Data analysis and Geostatistics - lecture VI

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Data analysis and Geostatistics - lecture VI Statistical testing with population distributions Statistical testing - the steps 1. Define a hypothesis to test in statistics only a hypothesis rejection is a strong statement: have to choose your hypothesis carefully (example: white swans - black swans) 2. Decide on a confidence level you cannot be 100% certain, because the chance of an unlikely event is small, but never zero: have to select a desired level of confidence at α = 5%, you accept to reach the wrong conclusion in 1 out of 20 cases at α = 2%, it is 1 out of 50 cases so what do you choose? depends very much on the situation identifying cheating schoolteachers: you have to be very certain!

Statistical testing - confidence levels For example: a mining company measures a property P (for example As content). when P belongs to population π the prospect is barren when P exceeds π: $$$$ The cut-off level is controlled by the confidence level alpha Let s set alpha = 5% If P has a value in the green domain: we assume that it does not belong to the red, barren distribution, but comes from a separate distribution that describes the ore deposit However, there is a 5% chance that it is still part of the red distribution: type I error population π mean of π BAR REN domain A 95% of π stdev of π P domain B 5% of π Statistical testing - confidence levels For example: a mining company measures a property P (for example As content). when P belongs to population π the prospect is barren population π when P exceeds π: $$$$ The cut-off level is controlled by the confidence level alpha mean of π P B3 B2 B1 Let s set alpha = 5% If P has a value in the red domain: we assume that it belongs to the red, barren distribution and will not drill it. However, there is a chance that it is part of the ore distribution, because we don t know what it s distribution looks like: type II error BAR REN domain A 95% of π stdev of π domain B 5% of π

Statistical testing - confidence levels For example: a mining company measures a property P (for example As content). when P belongs to population π the prospect is barren when P exceeds π: $$$$ The cut-off level is controlled by the confidence level alpha and varies: population π mean of π B3 B2 B1 when mining is doing well: alpha = 80% when mining is under stress: alpha = 5% stdev of π why? BAR REN at low alpha rarely wrong, but you don t find much. at high alpha, will find everything, but are commonly wrong domain A 95% of π domain B 5% of π Statistical testing - the steps 3. Compare the test property against a certain probability distribution the expected distribution defines the probability of finding a certain observation: can find these values in tables, for example the normal and student-t distributions 1 f(x) = e ( - 1 x-m ( ) 2 2 ) s 2p s total surface area = total probability = 1 ±1 stdev = 2/3 of data ±2 stdev = 95% of data -5.0-4.0-3.0-2.0-1.0 0.0 1.0 2.0 3.0 4.0 5.0

Statistical testing - starting with the basics every value or data point is derived from a population So, for a set of measurements: all same population all different population grouped in populations each population has a mean and stdev composition Statistical testing - population probability every value or data point is derived from a population population A population B mean of B probability wide and shallow: large spread mean of A stdev of A stdev of B narrow and steep: small spread differences in shape reflect differences in probability

Statistical testing - population probability within a population there is a prob for occurrence of each value every random sample has a: high prob of being close to the mean low prob of being far removed from the mean A B normally distributed this probability is known if: mean is known variance is known Statistical testing - population probability outliers are values that have a low probability of occurrence values beyond 3 stdev: highly unlikely B prob of belonging to population less than 0.5%: regarded as outlier A However, possibility is not zero! Identical for populations A and B, but a given deviation from the mean will be less likely to be an outlier in case A where the spread is larger.

