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Probability & Statistical Inference Lecture 4 MSc in Computing (Data Analytics) Lecture Outline Modern statistics uses a number of mathematical results to relate descriptive statistics and probability theory. These can be divided (roughly) under three headings: - Central Limit theorem (large samples) - Maximum Likelihood Methods (large samples) - Small sample results Although the mathematical details are quite different in each case – the end results and the reasoning used are almost identical. We will look in detail at the Central Limit Theorem but without the higher mathematics. If you can understand the working of the Central Limit Theorem – then you also get the essential understanding of the other methods as well. Sampling Theory – Statistical Models Central Limit Theorem (CLT) – A description How many voters will give F.F. a first preference in the next general election? We have 2 different estimates 1. Researcher A (10 people) => 40% 2. Researcher B (100 people) => 25% How much 'better' is estimate B than estimate A ? Real Question: What makes a 'good' estimate ? • • • unbiased low variability i.e. if the survey was repeated should get 'similar' answer Example Suppose an engineer wants to estimate the lifetime of a electronic component. using simple random sampling they select a sample and test. The sample is taken so that the component lifetimes can be considered to be independent of each other. Gods eye view: mean lifetime, µ= 4,900 hours σ = 3959 hours (you would never know this in practice however) • This is the population • Note: it is highly skewed and is NOT normal • What would happen if we took repeated samples of the same size and calculated their means? Example Continued Experiment: take a sample of size 2 from this population and get the mean of the sample Repeat this 2,000 times Now have 2,000 means - what would the histogram of all these means look like? What would happen if you did the same experiment, but with samples of sizes 10, 20 and 30? Distribution of the Sample Means varying the sample size Original Distribution Note that the histogram become more Normal as the sample size increases Same result but plotted on same scale Note the spread decreases with increasing sample size Central Limit Theorem What has happened? As the sample sizes increased the shape of the histogram of means => normal As the sample sizes increased the spread (standard deviation) between the sample means decreased These histograms are pictures of The Sampling Distribution of the Mean This phenomenon will happen in ALL cases The proof of this is called the Central Limit Theorem (CLT) The CLT involves some fairly non-trivial mathematics Central Limit Theorem Since bigger samples are more representative, two means from samples of size=100 are more likely to be closer together than two means from samples of size=10 The larger the sample size is the more the sample means will tend to agree, so the standard deviation of the Sampling Distribution of the Mean will decrease When the sample size is sufficiently large, the Sampling Distribution of the Mean will be Normally distributed Central Limit Theorem If a random sample is taken from a population, where: Each member of the sample can be considered to be independent of each other The are all members of the same population That population has a mean value μ and a standard deviation σ Then, A sample mean ( X ) can be considered a random variable sampled from a probability distribution of possible sample means of the same size called the Sampling Distribution of the Mean. ___ Definition: Central Limit Theorem continued… The sampling distribution of the mean has a average value = (the population mean). The sampling distribution of the mean has a standard deviation = n Where σ is the population standard deviation, and n is the sample size taken. This value is called the standard error of the mean. The Sampling Distribution of the Mean will be a Normal distribution if the sample size is large. CLT - Summary When the sample size is sufficiently large, the Sampling Distribution of the Mean will be normally distributed with a mean = , and a standard deviation (i.e. standard error) = n From the simulation above; For a sample size of 2, the standard error of the mean should be = 3959 / √2 = 2,799 Population Size = 2 Mean from 2,000 samples 4,900 5,017 Size = 10 Size = 20 Size = 30 4,899 4,915 4,934 Standard Actual Deviation Standard predicted by CLT Deviation 2,799 1,251 885 722 3959 2,805 1,232 871 732 Practical use for the CLT continued… This avoids the necessity of specifying a complete statistical model for all the sampled data. All we have to do is specify a probability model for the sample mean. For any sample mean, calculated from a large independent random sample taken from ANY population with a mean μ and standard deviation σ, we know from the CLT, that this sample mean is a random variable from a Normal distribution with a mean = μ and a standard deviation = n Practical use for the CLT continued… Take a single sample and calculate This is an estimate of μ – the true (but unknown) population mean. But, how good is this estimate? ___ ___ X We assume that X is not exactly , but is somewhere near - but how near is it likely to be? Confidence Intervals Intoduction We would like to make probability statements as to ___ how close X is likely to be to . ___ If sample size is sufficiently large – then the estimate X can be considered as: a random variable from a Normal distribution, so probability statements are possible. This is how we use the CLT in practical data analysis. For a Normal distribution, we know that 95% of values will be within 1.96 Standard deviations of So, given one estimate we can say that this estimate is within 1.96 standard errors of the actual population mean , with 95% confidence 95% in shaded area •We can turn this knowledge on its head: given we can be 95% confident that the true mean is within 1.96 standard errors of it. Confidence Interval From this we can specify a range of values within which we are 95% confident that the population mean () lies This is called a confidence interval 95% Confidence Interval for a population mean (from large enough sample): __ x 1 . 