1. Slide Slide 1 Chapter 8 Sampling Distributions Mean and Proportion

2. Slide Slide 2 2 Goal of Statistical Analysis: Find Parameters of Population from Statistics on Sample Sample Population Random Sampling: every unit in the population has an equal chance to be chosen The quality of all statistical analysis depends on the quality of the sample data

3. Slide Slide 3  Parameter: A number describing a population.  Statistic: A number describing a sample. 1.A random sample should represent the population well, so sample statistics from a random sample should provide reasonable estimates of population parameters. 2.All sample statistics have some error in estimating population parameters. 3.If repeated samples are taken from a population and the same statistic (e.g. mean) is calculated from each sample, the statistics will vary, that is, they will have a distribution. 4.A larger sample provides more information than a smaller sample so a statistic from a large sample should have less error than a statistic from a small sample.

4. Slide Slide 4 Sampling distributions for:  Mean (8.1) EVave (mean of a parameter in a population or EVave )  Proportion (8.2) EV% (percentage of a parameter in a population or EV% )

5. Slide Slide 5 8.1 Distribution of the Sample Mean

6. Slide Slide 6 Statistics such as are random variables since their value varies from sample to sample. So they have probability distributions associated with them. In this chapter we focus on the shape, center and spread of statistics such as.

7. Slide Slide 7 8-7 The sampling distribution of a statistic is a probability distribution for all possible values of the statistic computed from a sample of size n. The sampling distribution of the sample mean is the probability distribution of all possible values of the random variable computed from a sample of size n from a population with mean  and standard deviation .

8. Slide Slide 8 8-8 The weights of pennies minted after 1982 are approximately normally distributed with mean 2.46 grams and standard deviation 0.02 grams. Approximate the sampling distribution of the sample mean by obtaining 200 simple random samples of size n = 5 from this population. Example 1: Sampling Distribution of the Sample Mean-Normal Population

9. Slide Slide 9 8-9 The data on the following slide represent the sample means for the 200 simple random samples of size n = 5. For example, the first sample of n = 5 had the following data: 2.493 2.466 2.473 2.4922.471 Note: =2.479 for this sample

10. Slide Slide 10 8-10 Sample Means for Samples of Size n =5

11. Slide Slide 11 8-11 The mean of the 200 sample means is 2.46, the same as the mean of the population. The standard deviation of the sample means is 0.0086, which is smaller than the standard deviation of the population. The next slide shows the histogram of the sample means.

12. Slide Slide 12 8-12

13. Slide Slide 13 8-13 What role does n, the sample size, play in the standard deviation of the distribution of the sample mean?

14. Slide Slide 14 8-14 What role does n, the sample size, play in the standard deviation of the distribution of the sample mean? As the size of the sample gets larger, we do not expect as much spread in the sample means since larger observations will offset smaller observations.

15. Slide Slide 15 Suppose that a simple random sample of size n is drawn from a large population with mean  and standard deviation . The sampling distribution of will have mean and standard deviation. The standard deviation of the sampling distribution is called the standard error of the mean and is denoted. The Mean and Standard Deviation of the Sampling Distribution of

16. Slide Slide 16 Notation the mean of the sample means the standard deviation of sample mean  (often called the standard error of the mean) µx = µµx = µ n x =x = 

17. Slide Slide 17 Sampling from Normal Populations

18. Slide Slide 18 The weights of pennies minted after 1982 are approximately normally distributed with mean 2.46 grams and standard deviation 0.02 grams. What is the probability that in a simple random sample of 10 pennies minted after 1982, we obtain a sample mean of at least 2.465 grams? Example – Weight of pennies

19. Slide Slide 19 is normally distributed with =2.46 and.. P(Z>0.79)=1-0.7852 =0.2148. Solution On CALCULATOR: P(x>2.465)=normalcdf(2.465,10^99,2.46, 0.0063)=0.2148

20. Slide Slide 20 Given the population of passengers has normally distributed weights with a mean of 172 lb and a standard deviation of 29 lb, a) if one man is randomly selected, find the probability that his weight is greater than 175 lb. b) if 20 different men are randomly selected, find the probability that their mean weight is greater than 175 lb Another Example – Water Taxi (work on your own)

21. Slide Slide 21 Or use table: z = 175 – 172 = 0.10 29 a) if one man is randomly selected, find the probability that his weight is greater than 175 lb: CALCULATOR: P(X>175)=normalcdf(175,10^99,172,29) Ans

22. Slide Slide 22 b) if 20 different men are randomly selected, find the probability that their mean weight is greater than 172 lb. CALCULATOR: P(X>175)=normalcdf(175,10^99,172,29/ 20 ) Ans – cont Or use table: z = 175 – 172 = 0.46 29 20

23. Slide Slide 23 b) if 20 different men are randomly selected, their mean weight is greater than 175 lb. P(x > 175) = 0.3228 It is much easier for an individual to deviate from the mean than it is for a group of 20 to deviate from the mean. a) if one man is randomly selected, find the probability that his weight is greater than 175 lb. P(x > 175) = 0.4602 Ans - conclusion

