1.  The Sampling Distribution of the Sample Mean  The Sampling Distribution of the Sample Proportion Chapter 6 Sampling Distributions( 样本分布 )

2. Chapter 6 Sampling Distributions Sample Mean  Let there be a population of units of size N  Consider all its samples of a fixed size n (n<N)  For all possible samples of size n, we obtain a population of sample means. That is, is a random variable which may have all these means as its values  Before we draw the sample, the sample mean  Before we draw the sample, the sample mean is a random variable.  We consider the probability distribution of the random variable, i.e., the probability distribution for the population of sample means

3. Chapter 6 Sampling Distributions Section 6.1 The Sampling Distribution of the Sample Mean The sampling distribution of the sample mean is the probability distribution of the population of the sample means obtainable from all possible samples of size n from a population of size N.

4. Chapter 6 Sampling Distributions Example 6.1 Example 6.1 The Stock Return Case We have a population of the percent returns from six stocks – –In order, the values of % return are: 10%, 20%, 30%, 40%, 50%, and 60% Label each stock A, B, C, …, F in order of increasing % return The mean rate of return is 35% with a standard deviation of 17.078% – –Any one stock of these stocks is as likely to be picked as any other of the six Uniform distribution with N = 6 Each stock has a probability of being picked of 1/6

5. Chapter 6 Sampling Distributions The Stock Return Case ﹟ 2

6. Chapter 6 Sampling Distributions Now, select all possible samples of size n = 2 from this population of stocks of size N = 6Now, select all possible samples of size n = 2 from this population of stocks of size N = 6 –That is, select all possible pairs of stocks How to select?How to select? –Sample randomly –Sample without replacement –Sample without regard to order The Stock Return Case ﹟ 3

7. Chapter 6 Sampling Distributions Result: There are 15 possible samples of size n = 2Result: There are 15 possible samples of size n = 2 Calculate the sample mean of each and every sampleCalculate the sample mean of each and every sample For example, if we choose the two stocks with returns 10% and 20%, then the sample mean is 15%For example, if we choose the two stocks with returns 10% and 20%, then the sample mean is 15% The Stock Return Case ﹟ 4

8. Chapter 6 Sampling Distributions The Stock Return Case ﹟ 5

9. Chapter 6 Sampling Distributions The population of N = 6 stock returns has a uniform distribution. But the histogram of n = 15 sample mean returns: 1. 1.Seems to be centered over the same mean return of 35%, and 2. 2.Appears to be bell-shaped and less spread out than the histogram of individual returns Observations

10. Chapter 6 Sampling Distributions Example 6.2 Example 6.2 Sampling all the stocks Consider the population of returns of all 1,815 stocks listed on NYSE for 1987 – –See Figure 6.2(a) on next slide – –The mean rate of return was –3.5% with a standard deviation of 26% Draw all possible random samples of size n = 5 and calculate the sample mean return of each – –Sample with a computer – –See Figure 6.2(b) on next slide

11. Chapter 6 Sampling Distributions

12. Observations – –Both histograms appear to be bell-shaped and centered over the same mean of –3.5% – –The histogram of the sample mean returns looks less spread out than that of the individual returns Statistics – –Mean of all sample means:  =  = -3.5% – –Standard deviation of all possible means:

13. Chapter 6 Sampling Distributions   If the population of individual items is normal, then the population of all sample means is also normal   Even if the population of individual items is not normal, there are circumstances that the population of all sample means is normal (see Central Limit Theorem( 中心极限定理 ) later) General Conclusions

14. Chapter 6 Sampling Distributions The mean of all possible sample means equals the population meanThe mean of all possible sample means equals the population mean –That is, –That is,  =  The standard deviation s x of all sample means is less than the standard deviation of the population – –That is,  <  Each sample mean averages out the high and the low measurements, and so are closer to m than many of the individual population measurements General Conclusions

15. Chapter 6 Sampling Distributions The empirical rule holds for the sampling distribution of the sample meanThe empirical rule holds for the sampling distribution of the sample mean –68.26% of all possible sample means are within (plus or minus) one standard deviationof –68.26% of all possible sample means are within (plus or minus) one standard deviation  of  – of –95.44% of all possible observed values of x are within (plus or minus) two  of  In the example., 95.44% of all possible sample mean returns are in the interval [-3.5 ± (2  11.63)] = [-3.5 ± 23.26] That is, 95.44% of all possible sample means are between -26.76% and 19.76% – –99.73% of all possible observed values of x are within (plus or minus) three  of 

