item (each with 5-point scale) tool: each item is categoric, but the sum of them, which will give the job satisfaction score, will be measurement Alcohol consumption; likely to be categoric (e.g. never, occasional, frequent, etc., with each category defined as a range of number of drinks per week) Absenteeism (# days of worked missed in a year): measurement (but will probably be recoded into categories for analysis) 2. Two Areas of Statistical Analysis Descriptive Statistics includes tables and graphical summaries; and numerical summaries concerns the presentation, organization and summarization of data involves only the data at hand Inferential Statistics includes estimation, hypothesis testing, and model-building involves making inferences or generalizations from the sample information to the larger population uses the data at hand to comment on future data Descriptive Statistics form an extremely valuable part of statistical analysis. Only after a thoughtful descriptive analysis should you proceed to inferential analysis. Perhaps the most common error in data analysis is to jump directly to hypothesis testing before summarizing the data. 3. Graphical Summaries Graphical summaries include tables, charts, and graphs. All provide compact and visually appealing ways of summarizing a mass of numerical information or data. Every statistical analysis of data should include graphical summaries. These summaries give good general impressions of the data, including trends, relationships among variables, the likely distribution of the population, possible outliers, etc. They suggest which numerical summaries will be useful. They are also used for checking the validity of assumptions needed to properly use specific formal statistical procedures. There are many different types of available graphs. Check the MS Excel Chart Wizard for illustrations. Some very familiar types include: bar chart, pie chart, line graph, histogram, stem-and-leaf plot, stacked bar chart, etc. [XX Provide illustrations] Here is a useful rule of thumb to determine an appropriate level of summary. Use sentence structure for displaying 2 to 5 numbers, tables for more numerical information and graphs for more complex relationships. Example: a) Sentence structure: The blood type of the Belltown population is approximately 40% A, 11% B, 4% AB and 45% O. b) Table: Blood type of the Belltown population: O 45% A 40% B 11% AB 4% The difference between a bar chart and a histogram: bar charts are for categoric data where the categories are not contiguous; histograms are for measurement data where the underlying continuum is divided into contiguous class intervals. Although pie charts are the most ubiquitous of graphs they are rarely the best choice of graphical display. A pie chart has very low data density. The percentage in it can be better presented in a table. The only thing worse than one pie chart is many pie charts. See Tufte (1983) for further discussion. (Note: Edward Tufte is a Professor of Visual Arts at Yale University. He has made his mark by describing principles by which graphical representations can BEST communicate quantitative relationships. At some point, every researcher has read his classic “The Visual Display of Quantitative Data” As opposed to statistical books, this book is a pleasure to read and skim. The link above is to its Amazon listing. Tufte provides three criteria for graphical excellence. “Graphical excellence is the well-designed presentation of interesting data – a matter of substance, of statistics, and of design.” ”Graphical excellence consists of complex ideas communicated with clarity, precision and efficiency.” “Graphical excellence is that which gives to the viewer the greater number of ideas in the shortest time with the least ink in the smallest space.” Advice on Tables – Rules of Thumb Arrange rows and columns in the most meaningful way Limit the number of significant digits Use informative headings Use white space and lines to organize rows and columns Make the table as self-contained as possible Example: (Ref: Gerald van Belle, Statistical Rules of Thumb, Wiley, 2002) Number of Active Health Professionals According to Occupation in 1980, US (Source: National Center for Health Statistics, 2000) A Poor Display: Occupation1980Chiropractors25,600Dentists121,240Nutritionists/Dietitians32,000Nurses, registered1,272,900Occupational therapists25,000Optometrists22,330Pharmacists142,780Physical therapists50,000Physicians427,122Podiatrists7,000Speech therapists50,000 The listing is in alphabetical order, which is only helpful in the telephone directory. Note also that the frequencies for some occupations display spurious accuracy; for example, are there exactly 427,122 physicians? Aren’t we only interested in the frequencies to the nearest thousand? Here is a better display, with frequencies reporting in “thousands” and with the rows grouped by size. The blank lines set off groups of occupations with similar sizes. Occupation1980 (in 1000’s)Nurses, registered1,273Physicians427Pharmacists143Dentists121Physical therapists50Speech therapists50Nutritionists/Dietitians32Chiropractors26Occupational therapists25Optometrists22Podiatrists7 Can knowledge of statistics mean the difference between life and death? Yes! Here is an example of the power of a good graphical analysis. Example: Jan 28, 1986 – Space Shuttle Challenger Disaster (Ref: A Casebook for a First Course in Statistics and Data Analysis; Chatterjee, S, Handcock, MS, and Simonoff, JS) On Jan. 28, 1986 the Challenger took off, the 25th flight in the NASA space shuttle program. Two minutes into the flight, the spacecraft exploded, killing all on board. A presidential commission, headed by former Secretary of State William Rogers, and including the last Nobel-prize-winning physicist Richard Feynman determined the cause of the accident and wrote a two-volume report. Background: To lift it into orbit the shuttle uses two booster rockets; each consists of several pieces whose joints are sealed with rubber O-rings, designed to prevent the release of hot gases during combustion. Each booster contains three primary O-rings (for a total of 6 for the craft). In 23 previous flights for which there were data, the O-rings were examined for damage. The forecasted temperature on launch day was 31 ŗF. The coldest previous launch temperature was 53 ŗF. The sensitivity of O-rings to temperature was well-known; a warm O-ring will quickly recover its shape after compression is removed, but a cold one will not. Inability of the O-ring to recover its shape will lead to joints not being sealed and can result in a gas leak. This is what caused the Challenger explosion. Could this have been foreseen? Engineers discussed whether the flight should go on as planned (no statisticians were involved). Here is a simplified version of one of the arguments. The following table gives the ambient temperature at launch and # of primary field joint O-rings damaged during the flight. Ambient temp. at launch (ŗF): 53ŗ 57ŗ 58ŗ 63ŗ 70ŗ 70ŗ 75ŗ Number of O-rings damaged: 2 1 1 1 1 1 2 The table and a scatterplot (graph it yourself) shows no apparent relationship between temperature and # O-rings damaged; higher damage occurred at both lower and higher temperatures. Hence, just because it was cold the day of the flight doesn’t imply that the flight should have been scrubbed. This is an inappropriate analysis! It ignores the 16 flights when zero O-rings were damaged. When those are included the scatterplot looks quite different, and in fact, shows a strong relationship between # O-rings damaged and temperature. Here is the complete data set: Temp.(ŗF): 53 57 58 63 66 67 67 67 68 69 70 70 70 70 72 73 75 75 76 76 78 79 81 # damaged: 2 1 1 1 0 0 0 0 0 0 0 0 1 1 0 0 0 2 0 0 0 0 0  Except for the one observation in the upper right, there is a clear inverse relationship between the probability of O-ring damage and ambient temperature. Unfortunately, this plot was never constructed! The flaw in the analysis was not to include flights in which there was no O-ring damage. On Jan. 28, 1986, the ambient (outside) temperature was 31 ŗF. (Note that this is off the scale of available data so extrapolation is needed.) “Logistic regression” estimates the probability of an O-ring failure at 31 ŗF to be 0.96! One of the Commission’s recommendations was that a statistician must be part of the ground control team for all flights. Morals: 1) Look at all available data. 2) Include a statistician on any research team. 3) Good statistical thinking can save lives! 