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19 – 1 INTRODUCTION TO STATISTICAL METHODS 1. Introduction Statistics is the branch of scientific method that is concerned with collecting, arranging, and using numerical observations or data that arise from natural phenomena or experiment. For our purposes, statistical methods will be useful in one or more of the following areas: 1. Reduction of data. Much numerical information may often be "condensed" into a simple relationship, together with a statement as to the confidence we may place in the relationship. 2. Estimates and significance tests. From experimental data, certain population parameters (such as a mean) can be estimated. It is usually possible to determine whether or not these estimates differ significantly from certain preconceived values. 3. Reliability of inferences depending on one or more variables. For example, the total impurity content of a bulk shipment is predicted from samples taken throughout the shipment. We then ask, what reliability can be attached to the prediction of total impurity content? This leads to uncertainty analysis}. 4. Relationships between two or more variables. Suppose that some measurable quantity depends on one or more separate factors. Then, if it is possible by experimental design to control the separate factors at a series of fixed levels and to observe the measurable quantity at each level, the technique known as analysis of variance (ANOVA) is used to evaluate the dependency. 2. Definitions and Notation The following definitions will be important in later sections: • Statistic-—an item of information deduced from the application of statistical methods. • Population-—a collection of objects (or measurements) having in common some observable or measurable characteristic known as a variate. • Individual-—a single member of a population. • Sample-—a group of individuals drawn from a population, usually at random, so that each individual is equally likely to be selected. • Random variable —a numerical quantity associated with the variate. The value of the random variable for a given individual is determined by the value or nature of the variate for that individual. For example, if the variate were the color of an object, values 1, 2, 3, etc. might be assigned to the colors red, green, blue, etc. • Continuous variable-—one that can assume any value within some continuous range, such as the temperature of a reactor. • Discontinuous or discrete variable-—one that can assume only certain discrete values, such as the number of days in a month. 19 – 2 • Probability-—P, that a random variable x, belonging to an individual drawn from a population, shall have some particular value A, equals the fraction of individuals in the population having that value A associated with them. Example A deck of 52 cards contains 13 cards in each of the spade (♠), heart (♥), diamond (♦), and club (♣) suits. What is the probability that two cards dealt randomly from the deck will both} be spades? Solution By definition, the probability that the first card dealt will be a spade equals the fraction of cards in the deck that are spades, namely, 13/52 = 1/4. Likewise, the probability that the second card dealt will be a spade is 12/51, since there are now only 12 spades in the deck, whose overall number has been reduced from 52 to 51. The combined probability that both events will occur is obtained by multiplying the individual probabilities, and is P = 13 52 × 12 51 = 3 51 = 1 17 = 0.0588 x1 x2 f(x) x dx x0 f(x0) Fig. 1 Frequency function for continuous distribution. Frequency function  continuous distribution. In general, the probability that a continuous random variable x will have a specified value is infinitesimally small, since an infinite range of values is possible. However, we can define the probability that the random variable will lie in the very narrow range between x0 and x0 + dx, as