16. Chapter 6 Sampling Distributions Properties of the Sampling Distribution of the Sample Mean #1 If the population being sampled is normal, then so is the sampling distribution of the sample meanIf the population being sampled is normal, then so is the sampling distribution of the sample mean, The mean of the sampling distribution of is  =  That is, the mean of all possible sample means is the same as the population mean

17. Chapter 6 Sampling Distributions The variance  2 of the sampling distribution of isThe variance  2 of the sampling distribution of is  That is, the variance of the sampling distribution of is  directly proportional to the variance of the population, and  inversely proportional to the sample size Properties of the Sampling Distribution of the Sample Mean #2

18. Chapter 6 Sampling Distributions Properties of the Sampling Distribution of the Sample Mean #3 The standard deviation  of the sampling distribution of is   That is, the standard deviation of the sampling distribution of is   directly proportional to the standard deviation of the population, and   inversely proportional to the square root of the sample size

19. Chapter 6 Sampling Distributions  is more clustered to the population  is the point estimate of , and the larger the sample size n, the more accurate the estimate, because when n increases,  decreases, is more clustered to the population –In order to reduce –In order to reduce , take bigger samples! Notes

20. Chapter 6 Sampling Distributions Example 6.3 Example 6.3 Car Mileage Case Population of all midsize cars of a particular make and model – –Population is normal with mean  and standard deviation  –Draw all possible samples of size n –Then the sampling distribution of the sample mean is normal with mean = –Then the sampling distribution of the sample mean is normal with mean  =  and standard deviation – –In particular, draw samples of size: n = 5 n = 49

21. Chapter 6 Sampling Distributions So, all possible sample means for n=49 will be more closely clustered around  than the case of n =5

22. Chapter 6 Sampling Distributions Recall from Chapter 2 mileage example, = 31.5531 mpg for a sample of size n=49 –With s = 0.7992 Does this give statistical evidence that the population meanis greater than 31 mpg?Does this give statistical evidence that the population mean  is greater than 31 mpg? –That is, does the sample mean give evidence that is at least 31 mpg? –That is, does the sample mean give evidence that  is at least 31 mpg? = 31 mpgCalculate the probability of observing a sample mean that is greater than or equal to 31.5531 mpg if  = 31 mpg –Want –Want P( > 31.5531 if  = 31) Reasoning from Sample Distribution

23. Chapter 6 Sampling Distributions  Use s as the point estimate for  so that  Then  But z = 4.84 is off the standard normal table  The largest z value in the table is 3.09, which has a right hand tail area of 0.001

24. Chapter 6 Sampling Distributions = 31 Probability that > 31.5531 when  = 31

25. Chapter 6 Sampling Distributions z = 4.84 > 3.09, so P(z ≥ 4.84) 3.09, so P(z ≥ 4.84) < 0.001 That is, if samples have a mean at least as large as observedThat is, if  = 31 mpg, then fewer than 1 in 1,000 of all possible samples have a mean at least as large as observed Have either of the following explanations:Have either of the following explanations: is actually 31 mpg, then picking this sample is an almost unbelievable thing If  is actually 31 mpg, then picking this sample is an almost unbelievable thingOR is not 31 mpg  is not 31 mpg evidence that.Difficult to believe such a small chance would occur, so conclude that there is strong evidence that  does not equal 31 mpg. – is in fact larger than 31 mpg –  is in fact larger than 31 mpg

26. Chapter 6 Sampling Distributions Central Limit Theorem ( 中心极限定理 ) #1  If the population is non-normal, what is the shape of the sampling distribution of the sample means?  In fact the sampling distribution is approximately normal if the sample is large enough, even if the population is non-normal by the “Central Limit Theorem”

27. Chapter 6 Sampling Distributions   No matter what is the probability distribution that describes the population, if the sample size n is large enough, then the population of all possible sample means is approximately normal with mean and standard deviation   Further, the larger the sample size n, the closer the sampling distribution of the sample means is to being normal – –In other words, the larger n, the better the approximation