4) Good statistical thinking isn’t “rocket science”, but in this case it should have been! Quicklink for further reading: Statistics at Square One -- “Data Display” http://bmj.bmjjournals.com/collections/statsbk/1.shtml 4. Numerical Summaries A graph or table summarizes the data. A numerical summary summarizes key features of the graph or table. Categoric data are usually summarized by counts or frequencies, and proportions or percentages. Example for Categoric data – a frequency table Does C&Wour hospital see more male or more female patients? An exit survey is done and gives following results. Count Percent Valid Percent Male 80 40.0 43.0 Female 106 53.0 57.0 Missing(*) 14 7.0 ------ Total 200 100.0 100.0 The 14 respondents in the “Missing” category refused to complete this question on the exit survey. Note that there are two Percent columns. The first Percent column uses all 200 as the denominator, while the second, Valid Percent, columns uses only those cases whether a legitimate answer was given. This is analogous to opinion polling where a certain portion of voters claims to be “undecided” and so the reporting includes a computation of support among “decided” voters. Exercise 1: At Belltown Elementary School, the ethnic mix of the 600 students is as follows: 270 are Caucasian, 150 are African-American, 100 are Hispanic, 75 are Asian, and 25 have unknown ethnicity. Construct a frequency table. What proportion is not Caucasian? [LINK TO ANSWER] Answer to Numerical Summary Exercise 1: Here is the frequency table; Valid Percent is computed based on the 575 for whom ethnicity is known. Of those for whom ethnicity is known, 53.0% (i.e. just over one-half) are not Caucasian. Count Percent Valid Percent Caucasian 270 45.0 47.0 African-American 130 21.7 22.6 Hispanic 100 16.7 17.4 Asian 75 12.5 13.0 Other/Unknown 25 4.2 ----- ---------------------- ------ --------- --------- Total 600 100.0 100.0 Measurement data lead to many more numerical summaries. And they are of two kinds: measures of location (i.e. where is the “middle of the data distribution) and measures of scale (i.e. what kind of spread does the distribution have). Note that gaps in the distributions and extreme values (i.e. outliers) cannot be easily detected using numerical summaries; that is a job for graphs! Location: Mean (denoted by  EMBED Equation.3  ) = sum of the data values divided by the number of data values Median: the middle value, after arranging the data in ascending order Mode: the most frequently occurring value (by far the least useful of the three summaries) The mean can be thought of as the “centre of gravity” of the distribution. That is, if the histogram (since we have measurement data!) were sitting on a seesaw, the mean would be the point at which the seesaw was in balance. The median can be though of as the middle of the data; 50% are on each side of the median. Note; If there is an even number of data values, the median is the average of the two middle values. For symmetric distributions, the mean equals the median. For asymmetric or skewed distributions, the mean and the median are not equal. If the distribution has a long right-hand tail (i.e. skewed right) the mean is greater than the median. If the distribution has a long left-hand tail (i.e. skewed left), the mean is less than the median. The median is a more “robust” measure than the mean; that is, it is not highly affected by the presence of outliers. Example: For data values: 1,2,3,4,5,6,7,8,9, both the mean and median are 5; For data values: 1,2,3,4,5,6,7,8,90, the mean is 14 but the median is still 5. Example (adapted from Martin Gardner): Belltown Smelter has a long history of nepotism in its hiring. In the early days of the company, the management team consists of Mr. X, his brother and six relatives. The work force consisted of five forepersons and ten workers. When the company was ready to expand, Mr. X interviewed applicants and told them that the average weekly salary at the company was $600. A new worker, Y, was hired and, after a few days, complained that Mr. X had misled him. Y had checked with the other workers and discovered that none was getting more than $200 a week, so how could the average be $600. Mr X explained the salary structure: Each week Mr. X gets $4800, his brother gets $2000, each of the six relatives makes $500, the forepersons each make $400 and the ten workers each get $200. That makes a weekly total payroll of $13,800 for 23 people, leading to an average of $600. What about the median and mode? The median is the middle value, in this case, the 12th value, which is $400. The mode is the most frequently occurring value, which is $200. So here we have three measures of location: $200, $400 and $600. Which is the preferred one.Which is the preferred one? Since the distribution is so skewed and also has a large extreme value, the median is preferred. Management would, of course, want to report the mean, but the Union would probably prefer to report the mode during contract negotiations! Scale: Range = Maximum value – Minimum value Variance (s2) and Standard Deviation (s or SD) s =  EMBED Equation.3  The standard deviation (sometimes referred to as the SD) can best be interpreted as “the typical distance from a data value to the mean”. Neither the mean nor the standard deviation is resistant to outliers. They are also a poor choice of summary if the distribution is highly skewed. Percentiles: The kth percentile is a value below which k% of the data values fall. Some percentiles have special names: 75th percentile = 3rd or upper quartile = Q3 25th percentile = 1st or lower quartile = Q1 50th percentile = Median The Interquartile Range (IQR) = Q3 – Q1. Note: If there are an odd number of observations, the median is indeed the middle one; but if there are an even numberis an even number of observations the convention is to take the average of the two middle ones as the median. This can be extended to computing the quartiles; if a quartile lies between two observations, take the average. Beware, however, that different software packages have different conventions for computing the quartiles. But quite frankly, it doesn’t matter, since you are only using them as summaries! IMPORTANT: For symmetric distributions ( Mean and SD For skewed distributions ( Median and IQR Question: Why do people still use the mean and SD even when the data show strong skewness? Possible Answers: they forget, or don’t know how, to check for the symmetry of the distribution they have no idea that there are alternatives they think it doesn’t matter! they don’t have the ability to compute or explain alternative summaries sheer bloody-mindedness! because other papers in the field do! Rules of Thumb: s H" IQR/1.35 s H" Range/6 (for bell-shaped distributions and large n) s H" Range/4 (for bell-shaped distributions and small n, approx. 20) The Empirical Rule  for symmetric, bell-shaped distributions Mean ( 1 SD contains about 68% (2/3) of the data values. Mean ( 2 SD contains about 95% of the data values. Mean ( 3 SD contains about 99.7% (almost all) of the data values. [Aside: Chebyshev’s Inequality: Mean ± c SD contains at least 1 – [1/c2] of the data values, regardless of the shape of the distribution (for c > 1).] Boxplot (or box-and-whisker plot) – a graphical display of the median, quartiles, IQR, and outliers It is useful for assessing the general shape of the distribution (symmetric or skewed) and the presence of outliers. It is also useful for comparing distributions of several samples. Version 1: _________________ |------------|________|________|--------------------| _________________________________________________ Min Q1 Median Q3 Max Version 2: Compute the inner fences: Q1 – 1.5 IQR and Q3 + 1.5 IQR _________________ * |-------|________|________|------------| ** * _________________________________________________ Min Q1 Median Q3 Max In Version 2, each whisker extends to the last data value inside the inner fence; the asterisks represent the outlying values. There are also outer fences: Q1 – 3 IQR and Q3 + 3 IQR. Boxplots and fences were developed by John Tukey. He suggests using the fences as indicators of outliers. Moderate outliers are outside the inner fences; extreme outliers are outside the outer fences. Exercises (adapted from C. Chatfield, Problem-Solving: A Statistician’s Guide) Summarize each of the following data sets in whatever way you think is appropriate. a) Marks on a math exam (out of 100 and ordered by size) of 20 students in Kyle’s class. 30 35 37 40 40 49 51 54 54 55 57 58 60 60 62 62 65 67 74 89 b) The number of days of work missed by 20 workers at the Belltown Smelter in one year (ordered by size; i.e. the number of days, not the heights of the workers!) 0 0 0 0 0 0 0 1 1 1 2 2 3 3 4 5 5 5 6 45 c) The number of issues of the monthly magazine read by 20 people in a year: 0 1 11 0 0 0 2 12 0 0 12 1 0 0 0 0 12 0 11 0 d) The height (in metres) of 20 women who are being investigated for a certain medical condition: 1.52 1.60 1.57 1.52 1.60 1.75 1.73 1.63 1.55 1.63 1.65 1.55 1.65 1.60 1.68 2.50 1.52 1.65 1.60 1.65 [LINK TO ANSWERS] Answers to Exercise (Numerical Summaries): a) A histogram of exam scores shows a reasonably symmetric bell-shaped distribution. If you see skewness, look again. If your opinion is swayed by the value of 89, cover it up and look again. Suitable summary statistics: mean = 55, median = 56 (average of 55 and 57), std. dev. = 14. Note that the mean and median are almost identical, which is another sign of symmetry. Make sure you haven’t reported too many decimal places; since the data are recorded as integers, one decimal place is more than enough! Stem-Leaf Plot of question a: 3 0 5 7 4 0 0 9 5 1 4 4 5 7 8 6 0 0 2 2 5 7 7 4 8 9 Click here to see an Excel file with the data summarized and a Histogram of the data. {Sheet QuestionA} b) A bar chart of days of work missed is severely skewed to the right. The value of 45 is an outlier, but probably not an error. It will cause a problem in the construction of the bar chart, and it highly influences the mean, which is 4.2 days. The median is 1.5 days and the mode is 0 days. The standard deviation is little help here since the distribution is so skewed. Overall, the summary statistics have little value here and the bar chart is probably the best way to summarize the data. I would also investigate the value of 45; it deserves a special comment in your summary. 