shown by the tall thin shaded area at the left of Fig. 1. By definition of the frequency function f(x) (also known as the probability density function), this probability is: 19 – 3 P(x1 ≤ x ≤ x0 + dx) = f(x0 )dx (2.1) The probability that the random variable will lie in a finite range is obtained by integration under the frequency-function curve, also illustrated by the wider shaded area at the right of Fig. 1: P(x1 ≤ x ≤ x2 ) = f(x)dx. x1 x2∫ (2.2) A special case of Eqn. (2.2) is P(−∞ ≤ x ≤ ∞) = f (x)dx = 1, −∞ ∞∫ (2.3) since the probability that x must lie somewhere is one. In Eqn. (2.3), the extreme limits of infinity have been used to account for all possibilities. In practice, however, finite limits can often be used with the same effect. x1 x F(x) 1 0 Fig. 2 Cumulative frequency function for continuous distribution. The cumulative frequency function, F(x), illustrated in Fig. 2, is defined as the probability that the random variable x has a value less than or equal to a specified value such as x1 F (x1) = P(−∞ ≤ x ≤ x1) = f (x)dx. −∞ x1∫ (2.4) Clearly, F (∞) = 1.. It is also possible to consider functions of the random variable, such as y(x). For example, if x were the diameter of a sphere, then y could be its volume, πx3/6. In this event, the expected or arithmetic average value of y(x) is defined for a continuous distribution by weighting the function with f(x) and then integrating over the whole spectrum: 19 – 4 E( y)[= ave( y)] = y(x) f (x)dx, −∞ ∞∫ (2.5) The term "expected" is actually a misnomer, since the probability of any one value occurring is infinitesimally small, as already explained. However, the terminology has stuck, so we shall use it. For the special case of y being the random variable x itself, the result is the mean µ of the distribution: µ = E(x) = xf (x)dx. −∞ ∞∫ (2.6) In addition to the mean, there are two further statistics that give an idea of the "center of gravity" of the distribution: 1. The median xm is defined so that half the values of x lie above it, and half below: F (xm ) = 1 2 . (2.7) 2. The mode is the most frequently occurring value of x , and (for a singly peaked distribution, such as in Fig. 1), is the value of x corresponding to the peak of the distribution. For symmetrical distributions, the above statistics coincide. For the distribution of Fig. 1, which is "skewed" to the left, the mean and median would both lie to the right of the mode. The variance is a measure of the "spread" of a particular distribution and is defined for a continuous distribution as σ 2 = var(x) = E[(x − µ )2 ] = (x − µ )2 −∞ ∞∫ f (x)dx (2.8) The standard deviation σ is the square root of the variance: σ = var(x) . (2.9) A useful formula for E(x2) is obtained by first noting that σ 2 = E[(x − µ )2 ] = E(x2 − 2xµ + µ 2 ) = E(x2 ) − 2µE(x) + E(µ 2 ) = E(x2 ) − µ 2 , (2.10) from which it follows that E(x2 ) = µ 2 + σ 2 . (2.11) Frequency function—discontinuous random variable. With some modifications, the above definitions also carry through for a discontinuous random variable. The probability of observing a specified value of the random variable, such as x0, is 19 – 5 P(x = x0 ) = f (x0 ). (2.12) f(x) x1 x2 x Fig. 3 Frequency function for discontinuous distribution. The probability that x lies within a range is now obtained by summation rather than integration: P(x1 ≤ x ≤ x2 ) = f (x). x1 x2∑ (2.13) The arithmetic average for a function y(x) is defined as E( y) = y(x) f (x), −∞ ∞∑ (2.14) and the mean and variance of the distribution are similar to Eqns. (2.6) and (2.8), except that summation replaces integration. Finally, for discontinuous distributions, the mean and median may lie between two possible values of x. Linear combination. Later in these notes, we shall need to examine a linear combination or sum of n independent random variables, x1, x2 ,..., xn , each weighted by a corresponding coefficient: y = ai i=1 n∑ xi = a1x1 + a2x2 +...