28. Chapter 6 Sampling Distributions Random Sample (x 1, x 2, …, x n ) Population Distribution ( ,  ) (right-skewed) X as n  large Sampling Distribution of Sample Means (nearly normal)

29. Chapter 6 Sampling Distributions Example 6.4 Example 6.4 Effect of the Sample Size The larger the sample size, the more nearly normally distributed is the population of all possible sample means Also, as the sample size increases, the spread of the sampling distribution decreases

30. Chapter 6 Sampling Distributions How Large? How large is “large enough?”How large is “large enough?” If the sample size n is at least 30, then for most sampled populations, the sampling distribution of sample means is approximately normalIf the sample size n is at least 30, then for most sampled populations, the sampling distribution of sample means is approximately normal Refer to Figure 6.6 on next slide Refer to Figure 6.6 on next slide –Shown in Fig 6.6(a) is an exponential (right skewed) distribution –In Figure 6.6(b), 1,000 samples of size n = 5 »Slightly skewed to right –In Figure 6.6(c), 1,000 samples with n = 30 »Approximately bell-shaped and normal If the population is normal, the sampling distribution of is normal regardless of the sample sizeIf the population is normal, the sampling distribution of is normal regardless of the sample size

31. Chapter 6 Sampling Distributions Example: Central Limit Theorem Simulation

32. Chapter 6 Sampling Distributions Unbiased Estimates( 无偏估计 ) A sample statistic is an unbiased point estimate of a population parameter if the mean of all possible values of the sample statistic equals the population parameterA sample statistic is an unbiased point estimate of a population parameter if the mean of all possible values of the sample statistic equals the population parameter is an unbiased estimate of  because  =  – –In general, the sample mean is always an unbiased estimate of  – –The sample median is often an unbiased estimate of  But not alwaysBut not always

33. Chapter 6 Sampling Distributions The sample variance s 2 is an unbiased estimate of  2 if the sampled population is infinite – –That is why s 2 has a divisor of n–1 (if we used n as the divisor when estimating  2, we would not obtain an unbiased estimate) is not an unbiased estimate of However, s is not an unbiased estimate of  – –Even so, since there is no easy way to calculate an unbiased point estimate of  the usual practice is to use s as an estimate of 

34. Chapter 6 Sampling Distributions Minimum Variance Estimates ( 最小方差估计 ) Want the sample statistic to have a small standard deviation – –All values of the sample statistic should be clustered around the population parameter. Then, the statistic from any sample should be close to the population parameter Given a choice between unbiased estimates, choose one with smallest standard deviation Even though the sample mean and the sample median are both unbiased estimates of , the sampling distribution of sample means has a smaller standard deviation than that of sample medians

35. Chapter 6 Sampling Distributions The sample mean is a minimum-variance unbiased estimate ( 最小方差无偏估计 ) of  – than if we used any other sample statistic –When the sample mean is used to estimate , we are more likely to obtain an estimate close to  than if we used any other sample statistic –Therefore, the sample mean is the preferred estimate of –Therefore, the sample mean is the preferred estimate of 

36. Chapter 6 Sampling Distributions Section 6.2 The Sampling Distribution of the Sample Proportion( 样本比例 ) For a population of units, we select samples of size n, and calculate its proportion for the units of the sample to be fall into a particular category. is a random variable and has its probability distribution. is a random variable and has its probability distribution. The probability distribution of all possible sample proportions is called the sampling distribution of the sample proportion

37. Chapter 6 Sampling Distributions the sampling distribution of is   approximately normal, if n is large (meet the conditions that np≥5 and n(1-p)≥5)   has mean   has standard deviation where p is the population proportion for the category

38. Chapter 6 Sampling Distributions Example 6.5 Example 6.5 The Cheese Spread Case A food processing company developed a new cheese spread spout which may save production cost. If only less than 10% of current purchasers do not accept the design, the company would adopt and use the new spout. 1000 current purchasers are randomly selected and inquired, and 63 of them say they would stop buying the cheese spread if the new spout were used. So, the sample proportion =0.063. To evaluate the strength of this evidence, we ask: if p=0.1, what is the probability of observing a sample of size 1000 with sample proportion ≦ 0.063?