0 0 0 0 0 0 0 1 1 1 2 2 3 3 1 5 5 5 6 … 45 Click here to see an Excel file with the data summarized and a Histogram of the data. {Sheet QuestionB} c) A frequency distribution of number of issues read shows two modes, at zero and twelve. Summarizing a bimodal U-shape is even harder than summarizing a skewed distribution. Neither the mean nor standard deviation are useful here, and worse, are misleading. No one reads half the issues in a year. Rather the readers should be classified as “regular” or “not regular”. With this new categoric (and indeed, binary) version of the data, you should report the percentage of regular readers, which is 5/20 or 25%. 0 1 11 0 0 0 2 12 0 0 12 1 0 0 0 0 12 0 11 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 11 11 12 12 12 Click here to see an Excel file with the data summarized and a Histogram of the data. {Sheet QuestionC} d) Undoubtedly you found the egregious error in the data; the value of 2.50 is most certainly an error, not just an outlier. How should you deal with it? The safest thing to do is to omit it. Although you suspect it should be 1.50 you can’t just go around correcting data to what you think it should be. The worst thing to do is to leave it in uncorrected! After omitting it, the remaining data are reasonably symmetric, so you can construct a histogram and compute the mean and standard deviation. 1.52 1.60 1.57 1.52 1.60 1.75 1.73 1.63 1.55 1.63 1.65 1.55 1.65 1.60 1.68 2.50 1.52 1.65 1.60 1.65 * * * ** 1.56 1.57 * 1.58 1.59 1.60 **** 1.61 1.62 1.63 ** 1.64 1.65 **** 1.66 1.67 1.68 * 1.69 1.70 1.71 1.72 1.73 * 1.74 1.75 * X X X 2.50 * Click here to see an Excel file with the data summarized and a Histogram of the data. {Sheet QuestionE} An interesting piece of forensic statistics… Can you tell what country the data came from? If you examine the final digits you’ll see that some numbers keep recurring and others never appear. A likely explanation is that the observations were made in inches and then converted to metres. Thus a good hunch is that the study was done in the U.S. where the metric system is not used! Moral: Chatfield writes, “Descriptive statistics is not always straightforward. In particular, the calculation of summary statistics depends on the shape of the distributions and on a sensible treatment of errors and outliers. Example E: Norman Statson and his buddies at the Belltown pub decided to compare their vehicles’ gas mileage. Remarkably, each participant drove a different car model. For a two-month period they recorded their cars mileage and gas consumption. With help from his daughter Effie, Norman prepared the following spreadsheet with miles per gallon (MPG) for 24 Belltown automobiles. (Actually, the data are real, and come from an Environmental Protection Agency study of 24 1992-model cars!) CarMileageCarMileage124133122314323361530427162953817286131825731192881720219402135105022311137232512202440 Ordered Data: 13,17,20,21,23,24,25,25,27,28,28,29,30,31,31,31,32,35,36,37,38,40,40,50  EMBED Equation.3  = 29.6 s2 = 68.2 s = 8.2 Median = (29+30)/2 = 29.5 Min = 13; Max = 50 Range = 37 Number of observations = 24 Q1 = between the 6th and 7th observations; take the average of the two = 24.5 Q3 = between the18th and 19th observations; take the average of the two = 35.5 IQR = 11.0 (Note that MS Excel has a slightly different algorithm for computing quartiles; according to Excel, Q1 = 24.75, and Q3 = 35.25. This would mean that IQR = 10.5. Since the quartiles and IQR are used for descriptive purposes only, the lack of consistency in algorithms is only a problem if you are obsessive compulsive. In that case, counseling may be required!) Lower inner fence = 24.5 – 1.5 (11.0) = 8.0 Upper inner fence = 35.5 + 1.5 (11.0) = 52.0 ( No outliers in this data set. Check the Rules of Thumb and Empirical Rule: IQR H" 1.35 x s = 11.15 (compare with 11.0) s H" Range/4 = 37/4 = 9.25 (compare with 8.26)  EMBED Equation.3  ± s = (21.4, 37.9): contains 16/24 = 67% of the observations  EMBED Equation.3  ± 2s = (13.1, 46.2): contains 22/24 = 92% of the observations Click here to see an Excel file with the data summarized and a Histogram of the data. {Sheet ExampleE} Quicklink for Further Reading: Statistics At Square One – Means and Standard Deviation. http://bmj.bmjjournals.com/collections/statsbk/2.shtml 5. Normal Distribution It is useful to have a compact mathematical form (i.e. an equation) that can describe the shape of many commonly occurring distributions of measurement data as seen by histograms, for example. Many phenomena have the familiar bell-shaped curve; the mathematical function that best describes these shapes is called the normal curve or normal distribution. (Statisticians also call it the Gaussian distribution.) Although the normal curve is very easy to visualize and quite easy to draw, it is difficult to handle mathematically. The equation of the normal curve is:  EMBED Equation.3  where ¼ is the mean of the distribution and Ć is the standard deviation of the distribution. The normal curve really does describe a bewildering range of phenomena. My favorite example is the popping behaviour of microwave popcorn. The intensity of popping follows a normal curve. For the first minute or two you hear nothing, and then the occasional pop. The popping becomes more vigorous until it reaches its peak intensity, and then begins to quiet down just as it began. Of course, you should remove it before you burn the popcorn, but if you left it in, the second half of the process would be the mirror image of the first half of the process. Listen carefully the next time you pop a bag of popcorn! Area under the curve can be interpreted as relative frequency. The total area under the curve is 100% (i.e. a total probability or relative frequency of 1). The Empirical Rule applies to normal distributions, and can now be written as: ¼ ± Ć contains 68% of the area under the normal curve ¼ ± 2Ć contains 95% of the area under the normal curve ¼ ± 3Ć contains 99.7% of the area under the normal curve To compute areas under the normal curve we use a linear transformation to standardize to a standard normal curve. To standardize something in statistics means to remove the distraction of units. A standardized distribution of popcorn popping can be directly compared to a standardized distribution of heights (why you would do this is a separate question). See example 2 below. If X has a normal distribution with mean ¼ and standard deviation Ć, then standardize as follows: Z = (X  ¼) / Ć Then Z has a standard normal distribution with mean 0 and standard deviation 1. The Empirical Rule for Z now becomes: 68% of the Z-values are between  1 and 1 95% of the Z-values are between  2 and 2 99.7% of the Z-values are between  3 and 3. Question: I can hear you scratching your head already; why have we introduced new notation, namely, ¼ for the mean and Ć for the standard deviation? Isn t that what  EMBED Equation.3  and s represented? In Section 6 below,, Introduction to Inferential Statistics, we will discuss this further. For now the answer is that we are making the leap from real observed empirical data to a hypothetical collection of possible values. You can think of the normal curve as a “stylized” description of your real data. And since it is “stylized” it needneeds its own notation for mean and standard deviation. Here is an illustration of how we will use the normal curve. Suppose you are interested in the behaviour of a measurement variable such as the height of a population. A histogram of your collected data shows a symmetric bell-shape so that you think it appropriate to summarize the shape with a normal curve. Because the properties of the normal curve are known, you will now be able to compute the chance that the true height of your population exceeds a certain limit, or falls within a particular interval. First, however, we need a bit of proficiency at normal curve calculations. Here are some examples. Examples of Normal Calculations Example 1F. Consider the Wechsler IQ test; scores for the general population of are normally distributed with mean 100 and standard deviation 15. We assume that Belltown citizens exhibit the same intelligence levels as for the general population. a) What is the chance that randomly chosen Belltownian (or is that Belltowner, or Belltownite) has an IQ exceeding 120? Pr (X > 120) = Pr ((X  ¼) / Ć > (120  100)/15 ) = Pr(Z > 1.33) = 0.5  0.4082 = .0918 or 9.2% To see how to do this