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Topics: Statistics & Experimental Design The Human Visual System Color Science Light Sources: Radiometry/Photometry Geometric Optics Tone-transfer Function Image Sensors Image Processing Displays & Output Colorimetry & Color Measurement Image Evaluation Psychophysics Design of experiments Why is it important? • We wish to draw meaningful conclusions from data collected • Statistical methodology is the only objective approach to analysis Design of experiments • Recognize the problem • Select factor to be varied, levels and ranges over which factors will be varied • Select the response variable • Choose experimental design: • Sample size? • Blocking? • Randomization? • Perform the experiment • Statistical analysis • Conclusions and recommendations Let’s start easy • We would like to compare the output of two systems. • Design a testing protocol and run it several times Run SystemA SystemB 1 y1A y1B 2 y2A y2B 3 y3A y3B … … … Visualize data For small data sets: Scatter plot 18.5 Output 18 17.5 system_A system_B 17 16.5 16 0 1 2 3 4 5 Run # 6 7 8 9 10 Visualize data frequency, ni For larger data sets: Histogram • Divide horizontal axis into intervals (bins) • Construct rectangle over interval with area proportional to number (frequency) of observations Statistical inference Draw conclusions about a population using a sample from that population. • Imagine hypothetical population containing a large number N of observations. • Denote measure of location of population as 1 Population mean yi N i Statistical inference • Denote spread of population as variance 2 2 yi i N Statistical inference A small group of observations is known as a sample. • A statistic like the average is calculated from a set of data considered to be a sample from a population Run SystemA SystemB 1 y1A y1B 2 y2A y2B 3 y3A y3B … … … yA yB 1 n Sample average y y i n i 1 Statistical inference • Sample variance supplies a measure of the spread of the sample 0.275 0.25 0.225 0.2 n 0.175 0.15 s2 0.125 0.1 0.075 0.05 0.025 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 2 y y i i 1 n 1 Probability distribution functions f 1 0.75 P(axb) 0.5 0.25 0 -5 -2.5 0 2.5 5 x Probability distribution functions P(xi) P(x = xi) = p(xi) xi 0 px i 1 for all values of x i Px x i px i for all values of x i px 1 i xi Mean, variance of pdf • Mean is a measure of central tendency or location xp x y • Variance measures the spread or dispersion x px 2 2 y Normal distribution 1 f x e 2 y -10 -5 1 x 2 2 = standard deviation = √2 mean 0.5 y 0.5 y 0.5 0.375 0.375 0.375 0.25 0.25 0.25 0.125 0.125 0.125 0 5 10 -10 -5 0 x 3, 3 5 10 -10 -5 0 x 0, 2 5 10 x 3, 1 Normal distribution, N , 2 • From previous examples we can see that mean = and variance = 2 completely characterize the distribution. • Knowing the pdf of the population from which sample is draw determine pdf of particular statistic. Normal distribution • Probability that a positive deviation from the mean exceeds one standard deviation is 0.1587 1/6 = percentage of the total area under the curve. (Same as negative deviation) • Probability that a deviation in either direction will exceed one standard deviation is 2 x 0.1587 = 0.3174 • Chance that a positive deviation from the mean will exceed two = 0.02275 1/40 Normal distribution • Sample runs differ as a result of experimental error • Often can be described by normal distribution Standard Normal distribution, N(0,1) y 0.5 0.375 0.25 0.125 -10 -5 0 5 10 x z y Values for N(0,1) are found in tables. Standard Normal distribution, N(0,1) Standard Normal distribution, N(0,1) Example: Suppose the outcome of a given experiment is approximately normally distributed with a = 4.0 and = 0.3. What is the probability that the outcome may be 4.4? z yμ 4.4 4 1.33 σ 0.3 Look in table in previous page, to find that the probability is 9%. 2 c distribution Another sampling distribution that can be defined in terms of normal random variables. • Suppose z1, z2, …, zk are normally and independently distributed random variables with mean = 0 and variance 2 = 1 (NID(0,1)), then let’s define c z z z 2 1 2 2 2 k Where c follows the chi-square distribution with k degrees of freedom. 2 c 0.2 distribution k=1 k=5 0.15 k = 10 0.1 k = 15 0.05 0 0 5 10 15 20 25 Student’s t Distribution • In practice we don’t know the theoretical parameter • This means we can’t really use z y and refer to the result of the table of standard normal distribution • Assume that experimental standard deviation s can be used as an estimate of Student’s t Distribution Define a new variable y t s It turns out that t has a known distribution. It was deduced by Gosset in 1908 Student’s t Distribution k=100 k=10 0.3 k=1 0.2 0.1 0 -5 -2.5 0 2.5 5 Probability points are given in tables. The form depends on the degree of uncertainty in s2, measured by the number of degrees of freedom, k. Inferences about differences in means • Statistical hypothesis: Statement about the parameters of a probability distribution. Let’s go back to the example we started with, i.e., comparison of two imaging systems. We may think that the performance measurement of the two systems are equal. Hypothesis testing H 0 : 1 2 H 1 : 1 2 First statement is the Null hypothesis, second statement is the Alternative hypothesis. In this case it is a two-sided alternative hypothesis. How to test hypothesis? Take a random sample, compute an appropriate test statistic and reject, or fail to reject the null hypothesis H0. We need to specify a set of values for the test statistic that leads to rejection of H0. This is the critical region. Hypothesis testing Two errors can be made: • Type I error: Reject null hypothesis when it is true • Type II error: Null hypothesis is not rejected when it is not true • In terms of probabilities: Ptype I error Preject H 0 H 0 is true Ptype II error Pfail to reject H 0 H 0 is false Hypothesis testing • We need to specify a value of the probability of type I error . This is known as significance level of the test. • The test statistic for comparing the two systems is: yA yB t0 1 1 sp kA kB Where 2 2 k 1 s k 1 s A B B s2 A p kA kB 2 Hypothesis testing • To determine whether to reject H0, we would compare t0 to the t distribution with kA+kB-2 degrees of freedom. • If t 0 t / 2,k A k B 2 we reject H0 and conclude that means are different. We have: System A System B yA 16.76 yB 17.92 s 2A 0.1 s 2A 0.061 sA 0.316 s B 0.247 k A 10 k B 10 Hypothesis testing H 0 : 1 2 H 1 : 1 2 • We have kA + kB – 2 = 18 • Choose = 0.05 • We would reject H0 if t 0 t 0.05,18 t 0.025,18 2.101 Hypothesis testing 2 2 k 1 s k 1 s 9.01 90.061 A B B s2 A 0.081 kA kB 2 p 18 s p 0.284 t0 yA yB 16.76 17.92 9.13 1 1 1 1 sp 0.284 kA kB 10 10 Hypothesis testing Since t0 = -9.13 < -t0.025,18 = -2.101 then we reject H0 and conclude that the means are different. Hypothesis testing doesn’t always tell the whole story. It’s better to provide an interval within which the value of the parameter is expected to lie. Confidence interval. In other words, it’s better to find a confidence interval on the difference A - B Confidence interval y A y B t / 2 ,k 1 k 2 2s p 1 1 1 1 A B y A y B t / 2 ,k 1 k 2 2s p kA kB kA kB Using data from previous example: 16.76 17.92 2.1010.284 1 1 1 1 A B 16.76 17.92 2.1010.284 10 10 10 10 1.16 0.27 A B 1.16 0.27 1.43 A B 0.89 So the 95 percent confidence interval estimate on the difference in means extends from -1.43 to -0.89. Note that since A – B = 0 is not included in this interval, the data do not support the hypothesis that A = B at the 5% level of significance.