39. Chapter 6 Sampling Distributions If p=0.10, since n=1000, np≥5 and n(1-p)≥5, is approximately normal with

40. Chapter 6 Sampling Distributions So, if p=0.1, the chance of observing at most 63 out of 1000 randomly selected customers do not accept the new design is less than 0.001 But such observation does occur. This means that we have extremely strong evidence that p≠0.1, and p is in fact less than 0.1 Therefore, the company can adopt the new design

41.  z-Based Confidence Intervals for a Population Mean: Known  t-Based Confidence Intervals for a Population Mean: Unknown  Sample Size Determination( 样本 量计算 )  Confidence Intervals for a Population Proportion Chapter 7 Confidence Intervals( 置信区间 )

42. Chapter 6 Sampling Distributions Section 7.1 z-Based Confidence Intervals for a Population Mean The starting point is the sampling distribution( 样本分 布 ) of the sample meanThe starting point is the sampling distribution( 样本分 布 ) of the sample mean –Recall from Chapter 6 that if a population is normally distributed with mean –Recall from Chapter 6 that if a population is normally distributed with mean  and standard deviation , then the sampling distribution of is normal with mean  =  and standard deviation – –Use a normal curve as a model of the sampling distribution of the sample mean Exactly, because the population is normal Approximately, by the Central Limit Theorem for large samples( 大样本中心极限定理 )

43. Chapter 6 Sampling Distributions Recall the empirical rule, so… – –68.26% of all possible sample means are within one standard deviation of the population mean – –95.44% of all possible sample means are within two standard deviations of the population mean – –99.73% of all possible sample means are within three standard deviations of the population mean

44. Chapter 6 Sampling Distributions Example 7.1 Example 7.1 The Car Mileage Case Recall that the population of car mileages is normally distributed with mean  and standard deviation  = 0.8 mpg – –Note that  is unknown and is to be estimated but assumed that  = 31 mpg Taking samples of size n = 5, the sampling distribution of sample mean mileages is normal with mean  =  (which is also unknown) and standard deviation The probability is 0.9544 that will be within plus or minus 2 = 2 0.35777 = 0.7155 of 

45. Chapter 6 Sampling Distributions That the sample mean is within ±0.7155 of  is equivalent to… will be such that the interval [ ± 0.7115] contains  containsThen there is a 0.9544 probability that will be a value so that interval [ ± 0.7115] contains  – –In other words P( – 0.7155 ≤  ≤ + 0.7155) = 0.9544 – –The interval [ ± 0.7115] is referred to as the 95.44% confidence interval for  The Car Mileage Case ＃ 2

46. Chapter 6 Sampling Distributions The Car Mileage Case ＃ 3 95.44% Confidence Intervals for  Three intervals shown Two contain  One does not

47. Chapter 6 Sampling Distributions According to the 95.44% confidence interval, we know even before we sample that of all the possible samples that could be selected … … There is 95.44% probability that the sample mean that is calculated is such that the interval [ ± 0.7155] will contain the actual (but unknown) population mean  – –In other words, of all possible sample means, 95.44% of all the corresponding intervals will contain the population mean  – –Note that there is a 4.56% probability that the interval does not contain  The sample mean is either too high or too low

48. Chapter 6 Sampling Distributions In the example, we found the probability that  is contained in an interval of integer multiples of  More usual to specify the (integer) probability and find the corresponding number of  The probability that the confidence interval will not contain the population mean  is denoted by  – –In the example,  = 0.0456 Generalizing

49. Chapter 6 Sampling Distributions Generalizing Continued The probability that the confidence interval will contain the population mean  is denoted by  – –1 –  is referred to as the confidence coefficient – –(1 – )  100% is called the confidence level – –In the example, 1 –  = 0.9544 Usual to use two decimal point probabilities for 1 –  – –Here, focus on 1 –  = 0.95 or 0.99

50. Chapter 6 Sampling Distributions General Confidence Interval is contained in the intervalIn general, the probability is 1 –  that the population mean  is contained in the interval – –The normal point z /2 gives a right hand tail area under the standard normal curve equal to /2 – –The normal point - z /2 gives a left hand tail area under the standard normal curve equal to /2 – is 1 – –The area under the standard normal curve between -z /2 and z /2 is 1 – 

