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15.Math-Review

15.Math-Review. Statistics. Central Limit Theorem. Let us consider X 1 , X 2 ,…,X n , n independent identically distributed random variables with mean  and standard deviation  . And define:. Central Limit Theorem. The Central Limit Theorem (CLT) states:

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15.Math-Review

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  1. 15.Math-Review Statistics

  2. Central Limit Theorem • Let us consider X1, X2,…,Xn, n independent identically distributed random variables with mean  and standard deviation . • And define:

  3. Central Limit Theorem • The Central Limit Theorem (CLT) states: • If n is large (say n30) then Sn follows approximately a normal distribution with mean n and standard deviation • If n is large (say n30) then follows approximately a normal distribution with mean  and standard deviation

  4. Central Limit Theorem • Example:sums of a Bernoulli random variable

  5. Central Limit Theorem • Example: Averages of Bernoulli random variable

  6. Central Limit Theorem • Example: Compare a binomial random variable X~B(40,0.2) with its normal approximation: • What is the normal approximation? • Compare P(X10), P(X 20), P(X30) for the binomial and the normal approximation.

  7. Sampling • Let us consider the following example. • We work at a phone company and we would like to be able to estimate the shape of the demand. • We assume that monthly household telephone bills follow a certain probability distribution (continuous) • We 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.

  8. Sampling • Table:

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

  10. Sampling • A population (or “universe”) is the set of all units of interest. • A sample is a subset of the units of a population. • A random sample is a sample collected in such a way that every unit in the population is equally likely to be selected. • It is hard to ensure that a sample will be random.

  11. Sampling • In our example the population corresponds to all the households in our area of coverage. • The random sample selected were the 70 households (or their inhabitants) interviewed. • And for the random variables X1,X2,… ,Xn corresponding to households 1, 2,… , n we observed x1=$95.67, x2=$82.69,… , xn=$74.13. • Note that if we had chosen a different random set of households we would have observed a different collection of values.

  12. Sampling • To fix notation: • n will be our random sample size. • X1,X2,… ,Xn correspond to the random variables of unknown distribution f(x), which is common to our population, and what we want to study. • x1,x2,… ,xn are the observations obtained by observing the outcome of our random sample. These are numbers!! • We try to use these numbers to estimate the characteristics of f(x), for example what is the distribution, what is its mean, variance, etc.

  13. Sampling • To “look” at the shape of the distribution of X it is useful to create a frequency table and histogram of the sample values x1,x2,… ,xn.

  14. Sampling • A histogram can be obtained from excel, the output looks something like this:

  15. Sampling • From this analysis we can give the following description of the shape of this distribution (qualitative): • An 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 $125.00 and $155.00.

  16. Sampling • In order to answer the other relevant questions we can use the original data, and count favorable outcomes and divide by total possible outcomes (70): • P(X  45.00) = 5/70 = 0.07 • P (60.00  X  100.00) = 45/70 = 0.64 • Here we are approximating the continuous unknown distribution by the discrete distribution given by the outcomes of the sample

  17. Sampling • Sample mean, variance and standard deviation: • From our observed values x1,x2,… ,xn, we can compute:

  18. Sampling • In our example we have:

  19. Sampling • We will use these observed values to estimate the unknown mean , and standard deviation , of our unknown underlying distribution. • In other words: • Also note that if we pick a different sample of the population, our observed values will be different. • We can define the random variables: sample mean, sample standard deviation, of which x and s are observations.

  20. Sampling • Before the sample is collected, the random variables X1,X2,… ,Xn, can be used to define:

  21. Sampling • X and S are random variables • We distinguish between the sample mean X, which is a random variable, and the observed sample mean x, which is a number. • Similarly, the sample standard deviation S is a random variable, and the observed sample standard deviation s is a number.

  22. Sampling • Distribution of X • From the formula that defines the sample mean we see that according to CLT it should follow approximately a normal distribution (if n30) • The mean is E(X) =  • The standard deviation is E(X) = • In summary:

  23. Sampling • Example: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 $24.00. 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?

  24. Sampling • Example: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, • how confident are you that the average weigh of knobs is less than 0.45 pounds? • If 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? • Are these questions the same?

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