example in Excel click here: File ExampleF.xls b) To belong to Mensa, the high-IQ society, you need an IQ in the top 2% of the population. What IQ score on the Wechsler test is needed? Find z that has an area of .02 to the right; z = 2.055 Z = (X  ¼) / Ć ( X = ¼ + Z Ć Hence, X = 100 + (2.055 x 15) = 131 To see how to do this example in Excel click here: File ExampleF.xls Example 2. In course A you get 85%, the class average is 80% and SD is 5% In course B you get 75%, but the class average is 65% and the SD is also 5%. Which course did you do better in relative to your classmates? Assume that grades in both courses are normally distributed. To compare values from two normal distributions, convert to Z-scores by standardization. If X has a normal distribution with mean ¼ and standard deviation Ć, then standardize as follows: Z = (X  ¼) / Ć Then Z has a standard normal distribution with mean 0 and standard deviation 1. For course A, Z = (85-80) / 5 = 1 For course B, Z = ( 75-65) / 5 = 2 You were 2 SDs above the class average in course B but only 1 SD above the class average in course A. Quicklink for Further Reading: Statistics At Square One – Statements of Probability  HYPERLINK "http://bmj.bmjjournals.com/collections/statsbk/4.shtml" http://bmj.bmjjournals.com/collections/statsbk/4.shtml 6. Introduction to Inferential Statistics Inference – a synonym for “generalization”; that is, for drawing conclusions about a larger universe based on the available data Population – the collection of all “items” (real or hypothetical) which have a common characteristic of interest. Sample – a subset (representative and randomly chosen) which will represent the population. (Note: “random” means “use a chance process to select the sample.”) Parameter – characteristic of the population. Statistic/Estimate – characteristic of the sample that is analogous to the parameter of the population. A “clever” mnemonic device: Population (( Parameter Sample (( Statistic/Estimate Note that Population and Parameter both start with “P”; Sample and Statistic both start with “S”. And, here’s the brilliant part, even the first syllable of Estimate is pronounced “S”! Example: Suppose you are interested in the mean household income of all Canadian households. The population is all Canadian households; the parameter is the mean household income. Only CCRA has this information! However, a survey research firm contacts 1,000 households at random. These 1,000 households form the sample; the mean household income of the 1,000 households is the statistic/estimate. Notation for some common parameters and their estimates Parameter Estimate µ: population mean  EMBED Equation.3 : sample mean Ć: population standard deviation s: sample standard deviation (SD) p : population proportion  EMBED Equation.3 : sample proportion µ1  µ2  EMBED Equation.3    EMBED Equation.3  p1  p2  EMBED Equation.3    EMBED Equation.3  ø: generic parameter  EMBED Equation.3 : generic estimate Sampling Distribution: The sampling distribution is the theoretical distribution of values taken by a statistics or estimate, if a large number of samples of the same size were taken from the same population. The sampling distribution of many commonly used statistics is the normal distribution. The best way to understand this is to see it visually. Do the lab available at:  HYPERLINK "http://stat.tamu.edu/~henrik/sqlab/sqlab1.html" http://stat.tamu.edu/~henrik/sqlab/sqlab1.html In the lab you can play with simulation of sampling from a given distribution. You’ll see that when you sample over and over again, the resulting distribution of samples (the sampling distribution) is normally distributed. This is an important concept because it underlies many of our statistical tests. To run the simulation you might have to download and install the “Jave Runtime Plugin for Internet Explorer”. This is very easy to do at the Java site:  HYPERLINK "http://java.com/en/download/manual.jsp" http://java.com/en/download/manual.jsp Directions on how to complete the lab are available at: http://stat.tamu.edu/~henrik/sqlab/lab1hlp.html Example 1: The sample mean,  EMBED Equation.3 , is an estimate of (, the population mean. It is a “good” estimate because of two properties, neither of which will be proven here. First, it is an unbiased estimate, meaning that if you repeated the process of collecting a sample and computing the sample mean, the average of these sample means would converge to the true population mean (. That is, “in the long run”,  EMBED Equation.3  tends to (. Second, among all unbiased estimates of (,  EMBED Equation.3  has the smallest variation; in fact, the standard deviation of  EMBED Equation.3  is (/ EMBED Equation.3 . A famous theorem of statistics shows that the sampling distribution of  EMBED Equation.3  is approximately normal. Central Limit Theorem: The mean of a sample of random quantities from a population with mean ( and standard deviation ( is approximately normal with mean ( and standard deviation (/ EMBED Equation.3 , if n is large enough. This is graphically shown in the simulation lab above. Hence we say that  EMBED Equation.3  has an approximately normal sampling distribution with mean ( and standard deviation (/ EMBED Equation.3 . Example 2: For categoric data, the sample proportion,  EMBED Equation.3 , is an estimate of p, the population proportion;  EMBED Equation.3  = X/n = (# “successes”)/(total sample size)). Applying the Central Limit Theorem here means that  EMBED Equation.3  has an approximate normal sampling distribution with mean p and standard deviation  EMBED Equation.3 , where  EMBED Equation.3  is the true proportion in the population. Note: We have a special name for the standard deviation of an estimatethe sampling distribution; it is called the standard error (SE). SD represents the variability of individual observations (i.e. s and (); that is, how much does one observation differ from another. SE represents the variability of an estimate (for  EMBED Equation.3  this is s/ EMBED Equation.3  or (/ EMBED Equation.3 ); that is, how much would the estimate chance if you drew another sample and computed another estimate. Example: A study of all Belltown students who go on to get a university degree shows that the mean starting salary is $45,000 (= () with a standard deviation of $4,500 (= (). A histogram shows that salaries are normally distributed. a) Select one graduate at random. How far from $45,000 is his/her salary likely to be? Answer: About 1( or $4,500. b) Select 9 graduates at random. How far from $45,000 is the mean salary of these 9 likely to be? Answer: About 1 ( /(n = 4,500/(9 = $1,500. c) What is the probability that the mean starting salary of a random sample of 25 graduates is less than $43,000? Answer: Pr( EMBED Equation.3  < 43,000) = Pr{( EMBED Equation.3 – () / ( /(n < (43,000 – 45,000) / 4500/(25} = Pr(Z < -2.22) = .0132 or 1.32% Note: For an individual, Z = (X – () / ( For the mean of a group sample of size n, Z = ( EMBED Equation.3  – () / (/(n Quicklink for Further Reading: Statistics At Square One – Populations and Samples http://bmj.bmjjournals.com/collections/statsbk/3.shtml 7. Confidence Intervals Point estimates such as  EMBED Equation.3  and  EMBED Equation.3 , are based on samples, and hence have sampling variability. That is, if you sampled again, you would get different data and therefore different point estimates. Confidence intervals are an excellent way of quantifying this sampling variability and expressing it as a margin of error. The probability that a point estimate equals the true parameter value is actually zero, but the probability that an interval estimate “captures” the true parameter value can be made as high as you like, say, for example, 95%. For example, if a physician gives a pregnant woman her “due date” as a single day, the chance of a correct prediction is very small. But if, instead, the woman is given two-week range as a due date, the chance of a correct prediction is very high. Every opinion poll uses this strategy. There is always a disclaimer of the form, “the poll is estimated to be accurate to within 3 percentage points 19 times out of 20.” This is actually a 95% confidence interval. The pollster is quite sure (95% in fact) that his/her poll will not miss the true percentage by more than 3% – that’s the poll’s margin of error. The margin of error reminds you that effects may not be as large as the point estimate indicates; they could be much smaller, or they could be much larger. But they remind you not to get too excited about the accuracy of a point estimate. In Section 8, Hypothesis Testing, you will learn about the concept of statistical significance, namely, is a difference between two sample means real or can it be explained by sampling variability? Confidence intervals are often used in conjunction with hypothesis tests. Once a hypothesis test determines whether an effect is real, a confidence interval is used to express how large the effect is. So now that you know why you should compute confidence intervals, we’re ready to discuss how to compute them. A confidence interval is an interval estimate that has a high likelihood of containing the true population parameter. Using the Empirical Rule and the normal sampling distribution of  EMBED Equation.3  and  EMBED Equation.3  gives the following confidence intervals. 