+anxn . (2.15) The mean µy of the linear combination is the sum of the individual means µi, each weighted by the coefficient ai: 19 – 6 µ y = E( y) = E ai i=1 n∑ xi = ai i=1 n∑ E(xi ) = ai i=1 n∑ µ i . (2.16) The variance σ y 2 of the linear combination is given by σ y = E( y − µ y )2 = E ai (xi − µ i ) i=1 n∑   2 = E ai 2 (xi − µ i )2 + aiaj (xi − µ i )(xj − µ j) i, j =1 i≠ j n∑ i=1 n∑       . (2.17) But, for xi and xj independent, E x x E x E xi i j j i i j j( )( ) ( ) ( ) .− − = − − =µ µ µ µ 0 (2.18) Hence, Eqn. (2.17) yields σ y 2 = ai 2 i=1 n∑ E(xi − µ i )2 = ai2σ i2 i=1 n∑ , (2.19) so the variance of the linear combination is the sum of the individual variances, each weighted by the square of the corresponding coefficient ai. 3. Sample Statistics Consider a sample that comprises n independent observations x1,x2,...,xn on the random variable x. (For example, x could represent the populations of the states in which members of the ChE 360 class normally reside.) Within the limitations of the sample size (it is necessarily limited to a finite number of observations), we would like to estimate as accurately as possible some statistics about the population from which x is drawn, notably its mean and variance. We start by defining the sample mean as the arithmetic average of the values x1,x2,...,xn in the sample: x = 1 n (x1 + x2 +...+ xn ) = 1 n xi i=1 n∑ = 1ni=1 n∑ xi . (3.1) Thus, x is a linear combination of the xi , each of which has the same mean µ and variance σ 2, and each of which is weighted by the same coefficient, ai = 1/n. It immediately follows from Eqns. (2.16) and (2.19) that the mean µ x and variance σ x 2 of the sample mean are: µ x = 1 ni=1 n∑ µ = µ . (3.2) σ x 2 = 1 n2i=1 n∑ σ 2 = σ 2 n . (3.3) 19 – 7 Thus, the sample mean is an unbiased estimate of the population mean. That is, if a large number of samples is taken, their mean will accurately reflect the population mean. Also, the variance of the sample mean becomes smaller as the sample size is increased ("there is safety in numbers"); however, the decrease is proportional only to 1/n, so that a quadrupling of sample size would be needed to halve the variance of x . The sample variance is defined as s2 = 1 n − 1 (xi − x ) 2 i=1 n∑ , not 1 n (xi − x ) 2 i=1 n∑ . (3.4) The mean or expected value of the sample variance is µ s2 = E s2( ) = E 1 n − 1 (xi − x ) 2 i=1 n∑   = 1 n − 1 E xi 2 − nx2 i=1 n∑   . (3.5) But, from Eqn. (2.11), noting that the variances of x and x are σ 2 and [from Eqn. (3.3)] σ 2/n, respectively: E xi 2( ) = µ 2 + σ 2 , (3.6) and E x( )2 = µ 2 + σ 2 n . (3.7) Hence µ s2 = 1 n − 1 n µ 2 + σ 2( ) − n µ 2 + σ 2 n         = σ 2 . (3.8) Thus the sample variance as defined by the first expression in Eqn. (3.4) is an unbiased estimate of the population variance. If the second expression had been used, an underestimate would have resulted. 4. The Normal Distribution The general concept of the normal distribution is that, superimposed on the true value µ for any random variable x, there is a very large number n of very small errors δ . Each δ has an equal probability of being positive or negative. The corresponding frequency function can be shown to be f x( ) = 1 σ 2π exp − (x − µ )2 2σ 2     , (4.1) in which σ 2 = 2nδ 2 as n → ∞, δ → 0. 19 – 8 σ σ µ f(x) x0 Fig. 4 Normal distribution frequency function. Examples of variables likely to conform to the normal distribution are: 1. A particular experimental measurement subject to several random errors. 2. The time taken to travel to work along a given route. Here, the variability of traffic lights would afford the "errors"---some positive, some negative---that are superimposed on an average driving time. 3. The heights of women belonging to a certain race. (If "people" were substituted for "women," would you still expect the normal distribution to be obeyed?) 