24. Slide Slide 24 Sampling from a Population that is not Normal

25. Slide Slide 25 The following table and histogram give the probability distribution for rolling a fair die:  =3.5,  =1.708 Note that the population distribution is NOT normal Face on DieRelative Frequency 1 0.1667 2 3 4 5 6 EXAMPLE: Sampling from a Population that is Not Normal

26. Slide Slide 26 Estimate the sampling distribution of (average of n tosses of the die) by obtaining 200 simple random samples of size n=4 and calculating the sample mean for each of the 200 samples. Repeat for n = 10 and 30. Histograms of the sampling distribution of the sample mean for each sample size are given on the next slide.

27. Slide Slide 27

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29. Slide Slide 29

30. Slide Slide 30 Central Limit Theorem  The random variable x has a distribution (which may or may not be normal) with mean µ and standard deviation .  Simple random samples all of size n are selected from the population. (The samples are selected so that all possible samples of the same size n have the same chance of being selected.) Given: 1. The distribution of sample x will, as the sample size increases, approach a normal distribution. 2. The mean of the sample means is the population mean µ. 3. The standard deviation of all sample means is  n

31. Slide Slide 31 Key Points  The mean of the sampling distribution is equal to the mean of the parent population and the standard deviation of the sampling distribution of the sample mean is regardless of the sample size.  The shape of the distribution of the sample mean becomes approximately normal as the sample size n increases, regardless of the shape of the population: This is a result of The Central Limit Theorem. As the sample size increases, the sampling distribution of sample means approaches a normal distribution.

32. Slide Slide 32 Practical Rules 1.For samples of size n larger than 30, the distribution of the sample means can be approximated reasonably well by a normal distribution. The approximation gets better as the sample size n becomes larger. 2. If the original population is itself normally distributed, then the sample means will be normally distributed for any sample size n (not just the values of n larger than 30).

33. Slide Slide 33 Example : (Using the Central Limit Theorem) Suppose that the mean time for an oil change at a “10-minute oil change joint” is 11.4 minutes with a standard deviation of 3.2 minutes. (a)If a random sample of n = 35 oil changes is selected, describe the sampling distribution of the sample mean. (b) If a random sample of n = 35 oil changes is selected, what is the probability the mean oil change time is less than 11 minutes?

34. Slide Slide 34 Example : (Using the Central Limit Theorem) Suppose that the mean time for an oil change at a “10-minute oil change joint” is 11.4 minutes with a standard deviation of 3.2 minutes. (a)If a random sample of n = 35 oil changes is selected, describe the sampling distribution of the sample mean. (b) If a random sample of n = 35 oil changes is selected, what is the probability the mean oil change time is less than 11 minutes? Solution: is approximately normally distributed with mean=11.4 and std. dev. =. Solution:, P(Z<-0.74)=0.23.

35. Slide Slide 35 8.2 Distribution of the Sample Proportion

36. Slide Slide 36 Point Estimate of a Population Proportion Suppose that a random sample of size n is obtained from a population in which each individual either does or does not have a certain characteristic. The sample proportion, denoted (read “p-hat”) is given by where x is the number of individuals in the sample with the specified characteristic. The sample proportion is a statistic that estimates the population proportion, p.

37. Slide Slide 37 In a Quinnipiac University Poll conducted in May of 2008, 1,745 registered voters nationwide were asked whether they approved of the way George W. Bush was handling the economy. 349 responded “yes”. Obtain a point estimate for the proportion of registered voters who approved of the way George W. Bush was handling the economy. Example 1: Computing a Sample Proportion

38. Slide Slide 38 In a Quinnipiac University Poll conducted in May of 2008, 1,745 registered voters nationwide were asked whether they approved of the way George W. Bush was handling the economy. 349 responded “yes”. Obtain a point estimate for the proportion of registered voters who approved of the way George W. Bush was handling the economy. Example 1: Computing a Sample Proportion Solution:

39. Slide Slide 39 8-39 According to a Time poll conducted in June of 2008, 42% of registered voters believed that gay and lesbian couples should be allowed to marry. Describe the sampling distribution of the sample proportion for samples of size n=10, 50, 100. Example 2: Using Simulation to Describe the Distribution of the Sample Proportion

40. Slide Slide 40 8-40

41. Slide Slide 41 8-41

42. Slide Slide 42 8-42

43. Slide Slide 43 8-43 Key Points from Example 2  Shape: As the size of the sample, n, increases, the shape of the sampling distribution of the sample proportion becomes approximately normal.  Center: The mean of the sampling distribution of the sample proportion equals the population proportion, p.  Spread: The standard deviation of the sampling distribution of the sample proportion decreases as the sample size, n, increases.

