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Estimation and Confidence Intervals

This chapter explains how to define and construct point and interval estimators, as well as confidence intervals. It also covers desirable properties of estimators and interpreting confidence intervals.

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Estimation and Confidence Intervals

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  1. Estimation and Confidence Intervals Chapter 9

  2. 1 2 3 4 Chapter Goals When you have completed this chapter, you will be able to: Definea point estimator, a point estimate, and desirable properties of a point estimator such asunbiasedness, efficiency, and consistency. Definean interval estimator and an interval estimate Definea confidence interval, confidence level, margin of error, and a confidence interval estimate Constructa confidence interval for the population mean when the population standard deviation is known and...

  3. 5 6 7 8 Chapter Goals Constructa confidence interval for the population variance when the population is normally distributed Constructa confidence interval for the population mean when the population is normally distributed and the population standard deviation is unknown Constructa confidence interval for a population proportion Determinethe sample size for attribute and variable sampling

  4. Terminology Point Estimate …is a single value (statistic) used to estimate a population value (parameter) …states the range within which a population parameter probably lies Interval Estimate Confidence Interval …is a range of values within which the population parameter is expected to occur

  5. Desirable properties of a point estimator • efficient … possible values are concentrated close to the value of the parameter • consistent …values are distributed evenly on both sides of the value of the parameter • unbiased …unbiasedwhen the expected value equals the value of the population parameter being estimated. Otherwise, it is biased!

  6. Terminology s = n s s x x Standard error of the sample mean …isthe standard deviation of the sampling distribution of the sample means It is computed by where …is the symbol for the standard error of the sample mean s …is the standard deviation of the population n …is the size of the sample

  7. s s x = n Standard Error of the Means If is not known and n > 30, the standard deviation of the sample(s)is used toapproximate the population standard deviation Computed by…

  8. 1. The sample size, n 2. The variability in the population, usually estimated by s 3. The desired level of confidence Factors …that determine the width of a confidence interval are:

  9. s n z ± x α/2 Constructing Confidence Intervals IN GENERAL, A confidence interval for a mean is computed by: Interpreting…

  10. The Globe Interpreting Confidence Intervals Suppose that you read that “…the average selling price of a family home in York Region is $200 000+/- $15000 at 95% confidence!” This means…what?

  11. The Globe “…the average selling price of a family home in York Region is $200 000+/- $15 000 at 95% confidence!” Interpreting Confidence Intervals In statistical terms, this means: …that we are 95% surethat the interval estimateobtained contains the value of the population mean. Lower confidence limit is $185 000 ($200 000 - $15 000) Upperconfidence limit is $215 000 ($200 000 + $15 000) Also…

  12. The Globe Interpreting Confidence Intervals Your newspaper also reports that… You select a random sample of 36homes sold during the past year, and determine a 90% confidence intervalestimate for the population mean to be (31-39) days. “…the mean time to sell a family home in York Region is 40 days. Do your sample results support the paper’s claim?

  13. Interpreting Confidence Intervals Lower confidence limit is 31 days You select a random sample of 36homes sold during the past year, and determine a 90% confidence intervalestimate for the population mean to be (31-39) days. Upperconfidence limit is 39 days Our evidence does not support the statement made by the newspaper, i.e., the population mean is not 40 days, when using a 90% interval estimate There is a 10% chance (100%-90%) that the interval estimate does not contain the value of the population mean!

  14. Interpreting Confidence Intervals 90% Confidence Interval … 10% chance of falling outside this interval …or, focus on tail areas … i.e.  = 0.10 .05 .05 90% 39 31  is the probability of a value falling outsidethe confidence interval

  15. Try it! LocateArea on the normal curve 1 2 Look up a=0.46 in Table to get the corresponding z-score This is a 92% confidence interval Find the appropriate value of z: 0.92 0 -1.75 1.75 Search in the centre of the table for the areaof 0.46 Z = +/- 1.75

  16. Constructing Confidence Intervals s n z ± x α/2 Common Confidence Intervals Also, 95% of the sample means for a specifiedsample size will lie within 1.96standard deviations of the hypothesized population mean. About 95% of the constructed intervals will contain the parameter being estimated. 95% C.I. for the mean: 99% C.I. for the mean:

  17. Interval Estimates s s n n Use the t-table… z t ± ± x x α/2 α/2 Use the z table… If the population standard deviation is knownor n > 30 If the population standard deviation is unknown and n<30 More on this later…

  18. Q uestion A nswer The Dean of the Business School wants to estimate the mean number of hours worked per week by students. A sample of 49 students showed a meanof 24 hours with a standard deviation of 4 hours. What is the population mean? Our best estimate is 24 hours. This is a point estimate.

