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Session V: The Normal Distribution Continuous Distributions (Zar, Chapter 6) The subject of this chapter: Continuous Distributions The most important continuous distribution: The Normal distribution. 1) Originally defined by Laplace (~1799) 2) Developed by Gauss (~1850): often called the Gaussian distribution although some say he had little to do with it. 3) Named the “Normal Distribution” by Karl Pearson (1920) 4) Has nothing to do with normal vs. abnormal, but is the “common” distribution of many processes. Some Review: Universe Sample Space X3 X2 x1,x2, x3 , xn Xn X1 f (x) n 0 bin width Histogram Density x 1 100% F(x) n 0 bin width 0 x Distribution Function x F(x) f (u)du 0 Cummulative Histogram The “Normal” Distribution The Density Function N(m,s2): 1 x m 2 1 f ( x) e 2 s 2s 3.14159 ; e 2.71828 m xf ( x) dx s ( x m )2 f ( x)dx 2 Recall… n x i 1 n s2 i 1 s= s2 xi n estimates m ( xi x ) n 1 2 estimates s 2 estimates s If m 0 and s 1 "Standardized Normal" y x-m N(0,1) x ys m s If x ~ N(m , s 2 ) then y ~ N(0,1) From Table B.2 (App 17) Pr y> 0.0 0.5000 Pr y 0.0 Pr y> 1.0 0.1587 Pr y 1.0 Pr y> 2.0 0.0228 Pr y 2.0 Pr y> 2.57 0.0051 Pr y 2.57 Pr y> 3.0 0.0013 Pr y 3.0 34% 13.5% 2.5% Pr 0.0 y< 1.0 Pr 0.0 y Pr 1.0 y About 1/3 =.5000 - 0.1587 = .3413 Pr 1.0 < y <2.0 Pr 1.0 y Pr 2.0 y About 13.5 % = 0.1587 - 0.0228 = .1359 Pr y > 2.0 0.0228 About 2.5% 66.7% 95% A test of the normal distribution: Ho Bin Obs : : : N( x ,s2 ) (-,-2) (-2,-1) h1 h2 (-1,0) (0,1) h3 h4 (1,2) h5 (2,) h6 Prob : .025 .34 .135 .025 Expected : n.025 n.135 etc .135 .34 2 ? DF = 6 -1-1 1 3 for x for s 2 Moments Definitions: mp x p f ( x)dx ( Note : m1 m ) (estimated by x ) Central Moments: Kp ( x m ) p f ( x ) dx First Central Moment is Zero: K1 1 ( x m ) f ( x)dx xf ( x)dx m f ( x)dx mm 0 Second Central Moment is the Variance: K2 2 ( x m ) f ( x)dx s2 Estimated by s2 “Machine Calculation Formula” for the Variance: s 2 (x x) i Hint: 2 n 1 2 2 x nx i n 1 2 2 a b a 2 ab b ( ) 2 Moments (Cont.) Third Moment: Skewness K3 3 ( x m ) f ( x)dx K3 < 0 Mean < Median Skewed “to the left” K3 = 0 Mean = median Symmetric K3 > 0 Mean> Median Skewed “to the right” Normalized as: K3 1 s “Unit”-less 3 Estimated by: n k3 n ( xi x )3 i 1 (n 1)(n 2) and k3 g1 3 s “Machine” formula: k3 n xi3 3 xi xi2 2( xi )3 / n (n 1)(n 2) Fourth Moment: Kurtosis k4 4 ( x m ) f ( x) dx Normalized by 2 k4 s 4 3 “Unit”-less Estimated by ( xi x ) n(n 1) /(n 1) 3 ( xi x ) 4 k4 ( n 2)( n 3) and g2 2 k4 3 4 s 2 Why –3? Normal Distribution has uncorrected 4th moment = 3 K4 < 0 Lepto-kurtotic K4 = 0 Meso-kurtotic K4 > 0 Platy-kurtotic So, for the normal distribution, k4 s 4 X ~ N( m , s 2 ) 3 Moment = Symbol For normal Estimated by 1st moment (mean) = m m x 2nd moment (variance) = s 2 s2 s2 3rd moment (skewness) = 1 0 g1 4th moment (kurtosis) = 2 0 g2 Many distributions can be characterized by their first four moments. A system called the Pearson system (after E.S. Pearson) is such a system. Not used much any more. So… Why the “Normal” Distribution? Distribution of Means: Universe These are data points. If we selected them again they would be different. X1 Xn