96 standard __ x 1 . 96 error n Remarkably, this result holds for samples of size 30 or more. So, a large sample in this context, is a sample of 30 or more. Example One sample of size 30 from the electronic components yields a sample mean = 5,873 hours .We know = 3,959 so a 95% confidence interval would be; __ x 1 . 96 standard __ x 1 . 96 error 5873 1 . 96 n 3959 30 5873 1417 4456 to 7290 So, we would say that the average lifetime of all components (μ) is between 4,456 and 7,290 hours with 95% confidence Confidence Intervals Why is this any good? Before: one estimate, = 5,873 but no idea of how good or bad it was, i.e. how close to μ is was likely to be. Now: 95% confident that μ is between 4,456 and 7,290 hours. So, using CLT ~> Confidence Intervals ~> able to get an estimate with certain level of confidence that can be justified, i.e. it gives us an objective measure of the actual amount of information contained in our sample about the likely location of μ. General Confidence Interval for μ (σ known) The general formula is: __ CI 1- x z 1 / 2 n Where: • is between a value between 0-1, • (1-)×100% is the confidence level you want • Z1-/2 is a value from the Normal distribution table. • Example: for a 95% CI, = 0.05 (1-)×100% = 95% Z1-/2 = 1.96 Problem with σ All of the above assumes that the population standard deviation (i.e. ) is known. In practice this is not known (just like ). => So, we need to estimate as well as => we get this estimate from the standard deviation of the sample Sample Standard Deviation is called ‘s’ => Estimate by s, When sample size is large s x x n 1 2 Z-Values The value of Z1-/2 for other % confidence intervals are given in standard tables. Confidence Level α/2 Z1-/2 90% 0.05 (5%) 1.6449 95% 0.025 (2.5%) 1.96 99% 0.005 (0.5%) 2.5758 99.9% 0.0005 (0.05%) 4.4172 Example Using these we get the following results for the electronic component example: Confidence Level Z1-/2 CI 90% 1.6449 4681 to 7065 95% 1.96 4456 to 7290 99% 2.5758 4011 to 7735 99.9% 4.4172 2679 to 9067 Note as gets smaller the CI gets wider Also, at the same time as n gets bigger the CI narrows – So big samples leads to more precise estimates (i.e. narrower confidence intervals) What CI’s and sample sizes should I use? • You can’t control s – it is inherent in the data (population). • You can’t control x-bar either. • You can control Z1-/2 but in practice scientific convention sets this to reflect 90%, 95% or 99% confidence, with 95% being the accepted default. • You can choose n – but resources may limit you. • There is a whole topic called sample size determination which you may want to review before collecting data or starting research Assumptions for hypothesis testing about μ (large sample) and Calculation of CIs Sample size 30 or greater Experimental units are independent or each other Experimental units were randomly sampled The independence assumption requires that value of the variable for one experimental unit should not tell us anything about the value of another. e.g. in the rats experiment – different and unrelated rats should be used – not 1 rat tested 100 times. Randomness is required to avoid systematic bias in selection. Exercise Complete Exercise 1 & 2 Calculation of CIs for small samples What about small samples? In the case of CIs about a mean we can use the Student-t distribution. The process turns of to be very similar – but the CLT no longer works History of the Student t test William Gosset used the publishing pseudonym ‘Student’. He derived the correct sampling distribution for the mean of samples < 30 – and called it the ‘t distribution’. In his honour, it is often called the ‘Student t’ distribution. Gosset was a chief brewer for Guinness. The mathematical details are complicated, but, it turns out that we perform exactly the same calculations as before, with the one change that the t distribution instead of the normal distribution is used. Assumptions Student t’s result only referred to a mean where the distribution of the population was normally distributed with some mean μ and finite standard deviation σ. This is in contrast to the CLT for large samples that required no such assumption about normality. The t-test also requires the assumption regarding independence in the sample. Statistical Model for mean from small samples The experimental units are independently sampled from a population with mean=μ and standard deviation = σ The population is normally distributed (we don’t need this with large samples) So, to use the t-test for a small sample, you need to establish that data is sampled from a population that is normally distributed – you could look at the histogram of the sample and see if it is symmetric and bell shaped – or use other methods. The t - Statistic If Assumptions met: The statistic: ___ t X s n Can be shown to be distributed according to a (student) t-distribution. The t-distribution has one parameter, called ‘degrees of freedom’ (df). The t-Distribution The t-distribution itself is bell shaped and symmetric – just like the normal distribution but is ‘flatter’. There are many t distributions – one for each sample size. The rule used is: for a sample of size n – use the t distribution with degrees of freedom = n−1 Example: if the sample size is 15, then use a t distribution