1 / 108

Math 144

Math 144. Confidence Interval. In addition to the estimated value of the estimator, some statisticians suggest that we should also consider the variance of the estimator.

slade-mcbride
Download Presentation

Math 144

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Math 144 Confidence Interval

  2. In addition to the estimated value of the estimator, some statisticians suggest that we should also consider the variance of the estimator. Use the single value and the variance of the estimator to form an interval that has a high probability to cover the unknown parameter. This method including the variance of the point estimator is called interval estimation, or "confidence interval".

  3. Assume that and are two functions of a random sample and are determined by a point estimator of an unknown parameter such that Interval estimation where α is a known value between 0 and 1.

  4. After sampling, if the actual values of and are a and b, respectively, then the interval [a, b] is called a 100(1-α)% confidence interval (hereafter, C.I.) for θ. Interval estimation The quantity 1-αis called the confidence level associated with the confidence interval.

  5. By the definition, before sampling, we have a random interval estimation for the unknown parameter θ. Caution: After sampling, the confidence interval [a, b] is a fixed (not random) interval. Indeed, it depends on the particular sample observations.

  6. After sampling, we have observations a b P(a≤ θ≤b) = 0 θ [ [ Caution: Most importantly, the unknown parameter θ is either inside or outside the confidence interval [a, b].That is, P(a≤ θ≤b) = 0 or 1.

  7. After sampling, we have observations a b P(a≤ θ≤b) = 1 θ [ [ Caution: Most importantly, the unknown parameter θ is either inside or outside the confidence interval [a, b].That is, P(a≤ θ≤b) = 0 or 1.

  8. Recall that before sampling, we have Caution: Most importantly, the unknown parameter θ is either inside or outside the confidence interval [a, b].That is, P(a≤ θ≤b) = 0 or 1.

  9. Interpretation of C.I. The interpretation of a 100(1-α)% C.I. is that when we obtained N (sufficient large) independent sets of random sample and for each set of random sample, we construct one particular interval by using the same point estimator, then there are N(1-α) out of these N intervals will contain the true unknown parameter θ. However, we do not know which interval will contain θ and which will not contain θ, because θ is unknown.

  10. Interpretation of C.I. For instance, if we construct a random interval by drawing different sets of samples repeatedly, say 100 times, then 95% = 100(1-0.05)% C.I. for μmeans that μ is contained in 95 out of the 100 fixed intervals. Again, we do not know what these 95 intervals are, because µ is unknown.

  11. Steps to construct a confidence interval Step 1: Find a point estimator of θ Step 2: Find its EXACT (or approximate) distribution. Step 3: Based on the exact (or approximate) distribution found in Step 2 to construct the C.I. Throughout this course, we are only interested in how to construct confidence intervals of parameters µ and σ2 by the sample mean and sample variance S2. In the following, we will discuss the distributions of and S2, and then see how to obtain the confidence interval of µ and σ2 case by case.

  12. One sample Confidence Interval for µ with NORMAL population (known variance)

  13. Consider a random sample of size n, {X1, X2, …, Xn}, from a normal distribution with unknown mean µ and KNOWN variance σ2. That is, Then we have a result that the sampling distribution of the sample mean is Or equivalently, Confidence interval for µ Case I: Normal distribution with unknown mean and KNOWN variance:

  14. How to construct the interval? Define a quantity such that α zα

  15. How to construct the interval? By the symmetry of the standard normal distribution, we have Define a quantity such that

  16. How to construct the interval? z1-α/2 = -zα/2 1 - α α/2 α/2 zα/2

  17. How to construct the interval? By the symmetry of the standard normal distribution, we have θ Define a quantity such that

  18. How to construct the interval? or simply written as The margin of error After sampling, we can find an actual value of the sample mean, say . Thus, 100(1-α)% C.I for μ is that

  19. If all X1,…, Xn are observed, i.e. we have x1,…,xn , then 95% C.I for μ is that For example, if α = 0.05, then

  20. Note that μ is an unknown BUT fixed number, and and σ2 are known. Remark again that it does not mean that μ is inside this interval with a probability 0.95. So, μ is either inside or outside the fixed interval.

  21. Questions Page 12 Q1: Given a random sample of 100 observations from a normal distribution for which µ is unknown and σ = 8. Suppose that the sample mean is found to be 42.7 after sampling. Then what is the 95% C.I. for µ? Q2: A wine importer needs to report the average percentage of alcohol in bottles of French wine. From previous experience with different kinds of wine, the importer believes the alcohol concentration is normally distributed with standard deviation 1.2%. The importer randomly samples 60 bottles of the new wine and obtains a sample mean 9.3%. Find a 90% C.I. for the population average percentage.

  22. One sample Confidence Interval for µ with NORMAL population (unknown variance)

  23. Consider a random sample of size n, {X1, X2, …, Xn}, from a normal distribution with unknown mean µ and UNKNOWN variance σ2. That is, Then we have a result that the sampling distribution of the sample mean is Or equivalently, Confidence interval for µ Case II: Normal distribution with unknown mean and UNKNOWN variance:

  24. What is the sampling distribution of After sampling, we can find an actual value of the sample mean, say . Thus, 100(1-α)% C.I for μ is that However, σ is UNKNOWN. So, this interval is also unknown. Replace σ2 by the sample variance S2. However, the next problem is: Still normal? NO!

  25. Then the sampling distribution of has a Student t distribution (or simply t distribution) with n -1 degrees of freedom. Denote by where and Theorem Consider a random sample of size n, {X1, X2, …, Xn}, from a normal distribution with unknown mean µ and UNKNOWN variance σ2.

