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1 15.Math-Review Statistics

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2 zLet us consider X 1, X 2,…,X n, n independent identically distributed random variables with mean and standard deviation. zAnd define: Central Limit Theorem

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15.Math-Review3 Central Limit Theorem zThe Central Limit Theorem (CLT) states: yIf n is large (say n 30) then S n follows approximately a normal distribution with mean n and standard deviation yIf n is large (say n 30) then follows approximately a normal distribution with mean and standard deviation

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15.Math-Review4 Central Limit Theorem zExample: sums of a Bernoulli random variable

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15.Math-Review5 Central Limit Theorem zExample: Averages of Bernoulli random variable

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15.Math-Review6 Central Limit Theorem zExample: Compare a binomial random variable X~B(40,0.2) with its normal approximation: yWhat is the normal approximation? yCompare P(X 10), P(X 20), P(X 30) for the binomial and the normal approximation.

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15.Math-Review7 zLet us consider the following example. yWe work at a phone company and we would like to be able to estimate the shape of the demand. yWe assume that monthly household telephone bills follow a certain probability distribution (continuous) yWe have obtained the following data of monthly household telephone bills by interviewing 70 randomly chosen households (or their habitants rather) for the month of October. Sampling

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15.Math-Review8 zTable: Sampling

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15.Math-Review9 zFrom this information we would like to be able to estimate, for example: yWhat is an estimate of the shape of the distribution of October household telephone bills? yWhat is an estimate of the percentage of households whose October telephone bill is bellow $45.00 yWhat is an estimate of the percentage of households whose October telephone bill is between $60.00 and $100.00? yWhat is an estimate of the mean of the distribution of October household telephone bills? yWhat is an estimate of the standard deviation of the distribution of October household telephone bills? Sampling

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15.Math-Review10 zA population (or universe) is the set of all units of interest. zA sample is a subset of the units of a population. zA random sample is a sample collected in such a way that every unit in the population is equally likely to be selected. zIt is hard to ensure that a sample will be random. Sampling

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15.Math-Review11 zIn our example the population corresponds to all the households in our area of coverage. zThe random sample selected were the 70 households (or their inhabitants) interviewed. zAnd for the random variables X 1,X 2,…,X n corresponding to households 1, 2,…, n we observed x 1 =$95.67, x 2 =$82.69,…, x n =$ zNote that if we had chosen a different random set of households we would have observed a different collection of values. Sampling

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15.Math-Review12 zTo fix notation: yn will be our random sample size. yX 1,X 2,…,X n correspond to the random variables of unknown distribution f(x), which is common to our population, and what we want to study. yx 1,x 2,…,x n are the observations obtained by observing the outcome of our random sample. These are numbers!! yWe try to use these numbers to estimate the characteristics of f(x), for example what is the distribution, what is its mean, variance, etc. Sampling

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15.Math-Review13 zTo look at the shape of the distribution of X it is useful to create a frequency table and histogram of the sample values x 1,x 2,…,x n. Sampling

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15.Math-Review14 zA histogram can be obtained from excel, the output looks something like this: Sampling

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15.Math-Review15 zFrom this analysis we can give the following description of the shape of this distribution (qualitative): yAn estimate of the shape of the distribution of October telephone bills in the site area is that it is shaped like a Normal distribution, with a peak near $65.00, except for a small but significant group in the range between $ and $ Sampling

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15.Math-Review16 zIn order to answer the other relevant questions we can use the original data, and count favorable outcomes and divide by total possible outcomes (70): yP(X 45.00) = 5/70 = 0.07 yP (60.00 X ) = 45/70 = 0.64 zHere we are approximating the continuous unknown distribution by the discrete distribution given by the outcomes of the sample Sampling

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15.Math-Review17 zSample mean, variance and standard deviation: zFrom our observed values x 1,x 2,…,x n, we can compute: Sampling

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15.Math-Review18 zIn our example we have: Sampling

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15.Math-Review19 zWe will use these observed values to estimate the unknown mean, and standard deviation, of our unknown underlying distribution. zIn other words: Sampling zAlso note that if we pick a different sample of the population, our observed values will be different. zWe can define the random variables: sample mean, sample standard deviation, of which x and s are observations.

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15.Math-Review20 zBefore the sample is collected, the random variables X 1,X 2,…,X n, can be used to define: Sampling

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15.Math-Review21 zX and S are random variables zWe distinguish between the sample mean X, which is a random variable, and the observed sample mean x, which is a number. zSimilarly, the sample standard deviation S is a random variable, and the observed sample standard deviation s is a number. Sampling

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15.Math-Review22 zDistribution of X yFrom the formula that defines the sample mean we see that according to CLT it should follow approximately a normal distribution (if n 30) yThe mean is E(X) = yThe standard deviation is E(X) = zIn summary: Sampling

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15.Math-Review23 zExample: At two different branches of the G-Mart department store, they randomly sampled 100 customers on August 13. At Store 1, the average amount purchased was $41.25 per customer, with a sample standard deviation of $ At Store 2, the average amount purchased was $45.75 with a sample standard deviation of $34.00 Let X denote the amount of a random purchase by a single customer at Store 1 and let Y denote the amount of a random purchase by a single customer at Store 2. Assuming that X and Y satisfy a joint normal distribution, what is the distribution of X-Y? What is the probability that the mean of X exceeds the mean of Y? Sampling

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15.Math-Review24 zExample: In the quality control department of our company, knobs are inspected to make sure that they meet quality standards. Since it is not practical to test every knob, we draw a random sample to test. It is extremely necessary that our knobs weigh at least 0.45 pounds. If we know that the average weight is less than 0.45 pounds, we stop the production line and reset all the machines. In a day we produce 300,000 knobs, and draw a random sample of 1,000 knobs to test. If yesterday (Wednesday) the observed sample mean was 0.42 pounds, and observed sample standard deviation was 0.2, yhow confident are you that the average weigh of knobs is less than 0.45 pounds? yIf the average weight of knobs produced is 0.45 pounds, with standard deviation of 0.2, what is the probability that the average weight of the sample will be 0.42 or lower? yAre these questions the same? Sampling

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