Statistical testing - population probability to use Gaussian probabilities, have to standardize populations populations A and B are both normal, but different in shape: convert them to standardized form: A B Z-score: Zi = (xi - μ) / σ x i - μ x i - μ σ μ 0 0 μ σ μ + σ σ +σ 1 +1 Statistical testing - population probability Z-scores: standardized normal distribution can use the probabilities of the Gaussian distribution to determine the probability for a given value to occur: μ see table 2.2 on page 409 How likely to find a value < 14? Z = (14-12)/8 = 0.25 probability of Z = 0.25: 59% How likely to find a value > 14? probability of 100-59 = 41% 12 14 Given a population with a mean of 12 and a standard deviation of 8

Statistical testing - population probability Z-scores: standardized normal distribution can use the probabilities of the Gaussian distribution to determine the probability for a given value to occur: μ 4 12 20 see table 2.2 on page 409 How likely to find a value that is between 4 and 20? Z4 = (4-12)/8 = -1 Z20 = (20-12)/8 = +1 probability of Z = -1: 15.9% probability of Z = +1: 84.1% so prob = 84.1-15.9 = 68.2% Given a population with a mean of 12 and a standard deviation of 8 Statistical testing - population probability can be used as a criterion to classify a data point as an outlier How likely to find a value > 40? highly unlikely: outlier Z = (40-12)/8 = 3.5 probability: 100-99.98 = 0.02% must belong to another population: B problem with B: know nothing about it P

Statistical testing - population probability So to summarize these observations: if we can exclude something from population A: strong statement, exceeds our specified threshold of α will be wrong sometimes, but at least we know and can control it if we cannot exclude something from population A: there is still a possibility that it belongs to another population (e.g. B), but because we know nothing of B, cannot specify the prob of this weak statement: type II errors are worse you know your chances of failure, but not those of success... Statistical testing - population probability what if we know the properties of the other pop as well? for the ore sample example: population A: μ = 60, population B: μ = 130 population P: μ = 110, SEA,B = 20 (SE because comparing means) Zi = (μp - μ) / SE at α = 0.05: -1.96 < Z < 1.96 1) hypothesis: P part of A H0; μp = μa Z = 6.12, so it exceeds Z range: rejected 2) hypothesis: P part of B H0; μp = μb Z = 1.0, so it is within Z range: accepted

Statistical testing - population probability what if we know the properties of the other pop as well? another example: a well-established fossil population has length μ = 14.2 ± 4.7 mm now a researcher finds a mean of 30 mm from n = 10 can these belong to the same population? hypotheses: H0; μnew = μ HA; μnew μ Z = (μnew-μ) / (σ/ n) at α = 0.05: -1.96 < Z < 1.96 Z = (30-14.2)/(4.7/ 10) = 10.63 x Statistical testing - the t-distribution rarely know the population mean and stdev, rather sample stats In the previous examples we presumed to know the mean and stdev of the population, but in reality we rarely do: estimate them from a sample so, the test distribution should have a larger uncertainty and this has to depend on the number of samples (degrees of freedom): the t-distribution curve becomes wider and shallower as total area = 1 d.f. = d.f. = 25 d.f. = 15 d.f. = 5 values for the t statistic can be found in tables for a given confidence level and no. of d.f. e.g. 2.4 on page 411

The curse of low sample numbers The t-distribution elegantly shows the effect of small sample numbers on the probability of finding extreme values: the probability of finding a certain value depends on the number of samples: less samples means (ironically) a higher probability values for the t statistic can be found in tables for a given confidence level and no. of d.f. e.g. 2.4 on page 411 less than average d.f. = average number of cancer cases in Canada per 1000 d.f. = 25 d.f. = 15 d.f. = 5 more than average Statistical testing - t-distribution testing testing against the t-distribution is identical to that of Z-scores _ t = (x-μ) / (s/ n) using the means and SE, so independent of the type of distribution! Normally we do not test individual values against the t-distribution, but rather the mean derived from a sample against the mean of the population we think these values come from has the added advantage that we can ignore distribution (multi-modality...) Can use it for two very useful properties: - the confidence interval for a value or property by extracting μ - required sample size for specified confidence by extracting n