4. The logarithm of the diameter of particles in a powder is an example of a case in which one variable (diameter) does not conform to the normal distribution, but a transformation of that variable (the logarithm) does. In Eqn. (4.1), 1 / σ 2π can be viewed as a normalizing factor that automatically insures that f (x)dx = 1. −∞ ∞∫ (4.2) By integration—not particularly easy---µ and σ 2 may indeed be shown to be the mean and variance of the distribution: Mean: E(x) = xf (x)dx = µ . −∞ ∞∫ (4.3) Variance: E(x − µ )2 = (x − µ )2 −∞ ∞∫ f (x)dx = σ 2 . (4.4) A random variable that is normally distributed with mean µ and standard deviation σ is said to be an N (µ ,σ ) variable. Standardized normal distribution. To accommodate normal distributions with different means and variances, we can define a standardized random variable ξ by 19 – 9 ξ = x − µ σ , (4.5) which is the deviation of x from its mean µ, as a fraction of its standard deviation σ . The reader should check from Eqn. (4.1) that the corresponding standardized normal frequency function φ ξ( ) is given by φ ξ( ) = 1 2π e − 1 2 ξ 2 , (4.6) which is also illustrated in Fig. 5. Note that ξ is an N(0,1) variable. 0 Area beyond ξ is P ξ Area up to ξ is Φ(ξ) φ Fig. 5 Standard normal distribution. The cumulative frequency function Φ ξ( ) for the standardized normal variable is the probability that the variable has a value less than or equal to ξ : Φ ξ( ) = φ ξ( )dξ −∞ ξ∫ . (4.7) Table 1 tabulates the frequency function φ and the cumulative frequency function Φ for zero and positive values of the standardized normal variable ξ. By referring to Fig. 5, the reader should verify that negative values -ξ are accommodated by the formulas: φ −ξ( ) = φ ξ( ), (4.8) Φ −ξ( ) = 1− Φ ξ( ). (4.9) Table 1 Values of the Standardized Normal Distributionξ φ Φ ξ φ Φ 0.0 0.3989 0.5000 2.0 0.0540 0.9772 0.1 0.3970 0.5398 2.1 0.0440 0.9821 19 – 10 0.2 0.3910 0.5793 2.2 0.0355 0.9861 0.3 0.3814 0.6179 2.3 0.0283 0.9893 0.4 0.3683 0.6554 2.4 0.0224 0.9918 0.5 0.3521 0.6915 2.5 0.0175 0.9938 0.6 0.3332 0.7258 2.6 0.0136 0.9953 0.7 0.3122 0.7580 2.7 0.0104 0.9965 0.8 0.2897 0.7881 2.8 0.0079 0.9974 0.9 0.2661 0.8159 2.9 0.0060 0.9981 1.0 0.2420 0.8413 3.0 0.0044 0.9986 1.1 0.2178 0.8643 3.1 0.0033 0.9990 1.2 0.1942 0.8849 3.2 0.0024 0.9993 1.3 0.1714 0.9032 3.3 0.0017 0.9995 1.4 0.1497 0.9192 3.4 0.0012 0.9997 1.5 0.1295 0.9332 3.5 0.0009 0.9998 1.6 0.1109 0.9452 3.6 0.0006 0.9998 1.7 0.0940 0.9554 3.7 0.0004 0.9999 1.8 0.0790 0.9641 3.8 0.0003 0.9999 1.9 0.0656 0.9713 3.9 0.0002 1.0000 2.0 0.0540 0.9772 4.0 0.0001 1.0000 Table 2 Tail Areas of the Standardized Normal Distribution} P 0.20 0.10 0.05 0.02 0.01 0.005 0.002 0.001ξP 0.843 1.282 1.647 2.056 2.329 2.578 2.880 3.092 Example Referring to Table 1, what value of ξ has a 20% probability of being exceeded? Check your answer against Table 2. Solution We seek a value for ξ below which 80% of the population lies. Thus, the appropriate value of Φ is 0.8. Table 1 indicates that Φ (0.8) = 0.7881 and Φ (0.9) = 0.8159. By linear interpolation, the required value for ξ is approximately 0.8 + (0.9 − 0.8) × 0.8000 − 0.7881 0.8159 − 0.7881 = 0.8428, which is essentially the same as the first entry in Table 2, namely, ξP=0.2 = 0.843. Example The times taken to travel from your home to class on n=7 different days are: 30, 33, 26, 23, 30, 35, 27 min. (a) If you are willing to take a 1\% chance of being late for class, how long before the class starts should you set out? 19 – 11 (b) In this event, what is the probability of your being able to buy a cup of coffee and a donut (which take 12 minutes to consume) before class starts? Solution Since the exact population parameters are unknown, we must estimate them from the sample mean and variance, which are: x = 1 n xi i=1 n∑ = 30 + 33 + 26 + 23 + 30 + 35 + 277 = 29.14 min. (4.10) s2 = 1 n − 1 (xi − x ) 2 i=1 n∑ = 1n − 1 xi2 − nx2i=1 n∑   = 302 + 332 +...