Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

no text concepts found

Transcript

Confidence Interval Assignment # 2 Briefly describe following with the help of examples: Point Estimation of Parameters Confidence Intervals Students’ t Distribution Chi Square Distribution Testing of Hypothesis Hand Written assignments be submitted by 20 Dec 2012. Copying and Late Submissions will have appropriate penalty. Estimation Process Population Mean, , is unknown Sample Random Sample Mean X = 50 I am 95% confident that is between 40 & 60. Confidence Interval provides more information about a population characteristic than does a point estimate. It provides a confidence level for the estimate. Lower Confidence Limit Upper Point Estimate Width of Confidence Interval Confidence Limit Interval Estimates Provides range of values Takes into consideration variation in sample statistics from sample to sample Is based on observation from one sample Gives information about closeness to unknown population parameters Is stated in terms of level of confidence Never 100% certain Confidence Interval Estimates Confidence Intervals Variance Mean 2 Known 2Unknown Confidence Interval for µ ( б2 Known) Assumptions Population standard deviation is known Population is normally distributed If population is not normal, use large sample Confidence interval estimate c Conf ( x k x k )..where..k n Steps to Determine Confidence Interval Choose a Confidence Level γ (95%, 90%, etc) Determine the corresponding c [ Z(D) ] γ 0.90 0.95 0.99 0.999 C 1.645 1.960 2.576 3.291 Compute the mean x of the sample x1, x2,…, xn Confidence interval estimate c Conf ( x k x k )..where..k n Problem 1 Find a 95% Confidence Interval for the mean µ of a normal population with standard deviation 4.00 from the sample 30, 42, 40, 34, 48, 50 Confidence Interval for µ ( б2 Unknown) Choose a Confidence Level γ (95%, 90%, etc) Determine the solution C of the equation F(C) = ½ ( 1 + γ) [Table: t distribution with n-1 degrees of freedom] Compute the mean x and variance s2 of the sample x1, x2,…, xn Confidence interval estimate cs Conf ( x k x k )..where..k n Student’s t Distribution Standard Normal Bell-Shaped Symmetric ‘Fatter’ Tails t (df = 13) t (df = 5) 0 Z t Degrees of Freedom (df ) Number of observations that are free to vary after sample mean has been calculated df = n - 1 Example: Mean of 3 numbers is 2 degrees of freedom = n -1 = 3 -1 =2 Student’s t Distribution Let X1, X2,…, Xn be independent random variables with the same mean µ and the same variance б2. Then, the random variable x T S/ n has a t distribution with n-1 degrees of freedom n 1 2 2 S (X j X ) n 1 j 1 Student’s t Table Upper Tail Area df .25 .10 .05 Let: n = 3 df = n - 1 = 2 = .10 /2 =.05 1 1.000 3.078 6.314 2 0.817 1.886 2.920 / 2 = .05 3 0.765 1.638 2.353 t Values 0 2.920 t Problem 1 Find a 95% Confidence Interval for the mean µ of a normal population with standard deviation 4.00 from the sample 30, 42, 40, 34, 48, 50 Problem 9 Find a 99% Confidence Interval for the mean of a normal population. The length of 20 bolts with Sample Mean 20.2 cm and Sample Variance 0.04 cm2. Confidence Interval for the Variance Choose a Confidence Level γ (95%, 90%, etc) Determine the solutions C1 and C2 of the equation F(C1) = ½ ( 1 - γ) and F(C2) = ½ ( 1 + γ) [Table A10: Chi Square distribution with n-1 degrees of freedom] Compute (n-1)S2, where S2 is the variance of the sample x1, x2,…, xn Confidence Interval for the Variance Compute k1 = (n-1)S2 / C1 k2 = (n-1)S2 / C2 Confidence Interval Estimate Conf (k2 k1 ) 2 Chi Square Distribution Let X1, X2,…, Xn be independent random variables with the same mean µ and the same variance б2. Then, the random variable (n 1) With S 2 2 n 1 2 2 S (X j X ) n 1 j 1 has a Chi Square distribution with n-1 degrees of freedom Chi Square Distribution CDF PDF Problem 15 Find a 95% Confidence Interval for the variance of a normal population. The Sample has variance 0.0007. 