95% Confidence Interval for (:  EMBED Equation.3  ( 2(/ EMBED Equation.3  95% Confidence Interval for p:  EMBED Equation.3  ( 2 EMBED Equation.3  Note on interpretation: The 95% CI for ( is an interval has a 95% chance of containing (. This is not the same as saying that there is a 95% chance that ( is in the interval. The probability has to do with your ability, using your random sample, to correctly capture the parameter (. The parameter does not vary, but the confidence interval does. Note that by adjusting the multiplier, you can change the level of confidence. For example, using 3 instead of 2 in the above equations would give 99.7% confidence intervals. We can refine this to allow different multipliers and different levels of confidence. Let z±/2 be a value on the standard normal curve that has an area of ±/2 to the right of it. Then: Pr (-z±/2 < Z < z±/2) = 1  ±. This leads to calling  EMBED Equation.3  ± z±/2 ( / EMBED Equation.3  a 100(1  ±)% confidence interval for µ. Here are some common choices of ±: ± 1  ± ±/2 z±/2 .10 .90 .05 1.645 .05 .95 .025 1.96 .01 .99 .005 2.576 Since it is highly likely that ( is unknown, replace it with s, the sample standard deviation, in the formula for the CI for (. Similarly, in the formula for the CI for p, replace p by  EMBED Equation.3 . These give approximate confidence intervals. Hence, an approximate 100(1  ±)% confidence interval for µ is:  EMBED Equation.3  ± z±/2 s / EMBED Equation.3  This works well if n is large. How large? At least 30. Is it possible to make this an exact confidence interval? Yes, but we need to introduce the Student’s t distributions. Z = ( EMBED Equation.3 – () / [( / EMBED Equation.3 ] has a standard normal distribution t = ( EMBED Equation.3 – () / [s / EMBED Equation.3 ] has a t-distribution with n-1 degrees of freedom. Notice the difference between Z and t ; ( is replaced by s. As n gets larger, s becomes closer and closer to (. In the limit, as n approaches infinity, the t-distribution approaches the z-distribution. This can be seen graphically in StatConcepts Lab #: xx or by going to the “Sable Online Research Methods Course at:  HYPERLINK "http://simon.cs.vt.edu/SoSci/converted/T-Dist/" http://simon.cs.vt.edu/SoSci/converted/T-Dist/ Of specific interest is the simulator about half-way down the page entitled: “The t-Distribution as a Family of Sampling Distributions” It shows graphically how the t-distribution approximates the z-distribution at N>30. Therefore, an exact 100(1  ±)% confidence interval for µ is:  EMBED Equation.3  ± t±/2,n-1 s / EMBED Equation.3  Similarly, a 100(1  ±)% confidence interval for  EMBED Equation.3  is:  EMBED Equation.3  ± z±/2 EMBED Equation.3  These two are used widely as one-sample confidence intervals for a single mean or a single proportion. Extension to two populations Now we can extend this to two populations and the difference of two means or two proportions. An approximate 100(1  ±)% confidence interval for µ1  µ2 is: ( EMBED Equation.3    EMBED Equation.3 ) ± t±/2,n1+n2-2  EMBED Equation.3  This is essentially an arithmetic variation of the formulas for one sample. A little variation: If we can assume also that (12 = (22, then we can pool the two estimates of variances:  EMBED Equation.3  = pooled variance This is just a weighted average of s12 and s22 Then the exact 100(1  ±)% confidence interval for µ1  µ2 becomes ( EMBED Equation.3    EMBED Equation.3  ) ± t±/2,n1+n2-2  EMBED Equation.3  A nice description of this formula is available at:  HYPERLINK "http://www.socialresearchmethods.net/kb/stat_t.htm" http://www.socialresearchmethods.net/kb/stat_t.htm Rule of Thumb: We can safely assume that (12 = (22 if the larger s is not much more than twice the smaller s. Similarly a 100(1  ±)% confidence interval for p1  p2 is: ( EMBED Equation.3    EMBED Equation.3 ) ± z±/2  EMBED Equation.3  Note: To help keep things straight, remember that you use the t-distribution for means, and the z-distribution for proportions. {Where was this point first made? Let’s link to it} Summary of Confidence Intervals (so far) {I would put worked examples of each of these in Excel files as done above. Jonathan, if you give me textual descriptions and some dummy data, I’ll take it from there. } 1. A 100(1  ±)% confidence interval for µ is:  EMBED Equation.3  ± t±/2,n-1 s / EMBED Equation.3  2. A 100(1  ±)% confidence interval for  EMBED Equation.3 :  EMBED Equation.3  ± z±/2 EMBED Equation.3  3a. An approximate 100(1  ±)% confidence interval for µ1 - µ2 ( EMBED Equation.3    EMBED Equation.3 ) ± t±/2,n1+n2-2  EMBED Equation.3  If (12 = (22 , compute  EMBED Equation.3  3b. An exact 100(1  ±)% confidence interval for µ1 - µ2 is: ( EMBED Equation.3    EMBED Equation.3 ) ± t±/2,n1+n2-2  EMBED Equation.3  Rule of Thumb: We can safely assume that (12 = (22 if the larger s is not much more than twice the smaller s. 4. A 100(1  ±)% confidence interval for p1 - p2 is: ( EMBED Equation.3    EMBED Equation.3 ) ± z±/2  EMBED Equation.3  Cautions: The margin of error deals only with random sampling errors. Practical difficulties such as non-response or undercoverage in sample surveys can contribute additional errors, and these can be larger than the sampling errors! To see an interesting, take on this, go to:  HYPERLINK "http://www.cmaj.ca/cgi/content/full/165/9/1226?maxtoshow=&HITS=10&hits=10&RESULTFORMAT=&fulltext=CONFIDENCE+INTERVAL&andorexactfulltext=and&searchid=1107810091642_4824&stored_search=&FIRSTINDEX=0&sortspec=relevance&resourcetype=1&journalcode=cmaj" http://www.cmaj.ca/cgi/content/full/165/9/1226?maxtoshow=&HITS=10&hits=10&RESULTFORMAT=&fulltext=CONFIDENCE+INTERVAL&andorexactfulltext=and&searchid=1107810091642_4824&stored_search=&FIRSTINDEX=0&sortspec=relevance&resourcetype=1&journalcode=cmaj The professional medical literature has only recently embraced confidence intervals as an appropriate way to report results. See  HYPERLINK "http://bmj.bmjjournals.com/cgi/content/full/318/7194/1322" http://bmj.bmjjournals.com/cgi/content/full/318/7194/1322 for just one example. Many researchers are still unclear about why! Perhaps one day in the future confidence intervals will completely supersede hypothesis tests in the reporting of statistical analysis. Quicklink for Further Reading: Statistics At Square One – Confidence Intervals  HYPERLINK "http://bmj.bmjjournals.com/collections/statsbk/4.shtml" http://bmj.bmjjournals.com/collections/statsbk/4.shtml 8. Hypothesis Testing Hypothesis Tests (also called Tests of Significance) address the issue of which of two claims or hypotheses about a parameter (or parameters) is better supported by the data. For example, is a target mean being achieved or not; does a treatment group have a higher mean outcome than a control group; is there a greater proportion of successes in one group than another? Generally this boils down to assessing whether a perceived difference from the null hypothesis is due to random error (ie was it a fluke?) or is it due to, for example, the intervention under study. All tests have the following components: 1. Hypotheses: Ho: null hypothesis Ha: alternative hypothesis These are statements about parameters. The null hypothesis represents “no change from the current position” or the default position. The alternative hypothesis is the research hypothesis. The burden of proof is on the investigator to convince the skeptic to abandon the null hypothesis. 2. Test Statistic: Uses estimate(s) of the parameter(s), the standard error(s) of the estimate(s), and information in the null hypothesis and puts them together in a “neat package” with a known distribution. Certain values of this test statistic will support Ho while others will support Ha. 3. P-value: a probability that judges whether the data (via the value of the test statistic) are more consistent with Ho or with Ha. Note: The P-value assumes that Ho is true and then evaluates the probability of getting a value of the test statistic as extreme as or more extreme than what you observed. The P-value is not the probability that Ho is true. The smaller the P-value, the greater the evidence against the null hypothesis. 