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THE DISTRIBUTION OF THE SAMPLE MEAN CENTRAL LIMIT THEOREM A-) FROM ANY POPULATION

An importer of Herbs and Spices claims that average weight of packets of Saffron is 20 grams. However packets are actually filled to an average weight, μ =19.5 grams and standard deviation, σ =1.8 gram. A random sample of 36 packets is selected, calculate

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THE DISTRIBUTION OF THE SAMPLE MEAN CENTRAL LIMIT THEOREM A-) FROM ANY POPULATION

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  1. An importer of Herbs and Spices claims that average weight of packets of Saffron is 20 grams. However packets are actually filled to an average weight, μ=19.5 grams and standard deviation, σ=1.8 gram. A random sample of 36 packets is selected, calculate A- The probability that the average weight is 20 grams or more; B- The two limits within which 95% of all packets weight; C- The two limits within which 95% of all weights fall (n=36); D- If the size of the random sample was 16 instead of 36 how would this affect the results in (a), (b) and (c)? (State any assumptions made)

  2. THE DISTRIBUTION OF THE SAMPLE MEAN CENTRAL LIMIT THEOREM A-) FROM ANY POPULATION If x1,x2,…………,xn is a random variable of size n taken from any distribution with mean µ and variance σ2 then, for large n, the distribution of the sample mean x̄is approximately normal and x̄ ~ N(µ, σ2/n), where x̄ = (x1+x2+…………+xn )/n

  3. THE DISTRIBUTION OF THE SAMPLE MEAN CENTRAL LIMIT THEOREM B-) FROM A NORMAL POPULATION If x1,x2,…………,xn is a random variable of size n taken from a normal distribution with mean µ and variance σ2 such that X ~ N(µ, σ2), then the distribution of x̄is also normal and x̄ ~ N(µ, σ2/n), where x̄ = (x1+x2+…………+xn )/n The distribution of the sample mean (x̄) is known as the sampling distribution of means and the standard deviation of this distribution σ/ √n is known as the standard error of the mean

  4. Example for Correction Factor: What is the value of the finite population correction factor when a-) n= 20 and N=200 ? b-) n= 20 and N= 2000 ?

  5. Example 2: Tuition Cost The mean tuition cost at state universities throughout the USA is 4,260 USD per year (2002 year figures). Use this value as the population mean and assume that the population standard deviation is 900 USD. Suppose that a random sample of 50 state universities will be selected. A-) Show the sampling distribution of x̄ (where x̄ is the sample mean tuition cost for the 50 state universities) B-) What is the probability that the random sample will provide a sample mean within 250 USD of the population mean? C-) What is the probability that the simple random sample will provide a sample mean within 100 USD of the population mean?

  6. Example 1: A random variable of size 15 is taken from normal distribution with mean 60 and standard deviation 4. Find the probability that the mean of the sample is less than 58.

  7. Example 3: If a random sample of size 30 is taken from binomial distribution with n=9 and p= 0.5 Q: Find the probability that the sample mean exceeds 5.

  8. Example 4: Suppose we have selected a random sample of n=36 observations from a population with mean equal to 80 and standard deviation equal to 6. Q: Find the probability that x̄ will be larger than 82.

  9. Example 5: Ping-Pong Balls The diameter of a brand of Ping-Pong balls is approximately normally distributed, with a mean of 1.30 inches and a standard deviation of 0.04 inch. If you select a random sample of 16 Ping-Pong balls, A-) What is the sampling distribution of the sample mean? B-) What is the probability that sample mean is less than 1.28 inches? C-) What is the probability that sample mean is between 1.31 and 1.33 inches? D-) The probability is 60% that sample mean will be between what two values, symmetrically distributed around the population mean?

  10. Example 6: E-Mails Time spent using e-mail per session is normally distributed, with a mean of 8 minutes and a standard deviation of 2 minutes. If you select a random sample of 25 sessions, A-) What is the probability that sample mean is between 7.8 and 8.2 minutes? B-) What is the probability that sample mean is between 7.5 and 8.0 minutes? C-) If you select a random sample of 100 sessions, what is the probability that sample mean is between 7.8 and 8.2 minutes? D-) Explain the difference in the results of (A) and (C).

  11. Example 6: ELECTION A political pollster is conducting an analysis of sample results in order to make predictions on election night. Assuming a two-candidate election, if a specific candidate receives at least 55% of the vote in the sample, that candidate will be forecast as the winner of the election. If you select a random sample of 100 voters, what is the probability that a candidate will be forecast as the winner when A-) the population percentage of her vote is 50.1% ? B-) the population percentage of her vote is 60% ? C-) the population percentage of her vote is 49% (and she will actually lose the election)? D-) If the sample size is increased to 400, what are your answers to (A) through (C) ? Discuss.

  12. Types of Survey Errors • Coverage error • Non response error • Sampling error • Measurement error Excluded from frame Follow up on nonresponses Random differences from sample to sample Bad or leading question

  13. Population Distribution Sampling Distribution Standard Normal Distribution ? ? ? ? ? ? ? ? ? ? Sample Standardize ? ? Z X

  14. Sampling Distribution Properties As n increases, decreases Larger sample size Smaller sample size

  15. Sampling Distribution Properties (i.e. is unbiased) Normal Population Distribution Normal Sampling Distribution (has the same mean) Variation:

  16. How Large is Large Enough? • For most distributions, n ≥ 30 will give a sampling distribution that is nearly normal • For fairly symmetric distributions, n ≥ 15 • For normal population distributions, the sampling distribution of the mean is always normally distributed

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