51. Chapter 6 Sampling Distributions

52. z-Based Confidence Intervals for a Mean with  Known If a population has standard deviation  (known), and if the population is normal or if sample size is large (n  30), then … … a  ) 100% confidence interval for  is

53. Chapter 6 Sampling Distributions 95% Confidence Level For a 95% confidence level, 1 –  = 0.95  = 0.05  /2 = 0.025 For 95% confidence, need the normal point z 0.025 The area under the standard normal curve between - z 0.025 and z 0.025 is 0.95 Then the area under the standard normal curve between 0 and z 0.025 is 0.475 From the standard normal table, the area is 0.475 for z = 1.96 Then z 0.025 = 1.96

54. Chapter 6 Sampling Distributions The Effect of  on Confidence Interval Width z  /2 = z 0.025 = 1.96 z  /2 = z 0.005 = 2.575

55. Chapter 6 Sampling Distributions 95% Confidence Interval The 95% confidence interval is

56. Chapter 6 Sampling Distributions 99% Confidence Interval For 99% confidence, need the normal point z 0.005 Reading between table entries in the standard normal table, the area is 0.495 for z 0.005 = 2.575 The 99% confidence interval is

57. Chapter 6 Sampling Distributions Example 7.2 Example 7.2 The Car Mileage Case Given: = 31.5531 mpg  = 0.8 mpg n = 49 95% Confidence Interval: 99% Confidence Interval:

58. Chapter 6 Sampling Distributions The 99% confidence interval is slightly wider than the 95% confidence intervalThe 99% confidence interval is slightly wider than the 95% confidence interval –The higher the confidence level, the wider the interval Reasoning from the intervals:Reasoning from the intervals: –The target mean mileage should be at least 31 mpg –Both confidence intervals exceed this target –According to the 95% confidence interval, we can be 95% confident that the mileage is between 31.33 and 31.78 mpg –So we can be 95% confident that, on average, the mean mileage exceeds the target by at least 0.33 mpg and at most 0.78 mpg

59. Chapter 6 Sampling Distributions Section 7.2 t-Based Confidence Intervals for a Population Mean If  is unknown (which is usually the case), we can construct a confidence interval for  based on the sampling distribution of If the population is normal, then for any sample size n, this sampling distribution is called the t distribution(t 分布 )

60. Chapter 6 Sampling Distributions The t Distribution(t 分布 ) The curve of the t distribution is similar to that of the standard normal curveThe curve of the t distribution is similar to that of the standard normal curve –Symmetrical and bell-shaped –The t distribution is more spread out than the standard normal distribution –The spread of the t is given by the number of degrees of freedom( 自由度 ) Denoted by dfDenoted by df For a sample of size n, there are one fewer degrees of freedom, that is,For a sample of size n, there are one fewer degrees of freedom, that is, df = n – 1

61. Chapter 6 Sampling Distributions Degrees of Freedom and the t-Distribution As the number of degrees of freedom increases, the spread of the t distribution decreases and the t curve approaches the standard normal curve

62. Chapter 6 Sampling Distributions The t Distribution and Degrees of Freedom For a t distribution with n – 1 degrees of freedom, – –As the sample size n increases, the degrees of freedom also increases – –As the degrees of freedom increase, the spread of the t curve decreases – –As the degrees of freedom increases indefinitely, the t curve approaches the standard normal curve If n ≥ 30, so df = n – 1 ≥ 29, the t curve is very similar to the standard normal curve

63. Chapter 6 Sampling Distributions t and Right Hand Tail Areas Use a t point denoted by t  – –t  is the point on the horizontal axis under the t curve that gives a right hand tail equal to  –in a particular situation depends on the right hand tail area –So the value of t  in a particular situation depends on the right hand tail area  and the number of degrees of freedom df = n – 1df = n – 1    = 1 – , where 1 –  is the specified confidence coefficient

64. Chapter 6 Sampling Distributions

65. Rows correspond to the different values of df Columns correspond to different values of a See Table 7.3, Tables A.4 and A.20 in Appendix A and the table on the inside cover – –Table 7.3 and A.4 gives t points for df 1 to 30, then for df = 40, 60, 120, and ∞ On the row for ∞, the t points are the z points – –Table A.20 gives t points for df from 1 to 100 For df greater than 100, t points can be approximated by the corresponding z points on the bottom row for df = ∞ – –Always look at the accompanying figure for guidance on how to use the table Using the t Distribution Table