+272( ) − 7 × 29.142 7 − 1 = 17.34 min2. (4.11) The corresponding standard deviation is s = 17.34 = 4.14 min. Based on the above, and the knowledge that the travelling time is subject to many small uncertainties, assume that it is normally distributed, with µ = 29.14 min, σ = 4.14 min. (4.12) It is convenient to work in terms of the standardized normal variable, ξ = x − µ σ = x − 29.14 4.14 , (4.13) as shown in Fig. 6. For a tail\/ area of 1% = 0.01, observe from Table 2 that the value of the standardized normal variable having has a 0.01 probability of being exceeded is ξ = 2.329. Therefore, to incur this risk, you start at x = µ + ξσ = 29.14 + 2.329 × 4.14 = 38.78 min (4.14) before the scheduled class time. For part (b), which asks for the probability of arriving at least 12 minutes before class, i.e., at x = 38.78 - 12 = 26.78 (or earlier), the corresponding value of ξ is ξ = x − µ σ = 26.78 − 29.14 4.14 = −0.57, (4.15) for which [from Table 1, also invoking Eqn. (4.9) because ξ is negative] the fractional area in the tail is 0.285. 19 – 12 29.14x (min) ξ φ ξ = 26.78 - 29.14 4.14 = -0.570 Donut and coffee time 2.329 σ Fractional area = 0.01 Class starts here-0.57 0 Fractional area = 0.285 x = 29.14 + 2.329 × 4.14 = 38.78 26.78 2.329 Fig. 6 Arrival-time frequency function. 5. Student t Distribution In statistics, we often wish to test the hypothesis that a sample could conceivably have been taken from a certain population. One way of testing is to see if the sample mean x is realistically close (in terms of the standard deviation) to the population mean µ. Two different situations arise, depending on whether or not the population standard deviation σ is known a priori, or has to be estimated from the sample standard deviation s. Hypothesis A: "A sample, whose mean is x , comes from a normal distribution N(µ, σ) with mean µ and known variance σ2.'' In such an event, we first note from Eqn. (3.3) that the sample mean has a variance σ2/n and hence a standard deviation σ/ n . The hypothesis is then tested by computing the N(0,1) variable ξ µ σ = −( )x n (5.1) and checking it against tabulated values. The hypothesis is rejected if ξ is unduly large. Hypothesis B: "A sample , whose mean is x , comes from a normal distribution N(µ, s) with mean µ and variance σ2, the latter of which is estimated as being the same as the sample variance s2." 19 – 13 Table 3 Percentile Values (tp) for Student's t Distribution for Various Degrees of Freedom ν t. .55 t..60 t..70 t..75 t..80 t..90 t..95 t..975 t..99 t..995 1 .158 .325 .727 1.000 1.376 3.08 6.31 12.71 31.82 63.66 2 .142 .289 .617 .816 1.061 1.89 2.92 4.30 6.96 9.92 3 .137 .277 .584 .765 .978 1.64 2.35 8.18 4.54 5.84 4 .134 .271 .569 .741 .941 1.53 2.13 2.78 3.75 4.60 5 .132 .267 .559 .727 .920 1.48 2.12 2.57 3.36 4.03 6 .131 .265 .553 .718 .906 1.44 1.94 2.45 3.14 3.71 7 .130 .263 .549 .711 .896 1.42 1.90 2.36 3.00 3.50 8 .130 .262 .546 .706 .889 1.40 1.86 2.31 2.90 3.36 9 .129 .261 .543 .703 .883 1.38 1.83 2.26 2.82 3.25 10 .129 .260 .542 .700 .879 1.37 1.81 2.23 2.76 3.17 11 .129 .260 .540 .697 .876 1.36 1.80 2.20 2.72 3.11 12 .128 .259 .539 .695 .873 1.36 1.78 2.18 2.68 3.06 13 .128 .259 .538 .694 .870 1.35 1.77 2.16 2.65 3.01 14 .128 .258 .537 .692 .868 1.34 1.76 2.14 2.62 2.98 15 .128 .258 .536 .691 .866 1.34 1.75 2.13 2.60 2.95 16 .128 .258 .535 .690 .865 1.34 1.75 2.12 2.58 2.92 17 .128 .257 .534 .689 .863 1.33 1.74 2.11 2.57 2.90 18 .127 .257 .534 .688 .862 1.33 1.73 2.10 2.55 2.88 19 .127 .257 .533 .688 .861 1.33 1.73 2.09 2.54 2.86 20 .127 .257 .533 .687 .860 1.32 1.72 2.09 2.53 2.84 21 .127 .257 .532 .686 .859 1.32 1.72 2.08 2.52 2.83 22 .127 .256 .532 .696 .858 1.32 1.72 2.07 2.51 2.82 23 .127 .256 .532 .685 .858 1.32 1.71 2.07 2.50 2.81 24 .127 .256 .531 .685 .857 1.32 1.71 2.06 2.49 2.80 25 .127 .256 .531 .684 .856 1.32 1.71 2.06 2.48 2.79 26 .127 .256 .531 .684 .856 1.32 1.71 2.06 2.48 2.78 27 .127 .256 .531 .684 .855 1.31 1.70 2.05 2.47 2.77 28 .127 .256 .530 .683 .855 1.31 1.70 2.05 2.47 2.76