with degrees of freedom 15 − 1=14. Note the degrees of freedom often abbreviated to df. 0.4 The t-Distribution 0.0 0.1 0.2 0.3 Normal(0,1) t(df=4 t(df=1) -4 -2 0 2 4 The t probability density function with k degrees of freedom: f (x) k 1 / 2 k k 2 1 x 2 / k 1 ( k 1 ) / 2 General Confidence Interval for μ (small Samples) The general formula is: __ CI 1- x t ( 1 / 2 , n 1 ) s n Where (1-) 100% is the confidence level you want and t(n-1, /2) is a value from the t distribution with df=n-1, and with a specified level. What is t(n−1, 1−/2)? A value from the t distribution with n−1 df such that 100(1 − )% of values lie within that range around the mean. How do you find t(n−1, 1−/2)? from a table specifically designed to give it to you or use a computer Confidence Level /2 t(df=1) t(df=10) t(df=30) 90% 0.05 (5%) 6.314 1.812 1.697 95% 0.025 (2.5%) 12.71 2.228 2.042 99% 0.005 (0.5%) 63.66 3.169 2.750 99.9% 0.0005 (0.05%) 636.6 4.587 3.646 Note: as gets smaller then CI gets wider as df gets smaller then CI gets wider Example Internal temperature of autoclaved aerated concrete used in building. An engineer recorded the following data: 23.01, 22.22, 22.04, 22.62, 22.59 95% CI for the population mean? __ s CI 1- x t ( / 2 , n 1 ) 22 . 5 2 . 776 n 0 . 3793 5 22 . 5 0 . 4696 ( 22 . 03 , 22 . 97 ) Exercise Answer Questions 3-6 Confidence Intervals for Proportions (Large Samples) Proportions (including %) are often a statistic of interest Think of the proportion of defective items on a production line, the proportion of people who respond favourably to a survey question, to proportion of success versus failures in some experiment Proportions are also covered by the CLT - remember that a proportion is a different kind of average Confidence Intervals for Proportions (Large Samples) Take a sample of size n of electronic components coming off a production line, a test each one for defects. The statistic of interest is the proportion of defectives produced by the production process. The estimated proportion from the sample is, pˆ No of Defective s in the Sample n(the total sample size) where (p-hat) is the symbol used for the estimated proportion from the sample Confidence Intervals for Proportions (Large Samples) If the sample size is sufficiently large and we repeat the experiment a large number of times, then: The sampling distribution of the proportion will be normally distributed by the CLT The mean of this distribution will be p - i.e. the 'true' population proportion The standard deviation of the sampling distribution of the proportion, called the standard error of the proportion is estimated by S.E of proportion pˆ (1 pˆ ) n Example: A pharmaceutical company produces 400,000 capsules per day of a particular drug. They test 200 of the capsules for defects (too much/little active compound). If the population p = 0.05, and they take 10,000 repeated samples this is the histogram they would get Sample Size How big does the sample have to be for the CLT to work with proportions? The rule is different than the rule for means. Do the following test. A rule of thumb: the sample size is big enough if 1. 2. np > 5 and n(1-p) > 5 General Confidence Interval Formula for a Population Proportion (large Sample) CI pˆ z1 / 2 pˆ (1 pˆ ) n where = the confidence level and Z1-/2 = a value from the standard normal distribution such that 100(1-)% of values of a standard normal distribution lie within that range around the mean So the Z1-/2 values used for a population proportion are the same as those used for a population mean Example How many voters will give F.F. a first preference in the next general election ? There are 2 different estimates Researcher A (10 people) Researcher B (100 people) => 40% => 25% How much 'better' is estimate B than estimate A ? Step one: Can we use the formula for large numbers 1. 2. Researcher A: np = 10 * 0.4 = 4 => 4 is not greater than 5 therefore you cannot used the large number method Researcher B: np = 100 * 0.25 = 25 n(1-p) = 100 * (1-0 .25) = 75 both figures are greater than 5 therefore you can used the large number method Example Continued Researcher B - 95% Confidence Interval CI pˆ z 1 / 2 pˆ (1 pˆ ) CI 95 0 . 25 1 . 96 n 0 . 25 0 . 75 CI 95 0 . 25 1 . 96 0 . 04 CI 95 0 . 25 0 . 08 CI 95 0.17 to 0.33 So, the 95% CI is 17% to 33%. 100 Example Continued NB: If fact we can get a 95% CI for researcher A's findings using small sample theory (exact CI) - this is available in SAS and other software: Exact CI’s are often based on direct use of probability models. The method is based directly on calculations for the binomial distribution (see lecture 3) What do we have to do? Using the CLT, we found, that the 95% CI was composed of the set of values for the mean, such that an hypothesis test would not reject the null hypotheses for any of those values in the set using the α = 0.05 level. Using SAS we can calculate a 95% CI for Researcher A: CI 95% for Researcher A = 12% to 74% which is too wide to be informative anyway! If we use the same technique for researcher B we get: CI95 for Researcher B = 17% to 35% Which is virtually the same as before using the CLT. Exact CI and tests for population proportions These work for small samples as well as large samples With large sample will give essentially the same results as CLT Must be used for small samples, however Based on the binomial probability distribution. Difference between Exact and CLT based methods When sample sizes are ‘large’ they will give the same results – but exact tests can be very hard to compute even with modern PCs When sample sizes are small exact methods must be used The CIs from small samples tend to be very wide – there is no short cut from collecting as much high quality data as you can manage. Exercise Answer Question 7-9