  26. tk distribution • Similar to a standard normal distribution, it is also symmetric about 0, so • P(T ≤ -a) = 1 - P(T ≤ a) = P(T ≥ a), if T follows a t distribution. • Use a table of a t distribution to find a probability of a t-distributed random variable.

  27. How to construct the interval? Define a quantity such that By the symmetry of the t distribution, we have

  28. How to construct the interval? or simply written as After sampling, we can find the actual values of the sample mean and sample variance, say and s. Thus, 100(1-α)% C.I for μ is

  29. How to use the table of t distribution

  30. for the value of α

  31. For the value of the degree of freedom

  32. 2.353 = ?

  33. t 3, 0.05 α Degree of freedom first 2.353 = ?

  34. P(-t14, 0.025≤ T14 ≤ t14, 0.005) = P(T14 ≤ t14, 0.005) – P(T14≤ -t14, 0.025) By the symmetry of t distribution Questions Page 14 Q3 (i) Find P(-t14, 0.025≤ T14 ≤ t14, 0.005) = [1 - P(T14 > t14, 0.005)] – P(T14> t14, 0.025) = [1 – 0.005] – 0.025 = 0.97

  35. 0.045 = P( k ≤ T14 ≤ - 1.761) = P(T14 ≤ - 1.761) – P(T14≤ k) Questions Page 14 Q3 (ii) Find k such that P( k ≤ T14 ≤ - 1.761) = 0.045 = P(T14≥ 1.761) – P(T14≥ - k) By the symmetry of t distribution

  36. Questions Page 14 Q3 (ii) Find k such that P( k ≤ T14 ≤ - 1.761) = 0.045 0.045 = P( k ≤ T14 ≤ - 1.761) = P(T14 ≤ - 1.761) – P(T14≤ k) = P(T14≥ 1.761) – P(T14≥ - k) By the symmetry of t distribution = P(T14≥t14, 0.05) – P(T14≥ - k) = 0.05 – P(T14≥ - k)  P(T14≥ - k) = 0.05 – 0.045 = 0.005

  37. Questions Page 14 Q3 (ii) Find k such that P( k ≤ T14 ≤ - 1.761) = 0.045 0.045 = P( k ≤ T14 ≤ - 1.761) = P(T14 ≤ - 1.761) – P(T14≤ k) = P(T14≥ 1.761) – P(T14≥ - k) By the symmetry of t distribution = P(T14≥t14, 0.05) – P(T14≥ - k) = 0.05 – P(T14≥ - k)  P(T14≥ - k) = 0.05 – 0.045 = 0.005 = P(T14≥ 2.977) k = - 2.977

  38. s2 = 56.424, n = 12, α = 0.05 Questions Page 14 Frequencies, in hertz (Hz), of 12 elephant calls: 14, 16, 17, 17, 24, 20, 32, 18, 29, 31, 15, 35 Assume that the population of possible elephant call frequencies is a normal distribution, Now a scientist is interested in the average of the frequencies, say µ. Find a 95% confidence interval for µ. Population variance is UNKNOWN So, use t distribution to construct the C.I. for µ. Finally, the 95% C.I. for µ is [17.557, 27.103]

  39. Remark: When n > 30, the difference of a t distribution with n -1 degrees of freedom and the standard normal distribution is small. So, we have Therefore, we can use to approximate the 100(1-α)% C.I for μwith unknown variance, as n > 30.

  40. Two samples Confidence Interval for µX - µY with NORMAL populations (known variances)

  41. Consider two independent random samples, and use to estimate µX - µY. Confidence interval for µX - µY Case I: Normal distributions with unknown means and KNOWN variances: Want to construct a C.I. for the mean difference µX - µY. First, choose a point estimator of the mean difference.

  42. How to construct the interval? Second, find the sampling distribution of . Indeed, we have a result that Or equivalently,

  43. How to construct the interval? or Similar to Case 1 in the one-sample case. After sampling, the 100(1-α)% C.I for μX - μY is given by

  44. then the 100(1-α)% C.I for μX - μY becomes Confidence interval for µX - µY Case I: Normal distributions with unknown means and KNOWN variances: In particular, if two variances are EQUAL, say σX2 = σY2 = σ2,

  45. Example Two kinds of thread are being compared for strength. Fifty pieces of each type of thread are tested under similar conditions. Brand A had an average tensile strength of 78.3 kilograms with a population standard deviation of 5.6 kilograms, while brand B had an average tensile strength of 87.2 kilograms with a population standard deviation of 6.3 kilograms. Construct a 95% confidence interval for the difference of the population means µA - µB.

  46. n = m = 50 α = 0.05 σX = 5.6 σY = 6.3 Example Two kinds of thread are being compared for strength. Fifty pieces of each type of thread are tested under similar conditions. Brand A had an average tensile strength of 78.3 kilograms with a population standard deviation of 5.6 kilograms, while brand B had an average tensile strength of 87.2 kilograms with a population standard deviation of 6.3 kilograms. Construct a 95% confidence interval for the difference of the population means µA - µB. Two samples Known variances

  47. Example Two kinds of thread are being compared for strength. Fifty pieces of each type of thread are tested under similar conditions. Brand A had an average tensile strength of 78.3 kilograms with a population standard deviation of 5.6 kilograms, while brand B had an average tensile strength of 87.2 kilograms with a population standard deviation of 6.3 kilograms. Construct a 95% confidence interval for the difference of the population means µA - µB. = [-11.24, -6.56]

More Related