30 values with Problem 2 Does the Interval in Problem 1 get longer or shorter, if we take γ = 0.99 instead of 0.95? By what factor? Problem 3 By what factor does the length of the Interval in Problem 1 change, if we double the Sample Size? Problem 5 What Sample Size would be needed for obtaining a 95% Confidence Interval (3) of length 2 б? Of length б? Problem 7 What Sample Size is needed to obtain a 99% Confidence Interval of length 2.0 for the mean of a normal population with variance 25? Problem 9 Find a 99% Confidence Interval for the mean of a normal population. The length of 20 bolts with Sample Mean 20.2 cm and Sample Variance 0.04 cm2. Problem 11 Find a 99% Confidence Interval for the mean of a normal population. The Copper Content (%) of brass is 66, 66, 65, 64, 66, 67, 64, 65, 63, 64 Problem 13 Find a 95% Confidence Interval for the percentage of cars on a certain highway that have poorly adjusted brakes, using a random sample of 500 cars stopped at a road block on a highway, 87 of which had poorly adjusted brakes. Problem 17 Find a 95% Confidence Interval for the variance of a normal population. The Sample is the Copper Content (%) of brass: 66, 66, 65, 64, 66, 67, 64, 65, 63, 64 Problem 19 Find a 95% Confidence Interval for the variance of a normal population. The Sample has Mean Energy (keV) of delayed neutron group (Group 3, half life 6.2 sec) for uranium U235 fission: 435, 451, 430, 444, 438 Problem 21 If X is normal with mean 27 and variance 16, what distribution do –X, 3X and 5X-2 have? Problem 23 A machine fills boxes weighing Y lb with X lb of salt, where X and Y are normal with mean 100 lb and 5 lb and standard deviation 1 lb and 0.5 lb, respectively. What percent of filled boxes weighing between 104 lb and 106 lb are to be expected? Confidence Interval for Variance Choose a Confidence Level γ (95%, 90%, etc) Determine the solution C1 and C2 of the equation F(C1) = ½ ( 1 - γ) and F(C2) = ½ ( 1 + γ) [Table: Chi Square distribution with n-1 degrees of freedom] Compute (n-1) s2, where s2 is variance of the sample x1, x2,…, xn Confidence Interval for Variance Compute k1 = (n-1) s2/C1 and k2 = (n-1) s2/C2 Confidence interval estimate Conf (k2 k1 ) 2 Elements of Confidence Interval Estimation Level of confidence Precision (range) Confidence in which the interval will contain the unknown population parameter Closeness to the unknown parameter Cost Cost required to obtain a sample of size n Level of Confidence Denoted by 100 1 % A relative frequency interpretation In the long run, 100 1 % of all the confidence intervals that can be constructed will contain the unknown parameter A specific interval will either contain or not contain the parameter No probability involved in a specific interval Interval and Level of Confidence Sampling Distribution of the _ Mean Z / 2 X Intervals extend from X Z X to X Z X /2 X 1 X Z / 2 X /2 X 1 100% of intervals constructed contain ; 100 % do not. Confidence IntervalsChap 7-Confidence Intervals-37 Factors Affecting Interval Width (Precision) Data variation Sample size Measured by X Intervals Extend from X - Z x to X + Z x n Level of confidence 100 1 % © 1984-1994 T/Maker Co. Chap 7-Confidence Intervals-38 Determining Sample Size (Cost) Too Big: Too small: • Requires too many resources • Won’t do the job Chap 7-Confidence Intervals-39 Determining Sample Size for Mean What sample size is needed to be 90% confident of being correct within ± 5? A pilot study suggested that the standard deviation is 45. 