X3 x1, x2 , x3 ,, xn X2 These are random variables that generate. x1 , x2 , x3 ,, xn Now n xn xi / n i 1 is a data point, but if we did the experiment over and over again calculating a new x each time, they would be different and have a distribution based upon the random variables n Xn Xi / n i 1 So if X1, X 2 , X 3 ,, X n is a random sample, what’s the distribution of X n ? Start easy! n=2 If X1 and X2 are independent random variables (like a random sample) with means m1 and m2 and 2 variances s 1 and s 2 , then Mean(aX1 bX 2 ) am1 bm2 Variance(aX 1 bX 2 ) a 2s 12 b 2s 22 For a general “n”: If X1 , X 2 , X 3 ,, X n is a random sample with mean m and s 2 , then Mean ( X ) Mean ( X i / n ) Mean( xi ) / n = m /n=nm /n=m and n V ariance( X ) V ariance( X i ) / n 2 i 1 s 2 / n2 ns 2 / n 2 s 2 / n Note: True for any random sample with any 2 m and s ! distribution with Distribution of Xn Central Limit Theorem: If X1 , , X n is a random sample with mean m and variance s 2 , then n Dist( X n ) N (m , s 2 / n) or more precisely X m n Dist n N (0,1) s/ n This means that if n is large enough, then X is normally distributed, or at least close enough. How large does nhave to be? Ex: The uniform distribution 1 1 if 0 x 1 f ( x) 0 otherwise f 0 0 x F ( x) f ( u ) du if x 0 0 x if 0 x 1 1 if x 1 1 x How many samples are necessary before X n is normally distributed Authority Number IBM 12 DAJ ~30 Biomedical considerations Typical sample is actually a large sum of independent elements all with about the same distribution. or, at worst, Sample is the sum of distinct clones Testing the sample for the Normal Distribution! Chi-Square Goodness of Fit Chi-Square we have observed and expected Frequencies. We then combined them in the Chi-Square statistics. For continuous distribution and sample x1 , x2 , , xn , form the histogram: h3 h2 h1 observed: h1 h2 h3, h4 h5 expected: ? From H h4 0 h5 Ex:6.1 Ho : m Height Class < 62.5 62.5 - 63.5 63.5 - 64.5 64.5 - 65.5 65.5 - 66.5 66.5 - 67.5 67.5 - 68.5 68.5 – 69.5 69.5 – 70.5* 70.5 – 71.5 71.5 – 72.5 72.5 – 73.5 73.5 – 74.5 74.5 – 75.5 75.5 – 76.5 76.5 – 77.5 >77.5 x 70.17 s 2 10.96 s 3.31 Height fi 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 0 2 2 3 5 4 6 5 8 7 7 10 6 3 2 0 0_ 70 ;s 2 Lower limit - x s -2.317 Tabled value .01025 -2.015 -1.713 -1.410 -1.109 - .807 - .505 - .202 .0997 .4018 .7039 1.006 1.308 1.610 1.912 2.215 .0219 .0434 .0791 .1338 .2099 .3069 .4198 .4603 .3459 .2407 .1572 .0954 .0537 .0279 .0134 ; f = N(m , s ) 2 Pˆ (Height ) .01025 .0117 .0215 .0357 .05547 .0761 .0970 .1129 .1199 .1164 .1032 .0835 .0618 .0417 .0258 .0134 2 .72 .82 1.51 2.50 3.83 5.33 6.79 7.90 8.39 8.15 7.22 5.85 4.33 2.92 1.81 2.75 .94 Ho : X ~ N(70.17,10.96) How to Calculate the Probability of a “bin”: 63 62.5 63.5 62.5 63.5 Pr(<63.5) – Pr(<62.5)= 70.17 62.5 70.17 Pr (63) = Pr 63.53.31 Pr 3.31 Est #(63) = Pr(63) ·70 * P(70) P(69.5 X 70.17) P(70.17 X 70.5) = .5 - .4198 + .5 -.4603 =.0802 + .0397 = .1199 2 Rule of Thumb: Not more than 20-25% of expected frequencies < 5 and no frequency < 1. Now back to the example! Ex:6.1 Ho : m Height Class Height fi < 62.5 62.5 - 63.5 63.5 - 64.5 64.5 - 65.5 65.5 - 66.5 66.5 - 