44. Slide Slide 44 For a simple random sample of size n with population proportion p: The shape of the sampling distribution of is approximately normal provided np(1-p)≥10. The mean of the sampling distribution of is The standard deviation of the sampling distribution of is Sampling Distribution of

45. Slide Slide 45 Sampling Distribution of The model on the previous slides requires that the sampled values are independent. When sampling from finite populations, this assumption is verified by checking that the sample size n is no more than 5% of the population size N (n ≤ 0.05N). Regardless of whether np(1-p) ≥10 or not, the mean of the sampling distribution of is p, and the standard deviation is

46. Slide Slide 46 According to a Time poll conducted in June of 2008, 42% of registered voters believed that gay and lesbian couples should be allowed to marry. Suppose that we obtain a simple random sample of 50 voters and determine which believe that gay and lesbian couples should be allowed to marry. Describe the sampling distribution of the sample proportion. Example 3:

47. Slide Slide 47 Solution The sample of n=50 is smaller than 5% of the population size (all registered voters in the U.S.). Also, np(1-p)=50(0.42)(0.58)=12.18≥10. The sampling distribution of the sample proportion is therefore approximately normal with mean=0.42 and standard deviation=

48. Slide Slide 48 According to the Centers for Disease Control and Prevention, 18.8% of school-aged children, aged 6-11 years were overweight in 2004. (a)In a random sample of 90 school children aged 6-11 years what is the probability that at least 19% are overweight? (b)Suppose in one random sample of 90 school children aged 6-11 years there were 24 overweight children. What might you conclude? Example 4: Compute Probabilities of a Sample Proportion

49. Slide Slide 49 n=90 is less than 5% of the population size np(1-p)=90(.188)(1-.188)≈13.7≥10 is approximately normal with mean=0.188 and standard deviation = (a)In a random sample of 90 school-aged children, aged 6-11 years, what is the probability that at least 19% are overweight? Or (CALCULATOR): Solution, P(Z>0.05)=1-0.5199=0.4801 P(X>0.19)=normalcdf(0.19,10^99,0.188,0.0412)=0.4801

50. Slide Slide 50 is approximately normal with mean=0.188 and standard deviation = 0.0412 (b)Suppose in one random sample of 90 school children aged 6-11 years there were 24 overweight children. What might you conclude? Solution, P(X>0.2667)= normalcdf(0.2667,10^99,0.188,0.0412)= 0.028. We would only expect to see about 3 samples in 100 resulting in a sample proportion of 0.2667 or more. This is an unusual sample if the true population proportion is 0.188

51. Slide Slide 51 Next: Practice problems – Answers included (re-work on your own): Problem 1: Sample Mean and probability for sample mean Problem 2: Sample proportion and probability for sample proportion

52. Slide Slide 52 Problem 1 – Sample mean : Flight search processing time Web application for a flight search: An investigator takes a sample of 100 flight searches and notes the web response time. Assume that the population average of ALL web searches is 15 sec with a standard deviation is 5 sec. Here are the summary statistics calculated by a Statistical Software for the sample of 100: Summary of web processing time The MEANS Procedure Analysis Variable : time N Mean Std Dev Minimum Maximum --------------------------------------------------------------- 100 14.9955626 5.2117790 2.2461204 25.7383955 --------------------------------------------------------------- The estimated processing time is 14.99 seconds (sample average) The standard error is equal to  /sqrt(n) = 5/sqrt(100)=0.5.

53. Slide Slide 53 What is the probability that in 100 flight searches, the average time to process the requests is less than 14 seconds? We can use the normal approximation: The sample average is normally distributed with mean equal to 15 and standard deviation equal to the standard error = 0.5. 14 15 time P(X<14) = normalcdf(-10^99,14,15,0.5) =0.0228 There is only about 2.3% chance that the average time to process 100 flight requests is less than 14 seconds.

54. Slide Slide 54 A study by a Federal Agency in 1983 concluded that polygraph (lie detector) tests given to truthful people have probability 0.2 of suggesting that the person is deceptive. A firm asks 20 job applicants about thefts from previous employers, using a polygraph to assess their truthfulness. All applicants were truthful. What is the chance that at least one will fail the test? Problem 2 – Sample proportion: Polygraph percentages Compute sample proportion and standard error for the sample proportion: 0.05 0.2 proportion Sample proportion is p=0.2 (same as population proportion), The standard error is sqrt(p*(1-p)/n) =sqrt(0.2*0.8/20)=0.09 Thus sample percentage is approximately normal with mean 0.2 and standard deviation 0.09.

55. Slide Slide 55 What is the chance that at least 1 in 20 will fail the test? Answer: First figure what prportion does 1 in 20 constitute: 1/20 = 0.05 So probability is: P(X>0.05) = normalcdf(0.05,10^99,0.2,0.09) = 0.952 0.05 0.2 percentage 95.2% Conclusion: The chance that at least one applicant out of 20 will fail the polygraph test is 95.2%. That is extremely high!