1.645 45 Z n 2 2 Error 5 2 2 2 2 219.2 220 Round Up Chap 7-Confidence Intervals-40 Determining Sample Size for Mean in PHStat PHStat | sample size | determination for the mean … Example in excel spreadsheet Chap 7-Confidence Intervals-41 Confidence Interval for ( Unknown) Assumptions Population standard deviation is unknown Population is normally distributed If population is not normal, use large sample Use student’s t distribution Confidence interval estimate X t / 2,n 1 S S X t / 2,n 1 n n Chap 7-Confidence Intervals-42 Example A random sample of n 25 has X 50 and S 8. Set up a 95% confidence interval estimate for S S X t / 2,n 1 X t / 2,n 1 n n 8 8 50 2.0639 50 2.0639 25 25 46.69 53.30 Chap 7-Confidence Intervals-43 Confidence Interval Estimate for Proportion p Assumptions Two outcomes (0;1) Number of ‘1’ in n trials follows B(n,p) Normal approximation can be used if np 5 and n 1 p 5 Confidence interval estimate pS Z / 2 pS 1 pS p pS Z / 2 n pS 1 pS n Chap 7-Confidence Intervals-44 Example A random sample of 400 voters showed 32 preferred candidate A. Set up a 95% confidence interval estimate for p. ps Z / 2 ps 1 ps p ps Z / 2 n ps 1 ps n .08 1 .08 .08 1 .08 .08 1.96 p .08 1.96 400 400 .053 p .107 Normal Table Chap 7-Confidence Intervals-45 Confidence Interval Estimate for Proportion in PHStat PHStat | confidence interval | estimate for the proportion … Example in excel spreadsheet Chap 7-Confidence Intervals-46 Determining Sample Size for Proportion Out of a population of 1,000, we randomly selected 100, of which 30 were defective. What sample size is needed to be within ± 5% with 90% confidence? Z p 1 p 1.645 0.3 0.7 n 2 2 Error 0.05 227.3 228 2 2 Round Up Chap 7-Confidence Intervals-47 Determining Sample Size for Proportion in PHStat PHStat | sample size | determination for the proportion … Example in excel spreadsheet Chap 7-Confidence Intervals-48 Confidence Interval for Population Total Amount Point estimate NX Confidence interval estimate NX N t / 2,n1 S n N n N 1 Chap 7-Confidence Intervals-49 Confidence Interval for Population Total: Example An auditor is faced with a population of 1000 vouchers and wants to estimate the total value of the population. A sample of 50 vouchers is selected with average voucher amount of $1076.39, standard deviation of $273.62. Set up the 95% confidence interval estimate of the total amount for the population of vouchers. Chap 7-Confidence Intervals-50 Example Solution N 1000 n 50 NX N t / 2,n 1 S n X $1076.39 S $273.62 N n N 1 273.62 1000 50 1000 1076.39 1000 2.0096 1000 1 100 1, 076,390 75,830.85 The 95% confidence interval for the population total amount of the vouchers is between 1,000,559.15, and 1,152,220.85 Chap 7-Confidence Intervals-51 Confidence Interval for Total Difference in the Population Point estimate n Where D difference ND D i 1 i is the sample average n Confidence interval estimate SD ND N t / 2,n1 n n Where SD N n N 1 D D i 1 2 i n 1 Chap 7-Confidence Intervals-52 Estimation for Finite Population Samples are selected without replacement Confidence interval for the mean ( unknown) X t / 2,n1 S n N n N 1 Confidence interval for proportion pS Z / 2 pS 1 pS n N n N 1 Chap 7-Confidence Intervals-53 Sample Size Determination for Finite Population Samples are selected without replacement When estimating the mean Z / 2 n0 2 e 2 2 When estimating the proportion Z / 2 p 1 p n0 2 e 2 Chap 7-Confidence Intervals-54 Ethical Considerations Report confidence interval (reflect sampling error) along with the point estimate Report the level of confidence Report the sample size Provide an interpretation of the confidence interval estimate Chap 7-Confidence Intervals-55 Chapter Summary Illustrated estimation process Discussed point estimates Addressed interval estimates Discussed confidence interval estimation for the mean ( known) Addressed determining sample size Discussed confidence interval estimation for the mean ( unknown) Chap 7-Confidence Intervals-56 Chapter Summary (continued) Discussed confidence interval estimation for the proportion Addressed confidence interval estimation for population total Discussed confidence interval estimation for total difference in the population Addressed estimation and sample size determination for finite population Addressed confidence interval estimation and ethical issues Chap 7-Confidence Intervals-57

Related documents