66. Chapter 6 Sampling Distributions Find t  for a sample of size n = 15 and right hand tail area of 0.025 – –For n = 15, df = 14 – –  = 0.025 Note that  = 0.025 corresponds to a confidence level of 0.95 – –In Table 7.3, along row labeled 14 and under column labeled 0.025, read a table entry of 2.145 – –So t  = 2.145 Example 7.3 Example 7.3

67. Chapter 6 Sampling Distributions t-Based Confidence Intervals for a Mean:  Unknown If the sampled population is normally distributed with mean , then a  )100% confidence interval for  is t  /2 is the t point giving a right-hand tail area of  /2 under the t curve having n – 1 degrees of freedom

68. Chapter 6 Sampling Distributions Example 7.4 Example 7.4 Debt-to-Equity Ratio Estimate the mean debt-to-equity ratio of the loan portfolio of a bank Select a random sample of 15 commercial loan accounts – –Box plot is given in figure below Know: = 1.34 s = 0.192 n = 15 Want a 95% confidence interval for the ratio Assume all ratios are normally distributed but  unknown

69. Chapter 6 Sampling Distributions Have to use the t distribution At 95% confidence, 1 –  = 0.95 so  = 0.05 and  /2 = 0.025 For n = 15, df = 15 – 1 = 14 Use the t table to find t  /2 for df = 14 t  /2 = t 0.025 = 2.145 for df = 14 The 95% confidence interval:

70. Chapter 6 Sampling Distributions Section 7.3 Sample Size Determination If  is known, then a sample of size Letting E denote the desired margin of error, so that is within E units of , with 100(1-  )% confidence.

71. Chapter 6 Sampling Distributions If  is unknown and is estimated from s, then a sample of size so that is within E units of , with 100(1-  )% confidence. The number of degrees of freedom for the t  /2 point is the size of the preliminary sample minus 1

72. Chapter 6 Sampling Distributions Example 7.5 Example 7.5 Car Mileage Case Given: = 31.5531 mpg s = 0.7992 mpg n = 49 t- based 95% Confidence Interval: where t  /2 = t 0.025 = 2.0106 for df = 49 – 1 = 48 degrees of freedom Note: the error bound B = 0.2296 mpg, within the maximum error of 0.3 mpg

73. Chapter 6 Sampling Distributions Section 7.4 Confidence Intervals for a Population Proportion If the sample size n is large*, then a  )100% confidence interval for p is * Here n should be considered large if both

74. Chapter 6 Sampling Distributions Example 7.6 Example 7.6 Phe-Mycin Side Effects Given: n = 200, 35 patients experience nausea. Note: so both quantities are > 5 For 95% confidence, z  /2 = z 0.025 = 1.96 and

75. Chapter 6 Sampling Distributions Determining Sample Size for Confidence Interval for p A sample size will yield an estimate, precisely within E units of p, with 100(1-  )% confidence Note that the formula requires a preliminary estimate of p. The conservative value of p = 0.5 is generally used when there is no prior information on p

76. Chapter 6 Sampling Distributions A Comparison of Confidence Intervals and Tolerance Intervals A tolerance interval contains a specified percentage of individual population measurements Often 68.26%, 95.44%, 99.73% A confidence interval is an interval that contains the population mean , and the confidence level expresses how sure we are that this interval contains  Often confidence level is set high (e.g., 95% or 99%) – –Because such a level is considered high enough to provide convincing evidence about the value of 

77. Chapter 6 Sampling Distributions Example 7.7 Example 7.7 Car Mileage Case Tolerance intervals shown: [ ± s] contains 68% of all individual cars [ ± 2s] contains 95.44% of all individual cars [ ± 3s] contains 99.73% of all individual cars The t-based 95% confidence interval for  is [31.32, 31.78], so we can be 95% confident that  lies between 31.23 and 31.78 mpg

78. Chapter 6 Sampling Distributions Summary: Selecting an Appropriate Confidence Interval for a Population Mean