Statistical testing - t-distribution testing Commonly, a company needs to guarantee certain specifications for a product. For example, that the concentration of the ore element is at a certain level, or the concentration of a contaminant below a certain level. Missing such targets can be very costly. So how do you decide what is a good, as in achievable, level? _ suppose x from μ1: prob high mean from μ2: prob lower from μ3: prob lower from μ4: prob low at some value of μ, will exceed the confidence level: too unlikely to come from a population with this mean: this is the upper μ μ5 μ1 μ2 μ3 μ4 similarly, will reach a lower μ when working down from the mean The confidence interval on the mean represent the range from this lower to the upper population mean for a given confidence level. Statistical testing - t-distribution testing the confidence interval for a mean at a given confidence level: 2.5% 2.5% mean can belong to populations for which the probability of occurrence of this mean is more than 0.5α μ- mean μ+ formulae:

Statistical testing - t-distribution testing example 1 the confidence interval for the concentration of phosphorus in iron ore Say we are required to supply iron ore with a bulk phosphorus content of less than 250 ppm, or the company has to pay a fine. The mean P content that you have determined is 215 ± 30.8 ppm based on 8 samples. the specifics: our mean: 215 ± 30.8 ppm from n = 8 the limit: 250 ppm desired confidence: 95% d.f. = n - 1 α = 0.05 tα;df = 2.365 What is the 95% confidence interval on the bulk concentration? 215 215 30.8 30.8 189 < mean < 241 ok Statistical testing - t-distribution testing example 2 All sample materials are heterogeneous on some scale and to overcome such issues you either have to take a large sample or multiple sample of the same material. But how many? the t-distribution allows you to determine the min sample size for a desired confidence level e.g. want to be within 1% of the real P content at a 95% confidence level: our pilot study gave s = 30.8 ppm at n = 8, so: d.f. = 8-1 α = 0.05 γ = 1% 30.8 n at least 53, so need another 53-8 = 45 samples

Statistical testing - significance of r So, let s now return to the correlation coefficient: t = r n - 2 1 - r 2 with d.f. = n - 2 and t α;d.f. Our hypotheses: H0: r = 0, if true, no significant correlation: domain A Ha r 0, cannot reject the absence of correlation: B α = 0.05 domain B- 2.5% of pop r = 0 domain A 95% of pop this is a t-distribution for a given α and d.f. domain B+ 2.5% of pop when r plots in domain B: prob of it belonging to the population r = 0 is lower than our threshold α: reject r = 0 and therefore conclude that the variables are correlated Statistical testing - significance of r What values of r are meaningful for a given confidence level t = r n - 2 1 - r 2 with d.f. = n - 2 and t α;d.f. Our hypotheses: H0: r = 0, if true, no significant correlation Ha r 0, cannot reject the absence of correlation Let s say: n = 25, so d.f. = 23 α = 0.05 r = -0.34 tcalc = -1.73 t 0.05;23 = 1.71 = -1.71 tcalc exceeds t 0.05;23 -> reject H0 in this example we can reject the H0: so we can make the strong statement that at the 5% confidence level, there is a significant correlation between the vars

Statistical testing - significance of r what if we want to be more certain? Our hypotheses: H0: r = 0, if true, no significant correlation Ha r 0, cannot reject the absence of correlation Let s say: n = 25, so d.f. = 23 α = 0.025 r = -0.34 tcalc = -1.73 t 0.05;23 = 1.71 = -1.71 t 0.025;23 = 2.07 = -2.07 tcalc exceeds t 0.05;23 -> reject H0 tcalc doesn t exceeds t 0.025;23 -> cannot reject H0 we can now only conclude that we have no reason to reject the absence of correlation, which is clearly not as strong a statement Statistical testing - significance of r The effect of degrees of freedom (n) on the significance: e.g. a data set like this: r = 0.42 is this correlation significant at α = 0.05 and the following n? at n = 5; t = 0.80 t0.05;4 = 2.353 at n = 10; t = 1.31 t0.05;4 = 1.860 at n = 25; t = 2.22 t0.05;4 = 1.717 x x v