4. Conclusion: a statement of decision in terms of the original question. It is not enough simply to write “reject Ho“ Example G: Suppose you are the manager of a walk-in medical clinic. Your clinic has a target waiting time of 30 minutes for non-emergency patients. To assess whether you are meeting the target you take a random sample of 25 patients and compute a sample mean of 38.1 minutes and a sample standard deviation of 10.0 minutes. A histogram shows that the waiting times are reasonably normally distributed. A 95% confidence interval for (, the true mean waiting time is:  EMBED Equation.3  ± t±/2,n-1 s / EMBED Equation.3  (see formulas above) where t.025,24 = 2.064 38.1 ± 2.064 x 10 / (25 = 38.1 ± 4.1 or [ 34.0, 42.2 ] The target waiting time of 30 minutes is not in the confidence interval, so there is evidence that the target is not being met. Click here to see an Excel spreadsheet with this example worked out (link to ExampleG.xls) Note that there are two competing claims or statements or hypotheses about the parameter. Target is being met: Ho: ( = 30 = (o null hypothesis Target is not being met: Ha: ( `" 30 alternative hypothesis Consider the quantity: ( EMBED Equation.3   (o) / s / EMBED Equation.3  If (o really is the true population mean waiting time, then ( EMBED Equation.3   (o) / s / EMBED Equation.3  should behave like a t-distribution and the value that you get when you compute it should be a reasonable value on the t-distribution. Here ( EMBED Equation.3 – (o) / s / EMBED Equation.3  = (38.1 – 30) / 10/(25 = 4.05 What is Pr (t > 4.05) = ? …. < .005 That is, what is the probability that a sample mean could be at least 4.05 standard deviations away from the hypothesized target population mean? This probability is called the P-value. In this case it’s way way less than 0.005. Since the P-value is so small, the probability that “chance” could account for the difference between the observed sample mean and hypothesized population mean is small. Therefore, we conclude that the hypothesized mean is not supported. The P-value assumes that Ho is correct and then evaluates the probability that “chance” can explain what’s going on. We will summarize and generalize this example into our first hypothesis test: One-sample t-test for (, the population mean Ho: ( = (o Test statistic: Ha: ( `" (o t = ( EMBED Equation.3   (o) / s / EMBED Equation.3  This test statistic has a t-distribution with n-1 degrees of freedom (Substitute in the values and compute the test statistic; call it tcalc.) P-value = 2 x Pr (t >| tcalc |) Notes: ( `" (o is a two-sided or two-tailed alternative hypothesis. There are also one-sided tests (i.e. based on one-sided or one-tailed alternative hypotheses). Ho: ( = (o Uses the same test statistic; but Ha: ( > (o P-value = Pr (t > tcalc) or Ho: ( = (o Uses the same test statistic; but Ha: ( < (o P-value = Pr (t < tcalc) The threshold P-value that is usually used to indicate statistical significance (i.e. rejection of Ho) is .05. Note that this means that observed differences are actually due to chance 1 time out of 20. What do you think of that rate? Note that a two-tailed P-value = 2 x (one-tailed P-value), so be careful about doing one-sided tests. You may get the contradictory result of significance for a one-sided alternative but not for a two-sided alternative, based on the same data! Quicklink for Further Reading: Statistics At Square One – T-tests http://bmj.bmjjournals.com/collections/statsbk/7.shtml Two-sample t-test for comparing two independent means µ1 – µ2 Ho: µ1 – µ2 = 0 (or D0) If D0 = 0 can also be written as: Ho: µ1 = µ2 Ha: µ1 – µ2 `" 0 (or D0) If D0 = 0 can also be written as: Ha: µ1 `" µ2 Test statistic: t = ( EMBED Equation.3    EMBED Equation.3 )  D0 /  EMBED Equation.3  separate or unequal variances version Example justifying unequal variances version: Comparing the mean heights of n1 (say 25) office workers with n2 (say 30) professional basketball players. In this case, you would anticipate that there would be less variability in the heights of the basketball players (assume they’re all tall and Mugsy Bogues had not been invented). In this case the population means are not expected to be equal (ie (12 `" (22 ) or: If (12 = (22 ,  EMBED Equation.3  = pooled variance Example justifying pooled variance: Comparing mean heights of n1 office workers with n2 school teachers. In this case, there is unlikely to be any difference in population variance (ie (12 = (22 ) so any difference in sample variances is likely due to random chance. If they ARE found to be different, we improve the situation by using a sort of average of the two variances (the “pooled” estimate). This gives us: t= ( EMBED Equation.3  –  EMBED Equation.3 ) – D0 /  EMBED Equation.3  pooled or equal variances version Note: t has (n1 + n2 – 2) degrees of freedom. P-value = 2 x Pr (t >| tcalc |) just as in the one-sample t-test! Note: One-sided alternative hypotheses are also possible. {? delete this as one-sided is easily confused with one-sample} Note: There is an equivalence between confidence intervals and two-sided hypothesis tests. If the hypothesized value of µ1 - µ2 is outside the 100(1  ±)% C.I., then the P-value from the hypothesis test is less than ±. Hence the null hypothesis would be rejected. Matched Pairs t-test (a.k.a. paired t-test) for two dependent means You may have heard the expression in study design that  each patient served as his/her own control”. What does that mean? If you have two measurements per subject, couldn’t you just compare the means of the two sets of measurements with the two-sample t-test just discussed? You could, but you would be ignoring an important feature of the data, and could easily miss finding significant differences. For example, suppose you wanted to test the effectiveness of a new fitness initiative in Belltown. Previous research has shown that dog owners get more walking exercise than people who do not have dogs. In this new initiative, people can borrow dogs from the Belltown SPCA free-of-charge so they have a reason to walk regularly. The question is whether the “mooch-a-pooch” program increases fitness levels of participants. A sample of 30 participants have their resting heart rate and body mass index (BMI) recorded at the start of the program and then again after six months. Program effectiveness would be seen in reduced heart rate and BMI. The design here is certainly different than in a two-sample t-test. In that situation you would have 30 subjects to begin and a different 30 subjects at six months – that would be silly since you are interested in change within an individual. Thus the need for a new test, the paired t-test for dependent means. If each experimental unit has two measurements (Before and After), or if the experimental units have been matched in pairs according to some characteristic, compute the differences and do a one-sample t-test on the differences. Ho: µ1 - µ2 = 0 (or D0) i.e. Ho: µd = 0 (or D0) Ha: µ1 - µ2 `" 0 (or D0) i.e. Ha: µd `" 0 (or D0) Compute the differences di = x1i  x2i Then compute the sample mean of the differences,  EMBED Equation.3  And the sample standard deviation of the differences, sd Then the test statistic is: t =( EMBED Equation.3  – D0) / sd/(n P-value = 2 x Pr (t >| tcalc | ) (exactly as before for the other t-tests) An Example of t-tests: A physician-cum-author has just published a new book entitled “Healthy Living for Dummies”. His publisher wants to test two types of displays in bookstores. 30 stores are chosen and matched on the basis of annual sales volume. In each pair, one store gets Display A and the other gets Display B. Sales for a one-week period are recorded. Pair # : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Display A: 46 39 40 37 32 26 21 23 20 17 13 15 11 8 9 Display B: 37 42 37 38 27 19 20 17 20 12 12 9 7 2 6 Diff.(di): 9 -3 3 -1 5 7 1 6 0 5 1 6 4 6 3 The correct way – a paired t-test Ho: µA - µB = 0 (i.e. µd = 0) Ha: µA - µB `" 0 (i.e. µd `" 0) Sample mean,  EMBED Equation.3  = 3.47 Sample std dev, sd = 3.31 Sample size, n = 15 Test statistic: t =( EMBED Equation.3   0) / sd/(n = (3.47  0) / 3.31/(15 = 4.05 P-value = 2 x Pr (t14 > 4.05) < .002 Conclusion: There is strong evidence that Display A has higher mean sales than Display B. Note: Although this is not exactly a Before/After design, the stores are matched so that each pair is equivalent to the same store being measured twice. The incorrect way – a two sample t-test Sample means:  EMBED Equation.3 A = 23.80 and  EMBED Equation.3 B =20.33 Sample SDs: sA = 12.33 and sB = 13.04 Pooled SD: sp = 12.691 Test statistic (pooled variance version): t = ( EMBED Equation.3  –  EMBED Equation.3 )  0 /  EMBED Equation.3  = (23.80  20.33) / 12.691 "(1/15 + 1/15) = 3.47 / 4.634 = 0.75 P-value >> .05, which suggests no evidence of a difference between displays What is the reason for such a difference in results? Too much variability in sales volume among stores hides the effect of the displays. Have a look at the Excel Spreadsheet located here. In it we have graphed the results showing the considerable variability in the data. Quicklink for Further Reading: Statistics At Square One – T-tests http://bmj.bmjjournals.com/collections/statsbk/7.shtml {Jonathan, is this where you want the Michael Jackson video?} Previously: Tests for one and two means of measurement data (independent and dependent) Next: Tests for categoric data. Parameters are: p (one population proportion), or p1 – p2 (the difference of two population proportions) One-sample z-test for p, the population proportion. {Jonathan, can you explain here why we can use a z-test here for sample data whereas we couldn’t in the case of the measurement variable tests and had to use a t-test? We medical types are disturbed when parallelism fails.