67.5 67.5 - 68.5 68.5 – 69.5 69.5 – 70.5* 70.5 – 71.5 71.5 – 72.5 72.5 – 73.5 73.5 – 74.5 74.5 – 75.5 75.5 – 76.5 76.5 – 77.5 >77.5 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 x 70.17 s 2 10.96 s 3.31 0 2 2 3 5 4 6 5 8 7 7 10 6 3 2 0 0_ 70 ;s 2 Lower limit - x s -2.317 Tabled value .01025 -2.015 -1.713 -1.410 -1.109 - .807 - .505 - .202 .0997 .4018 .7039 1.006 1.308 1.610 1.912 2.215 .0219 .0434 .0791 .1338 .2099 .3069 .4198 .4603 .3459 .2407 .1572 .0954 .0537 .0279 .0134 ; f = N(m , s ) 2 Pˆ (Height ) .01025 .0117 .0215 .0357 .05547 .0761 .0970 .1129 .1199 .1164 .1032 .0835 .0618 .0417 .0258 .0134 2 .72 .82 1.51 2.50 3.83 5.33 6.79 7.90 8.39 8.15 7.22 5.85 4.33 2.92 1.81 2.75 .94 Pooled 3.05 6.33 7.25 5.5 Ho : X ~ N(70.17,10.96) 25.88 d.f. = 11 – 1 –2 = 8 S70 x,s2 2 0.05,8 15.507 Accept H0 : X is N(70.17,10.96) Does this mean that Height is Normally Distributed? Yes, N(70.17,10.96)! Probits: A graphical way to test Normality The probit transformation: The value of x that corresponds to a probability if x is normal with mean 0 and variance 1 Sometimes with mean 3 (to make the 95% confidence positive) x P( x) x f ( x)dx s 1 2 y 1 2 e dy 2 -3 -2 -1 0 1 Use of probit paper: If you plot normally distributed data, you get a straight line: Use of probit transformation on SPSS Calculate the ranked data using the Syntax window: rank x. (Creates the ranks in variable rx) compute px=rx/#samples. (Normalizes to 1.00) compute probitx=probit(px). The new variable can be plotted against x. It should be close to a straight line if x is normally distributed. Statistics SAMPLE N Va lid Miss ing Mean 61 0 -9.8130 E-02 Std . Error of Mea n Medi an Mode .1 278 -.2 140 -2.74 Std . Deviation Va riance Skewness Std . Error of Skewnes s .9 984 .9 968 .0 15 .3 06 Kurtosis Std . Error of Ku rtosi s Rang e Mini mum .0 61 .6 04 5.14 -2.74 Maximu m Percentiles 25 50 2.40 -.8 258 -.2 140 75 .6 397 a. Multiple mod es e xi st. The smal lest value is shown a Kolomogorov-Smirnov goodness of fit test to the normal distribution Compare the frequency polygon to the hypothesized curve 1. 2. 3. 4. 5. Calculate the frequency polygon. H0: N(mean, variance) or other distribution Find or calculate the parameters (mean, variance, etc.) Calculate the distribution function using H0 Compare: ( Maximum ( F ( x ) )) d Minimumi F ( x(i ) ) Fˆ ( x( i ) ) d i (i ) ) Fˆ ( x( i ) D Maximum ( d , d ) where F ( x( i ) ) is the Hypothesized Distribution at order statistic x( i ) Fˆ ( x( i ) ) is the frequency polygon at x( i ) 1.0 .9 .8 .7 .6 .5 .4 .3 .2 .1 0.0 -4 -3 -2 -1 0 1 2 One-Sa mple Kolm ogorov-Smirnov Test N Norma l Para mete rs a,b Most Extreme Differences Mean Std . Deviation Ab solu te Pos itive Nega tive Kol mogo rov-Smirno v Z As ym p. Si g. (2 -tail ed) a. Te st d istri buti on i s Norma l. b. Calcula ted from data . SAMPLE 61 -9.8130 E-02 .9 984 .0 90 .0 90 -.0 70 .7 06 .7 01 3 4 But, these are tests of the original distribution! Do we necessarily care what that distribution is? We more often want to compare the parameters of the distribution. Has the distribution moved as a result of treatment? Different populations? Is the new treatment “better” than the old?