} Ho: p = po Ha: p `" po [Remember the 100(1  ±)% CI  EMBED Equation.3 :  EMBED Equation.3  ± z±/2 EMBED Equation.3  ] Test statistic: Z = [ EMBED Equation.3    EMBED Equation.3 ] /  EMBED Equation.3  This test statistic has a z-distribution (Substitute in the values and compute the test statistic; call it zcalc.) P-value = 2 x Pr (z >| zcalc |) If Ha: p > po Same test statistic; but P-value = Pr (z > zcalc ) If Ha: p < po Same test statistic; but P-value = Pr (z < zcalc ) Two-sample z-test for comparing two proportions p1 – p2 Ho: p1 – p2 = 0 Can also be written as: Ho: p1 = p2 Ha: p1 – p2 `" 0 Can also be written as: Ha: p1 `" p2 Test statistic: Z = ( EMBED Equation.3 1   EMBED Equation.3 2) / EMBED Equation.3  OR: a modification (using a pooled p)  EMBED Equation.3 1 = X1 / n1 and  EMBED Equation.3 2 = X2 / n2 so  EMBED Equation.3  = (X1 + X2 )/(n1 + n2) Then the test statistic is: Z = ( EMBED Equation.3 1– EMBED Equation.3 2) / EMBED Equation.3  And just as before, P-value = 2 x Pr (z >| zcalc |) We could set up the data for a two-sample z-test for two proportions as follows: Sample 1Sample 2YesX1X2Non1 – X1n2 – X2Column Totaln1n2 This two-by-two table is the basis for our non-parametric (ie non-measurement) tests. The z-tests listed above will handle hypothesis testing. Quicklink for Further Reading: Statistics At Square One – Differences Between Percentages and Paired Alternatives http://bmj.bmjjournals.com/collections/statsbk/6.shtml Chi-square test of independence of two categoric variables This is an extension of the two-sample z-test to more than two categories and more than two samples. Example I: Is socio-economic status (SES) independent of smoking status? SESHighLowSmoking StatusYesNo This is a 2x2 table; expand it to a 3x3 table as follows: SESHighMiddleLowSmoking StatusCurrentFormerNever This display of data in a table is called a crosstabulation or, more simply, a crosstab. It is also called a contingency table. Ho: Row and column classifications are independent Ha: Row and column classifications are dependent Here is the previous table, now with data: SESHighMiddleLowRow TotalsSmoking StatusCurrent512243116Former922128141Never6892299Column Totals2115293356 Ho: Smoking and SES are independent Ha: Smoking and SES are dependent (i.e. Smoking is related to SES) The data in the previous table are called observed frequencies and are denoted by: Oij , which stands for the observed frequency of the cell in the ith row and jth column. Now, compute, for each cell, the expected frequency (or count) Eij under the assumption of independence. e.g. Pr (‘Current’ and ‘High’) = (under independence) Pr (‘Current’) x Pr (‘High’) = (116/356) x (211/356) So the Expected Freq. of (‘Current’ and ‘High’) = 356 x (116/356) x (211/356) The general rule is, therefore: Eij = (ith Row Total) x (jth Column Total) / Overall Total E11 = (116 x 211) / 356 = 68.75 Here is the previous table with the observed frequencies and the expected frequencies (in parentheses) SESOij/ (Eij)HighMiddleLowRow TotalsSmoking StatusCurrent51 (68.75)22 (16.94)43 (30.31)116Former92 (83.57)21 (20.60)28 (36.83)141Never68 (58.68)9 (14.46)22 ( 25.86)99Column Totals2115293356 Next we compare the Oij‘s with the Eij‘s, using a distribution that Karl Pearson called the chi-square distribution.  EMBED Equation.3 =  EMBED Equation.3  where the sum is taken over all the cells in the table. This test statistic has a EMBED Equation.3  distribution with (r – 1)(c – 1) degrees of freedom; (r = number of rows, c = number of columns) P-value = Pr ( EMBED Equation.3 > EMBED Equation.3 calc) (Note that it is one-sided) In our example:  EMBED Equation.3 = (51 – 68.75)2 / 68.75 + … + (22 – 25.86)2 / 25/86 = 18.52 with degrees of freedom = (3 – 1)(3 – 1) = 4 P-value = Pr ( EMBED Equation.3 (4 df) > 18.52 ) < .001 To see the data graphically and to see how Excel handles this type of problem see Excel file “Example I” Conclusion: There is strong evidence that smoking is related to SES (i.e. smoking status and socio-economic status are dependent). Note: A 2x2 table can be tested using either a two-sample z-test for two proportions or a  EMBED Equation.3  table. The relationship is simple: a 2x2 table has 1 degree of freedom and a  EMBED Equation.3  with 1 d.f. is equal to the square of a Z. Quicklink for Further Reading: Statistics At Square One – The Chi Squared Tests  HYPERLINK "http://bmj.bmjjournals.com/collections/statsbk/8.shtml" http://bmj.bmjjournals.com/collections/statsbk/8.shtml Types of Error in Hypothesis Testing DecisionAccept Ho Reject Ho True StateHo TrueCorrectType I errorHo FalseType II errorCorrect Type I Error = Reject Ho when Ho is true (also known as a  false positive ) Pr (Type I Error) = ± = level of significance Type II Error = Accept Ho when Ho is false (also known as a  false negative ) Pr (Type II Error) = ² Power = 1  ² = Prob. of correctly rejecting Ho (i.e. reject Ho when Ho is false) Examples: Ho: defendant is not guilty Type I error  convict an innocent person Type II error  let a guilty person go free Ho: patient does not have the disease Type I error – a diagnostic test says a healthy person has the disease Type II error – a diagnostic test fails to detect that a sick person is indeed ill Ideally we want the chance of each type of error to be zero. Unfortunately, when one is very small, the other is very large, so we compromise by trying to make both of them “reasonably small”; that is, we strive for a low threshold for significance (± < .05 and the lower the better) and a high power (² < .20). Quicklink for Further Reading: Statistics At Square One  Differences between means: type I and type II errors and power  HYPERLINK "http://bmj.bmjjournals.com/collections/statsbk/5.shtml" http://bmj.bmjjournals.com/collections/statsbk/5.shtml Interpreting P-values – what exactly does the word “significant” mean? If P-value < .01 , highly statistically significant result (i.e. very sure that the effect was real and not just due to sampling variability) If P-value < .05, statistically significant If P < .10, marginally or weakly statistically significant Do you need to include the modifier “statistically” in front of the word “significant”? 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Statistically significant means “the effect is real and not due to chance”. That is, the effect would still be there is you sampled again. Clinically significant means “the effect is important.” An effect can be statistically significant without being clinically significant. For example, consider a new formulation of an analgesic extends the mean duration of pain relief from 8 hours to 8 hours and 10 minutes. Even if a two-sample t-test to compare the old and new formulations shows that the new formulation is statistically significantly longer, it wouldn’t be clinically significant – 10 minutes on 8 hours is not important, clinically. An effect could be clinically significant, but if it isn’t statistically significant too you can’t even conclude that the effect is real. In this case the apparent effect is just a tease; it may or may not mean anything. One last reminder: The P-value is NOT the probability that Ho is correct! Instead….I would explain it again… Questions to ask in pursuit of an appropriate hypothesis test 1. a. What kind of data do you have? Measurement or categoric b. What parameter(s) is(are) of interest? µ (for measurement) ( t-tests  EMBED Equation.3  (for categoric) ( z-tests and  EMBED Equation.3  tests 2. a. How many populations/samples do you have? One, two, or many ( one-sample, two-sample, etc. b. How many measurements/observations per sample? Is there ‘matching’ or ‘ pairing’ or ‘repeated measuring’ of the experimental units? We only have the tools to deal with this for means, in the t-test situation. That is, is it a two-sample t-test or a paired t-test? 3. Have you checked for the shape of distribution (i.e. normality for measurement data, especially with small samples) and for outliers? 9. Introduction to nonparametric hypothesis tests The confidence intervals and hypothesis tests for measurement data discussed previously are based on the assumption that the populations from which the data are drawn are normally distributed. The t procedures are very popular because they work well even with a moderate lack of normality, as long as the samples are reasonably large. What can be done with severe non-normality especially with small samples? First, check whether the apparent HYPERLINK "http://davidmlane.com/hyperstat/A11284.html" \o "Click for definition"skewness is really just a cluster of outliers. If so, try analyzing the data with and without the outliers. Do the results change in a meaningful way, or are the results quite stable?. Second, consider transformations of the data. For example, taking the logarithm of the data will pull in the long right-hand tail of right-skewed distributions, making the log-scale data approximately normal. Another useful transformation is to take the square root of the data. It may seem non-intuitive that you can just “transform” the data but it’s analogous to multiplying both sides of an equation by 6 – the size of the parameters change but the fundamental inter-relationships between them remain the same. Third, there are procedures based on distributions other than the normal; for example, equipment lifetimes (ie imagine the distribution of lightbulb survival). But these procedures are beyond the scope of this course (and not used widely enough in medical research, I’m afraid). Finally, there are procedures than make no assumptions about the distribution of the populations. These are called nonparametric methods, because there are no distributional parameters such as the mean or standard deviation of the normal. The tests are also known as rank tests, for reasons which will be seen shortly. These tests are somewhat less powerful than their parametric cousins and so may miss a difference that is actually there. When you have the choice, use the parametric version. We will consider two such nonparametric procedures which are analogous to the two-sample (independent means) t-test and the paired (dependent means) t-test. SettingNormal distribution testNonparametric (rank) testOne sampleOne-sample t-testWilcoxon signed-rank test (rarely used because …)Matched Pairs (Two dependent samples)Paired t-test (=one-sample t-test on the differences)Use the Wilcoxon signed-rank test on the differencesTwo independent samplesTwo-sample t-testWilcoxon rank sum test (a.k.a. Mann-Whitney U test** note the potential name confusion: the Wilcoxon signed-rank (SR) test is different from the Wilcoxon rank-sum (RS) test. Too bad they weren’t discovered by different people. We will begin with the last one, the Wilcoxon rank sum test, and consider two situations and types of data. 1. Suppose you have measurement data from a highly skewed distribution. Put the data values from the two independent samples together and arrange them from smallest to largest, but remember to keep an identifier so you can tell which observations came from which sample. Next, assign ranks to the observations. Ranks are simply the positions that the observations have in the ordered list, so the smallest data value will have rank 1, the second smallest rank 2, and so on. If there are  EMBED Equation.3  and  EMBED Equation.3  observations respectively in the two samples; the highest rank will be N =  EMBED Equation.3 . EMBED Equation.3  2. Suppose the data are already ranks. Streiner and Norman give an example of a trial for psoriasis using two different treatments; 10 subjects receive one treatment and 10 other subjects receive the other treatment. Afterward a dermatologist receives photos of the 20 subjects and ranks all 20 of them in order from best to worst in outcome. For each situation, the Wilcoxon rank sum statistic W is the sum of the ranks for the first sample. Exact P-values can be computed by software. Alternatively, approximate P-values can be computed by using a normal approximation for W. Small P-values reject the hypothesis that the two populations have identical distributions. If the two distributions turn out to have the same shape you can think of this test as a comparison of two medians; hence it is analogous to the two-sample t-test of two means. Example (Streiner and Norman): Two treatments (A and B) for psoriasis; A is the active treatment, B is the control Rank 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Treatment B B B B A B A B A A B A A B B A B A A A Here is the table again, with the ranks separated out for the two treatments Treatment B B B B A B A B A A B A A B B A B A A A A’s Ranks 5 7 9 10 12 13 16 18 19 20 B’s Ranks 1 2 3 4 6 8 11 14 15 17 The sum of the ranks for A is 129, and for B is 81. An exact two-tailed P-value from tables of the Wilcoxon test is .075. (See the table at:  HYPERLINK "http://bmj.bmjjournals.com/collections/statsbk/apptabe.shtml" http://bmj.bmjjournals.com/collections/statsbk/apptabe.shtml -- the smaller of the sums should be < 78 for p to be < 0.05). Hence there is weak evidence (0.075 but not <0.05) supporting the effectiveness of the active treatment. More testing is warranted. Notes: Sometimes when you undertake the ranking you discover that two data values are equal. To assign ranks in that case, simply assign all tied values the average of the ranks they occupy. For example, if the third and fourth observations were equal, they would each get a rank of 3.5. One last confusing thing about this test: It is also known as the Mann Whitney test. You can call it whatever your software calls it! Quicklink for Further Reading: Statistics At Square One – The Rank Tests http://bmj.bmjjournals.com/collections/statsbk/10.shtml Wilcoxon signed-rank test for matched pairs: As with the matched pairs t-test you begin by computing differences in responses within pairs. Next, rank the absolute values of these differences. The Wilcoxon signed rank statistic is the sum of the ranks of the positive differences. Note that a tie within a pair means a difference of zero. These cases must be dropped from the sample. Tied absolute differences are treated as before. Example Wilcoxon 2 {Link this to file Wilcoxon2.xls} (Adapted from Moore and McCabe) Twelve Belltown physicians/wannabe-golfers in Belltown play two rounds of golf at two different golf courses. As budding researchers they want a formal evaluation of whether scores were similar at the two courses. Here are the data. Player 1 2 3 4 5 6 7 8 9 10 11 12 Round 2 94 85 89 89 81 76 107 89 87 91 88 80 Round 1 89 90 87 95 86 81 102 105 83 88 91 79 Difference 5 –5 2 –6 –5 –5 5 –16 4 3 –3 1 Rearrange the absolute values of the differences in ascending order, keeping track of which were negative (in bold) and which were positive. Absolute difference 1 2 3 3 4 5 5 5 5 5 6 16 Rank 1 2 3.5 3.5 5 8 8 8 8 8 11 12 Ranks of neg. diffs. 3.5 8 8 8 11 12 Ranks of pos. diffs. 1 2 3.5 5 8 8 The sum of the ranks of the positive differences is 1+2+3.5+5+8+8=27.5. We could just have easily used the sum of the ranks of the negative differences (3.5+8+8+8+11+12=50.5) because it was arbitrary whether we subtracted Round 1 from Round 2 or Round 2 from Round 1. The null hypothesis is that the scores have the same distribution in both rounds. The alternative is that scores are systematically lower or higher in Round 2. Software gives the two-tailed P-value to be 0.38. Hence there is no evidence of a systematic difference in scores between rounds. Click here to see an Excel spreadsheet with this problem worked out. Quicklink for Further Reading: Statistics At Square One – The Rank Tests http://bmj.bmjjournals.com/collections/statsbk/10.shtml Concluding Notes about nonparametric tests: In a nonparametric test there is no measure of the size of the observed effect, and hence no confidence intervals. I prefer to use nonparametric tests in conjunction with their parametric counterparts. For example, in a series of two-group comparisons I will carry out two-sample t-tests of means AND Wilcoxon Rank Sum (a.k.a. Mann-Whitney) tests. This is a neat way of addressing the question of whether the distributions are skewed (rationale). If the results of the two tests are similar, the skewness issue becomes moot. There are many other procedures that can be considered “nonparametric”; for example, the sign test. In fact, the chi-square test of independence is also nonparametric because it makes no distributional assumptions. But when researchers use the term “nonparametric”, it is the tests we discussed (and a few others) that they are referring to. Although nonparametric tests seem like the poor cousins of normal-distribution tests, they are a useful addition to your statistical toolkit. What is coming in Part 2? Statistical Model-Building: Correlation and Regression (Simple and Multiple) One-Way Analysis of Variance Other Topics: Two-Way Analysis of Variance Analysis of Covariance Repeated Measures Analysis of Variance Logistic Regression Factor Analysis Here is a schematic for the techniques we have already discussed (1) and for those still to be discussed (2): Legend: X = Independent Variable (predictor, explanatory factor) Y = Dependent Variable (outcome, response) Variables are measurement or categoric; categoric variables can have either two categories only (i.e. binary) or more than two Statistical Test Dependent Variable  Independent VariableTwo-sample t-tests of meansOne measurement scale Y variableOne binary X variableSimple RegressionOne measurement scale Y variableOne measurement X variableMultiple RegressionOne measurement scale Y variableMultiple X variables; some measurement, some categoricLogistic RegressionOne binary Y variableMultiple X variables; some measurement, some categoricTwo-sample z-test of proportionsOne binary Y variableOne binary X variableChi-square test of independenceOne categoric Y variable (e"2 categories)One categoric X variable (e"2 categories)One-way analysis of varianceOne measurement scale Y variableOne categoric X variable (e"2 categories)Two-way analysis of varianceOne measurement scale Y variableTwo categoric X variables (e"2 categories) Quicklink for Further Reading: Statistics At Square One – Study Design and Choosing A Statistical Test http://bmj.bmjjournals.com/collections/statsbk/13.shtml >>> END OF MODULE – ANALYSIS OF QUANTITATIVE DATA Part 1     PAGE  PAGE 46 Where does this come from?  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