  19. 95 percent confidence Commonly denoted as 1- s z ± α/2 x n 4 S 49 olution Find the 95 percent confidenceinterval for the population mean. Q Mean = 24 SD = 4 N = 49 Z = +/- 1.96 95% Confidence Substitute values: +1.96 24 = 24 +/- 1.12 The Confidence Limitsrangefrom22.88 to 25.12

  20. Interval Estimates 90% confidence level 1- = 0.9 or  = 0.10 99% confidence level 1- = 0.99 or  = 0.010

  21. Student’s t-distribution ….used for small sample sizes Characteristics …like z, the t-distribution is continuous …takes values between –4and +4 …it is bell-shaped and symmetric about zero …it is more spread out and flatter at the centre than the z-distribution …for larger and larger values of degrees of freedom, the t-distribution becomes closer and closerto the standard normal distribution

  22. tdistribution Zdistribution Student’s t-distribution Chart 9-1 Comparison of The Standard Normal Distribution and the Student’stDistribution The t distribution should be flatter and more spread out than thezdistribution

  23. 0.10 t T -table Student’s t-distribution Example …with df = 9 and 0.10 area in the upper tail… t = 1.383

  24. 1.383 Student’s t-distribution Confidence Intervals 80% 90% 95% 98% 99% Level of Significance for One-Tailed Test 0.100 0.050 0.025 0.010 0.005 0.10 Level of Significance for Two-Tailed Test 0.20 0.10 0.05 0.02 0.01 9

  25. When? …to use the zDistribution or the tDistribution ? Population Normal? NO YES Population standard deviation known? n30 ormore? NO YES NO YES Use a nonparametric test (see Ch16) Use thetdistribution Use thez distribution Use thez distribution

  26. Q The Dean of the Business School wants to estimate the mean number of hours worked per week by students. A sample of only 12 students showed a meanof 24 hours with a standard deviation of 4 hours. Student’s t-distribution Find the 95 percent confidenceintervalfor the population mean. n is small so use the t- Distribution

  27. Q Data Formula = 24 n = 12 s = 4 df = 12-1 = 11  = 1 – 95% =.05 X S olution s t ± α/2 x n …sample of only 12 students …a meanof 24 hours …a standard deviation of 4 hours Looking up 5% level of significance for a two-tailed test with 11df, we find…

  28. Student’s t-distribution Confidence Intervals 80% 90% 95% 98% 99% Level of Significance for One-Tailed Test 0.100 0.050 0.025 0.010 0.005 Level of Significance for Two-Tailed Test 0.05 0.20 0.10 0.05 0.02 0.01 11 2.201

  29. Q Data Formula = 24 n = 12 s = 4 df = 12-1 = 11  = 1 – 95% =.05 X S olution s t 4 ± α/2 x n 24 ± 2.201 12 …sample of only 12 students …a meanof 24 hours …a standard deviation of 4 hours Looking up 5% level of significance for a two-tailed test with 11df, we find… t0.025 = 2.201 = 24 +/- 2.54 The confidence limits range from 21.46 to 26.54 Compare these with earlier limits of 22.88 to 25.12

  30. Student’s t-distribution Q The manager of the college cafeteria wants to estimate the mean amount spent per customer per purchase. A sample of 10 customers revealed the following amounts spent: $4.45 $4.05 $4.95 $3.25 $4.68 $5.75 $6.01 $3.99 $5.25 $2.95 Determine the 99% confidence interval for the mean amount spent.

  31. = $4.53 s = $1.00 X 1.00 10 S olution s t ± α/2 x n Student’s t-distribution $4.45 $4.05 $4.95 $3.25 $4.68 $5.75 $6.01 $3.99 $5.25 $2.95 Step 1 Determine the sample mean and standard deviation. Step 2 Enter the key datainto the appropriate formula. 10 df = 10 – 1 = 9  = 1-99% = .01 n = ± = 4.53 3.25 Formula = $4.53 +/- $1.03 We are 99%confident that the mean amount spent per customer is between $3.50 and $5.56

  32. Constructing Confidence Intervals for Population Proportions Formula A confidence interval for a population proportion is estimated by: where p …is the symbolfor the sample proportion Q ...

  33. Q Constructing Confidence Intervals for Population Proportions uestion A sample of 500 executives who own their own home revealed 175 planned to sell their homes and retire to Victoria. Develop a 98%confidence interval for the proportion of executives that plan to sell and move to Victoria.

  34. Constructing Confidence Intervals for Population Proportions Formula - . 35 ( 1 . 35 ) ± . 35 2 . 33 500 S olution ± . 35 . 0497 A sample of 500 executives who own their own home revealed 175 planned to sell their homes and retire to Victoria. Develop a 98%confidence interval for the proportion of executives… n = p = z = 500 2.33 175/500 = .35 98% CL =

  35. Q uestion Correction Factor s N - n Formula s = x N - 1 n Finite-Population Correction Factor Used when n/Nis 0.05 or more The attendance at the college hockey game last night was 2700. A random sample of 250 of those in attendance revealed that the average number of drinks consumed per person was 1.8 with a standard deviation of 0.40. Developa 90%confidence interval estimate for the mean number of drinks consumed per person.

  36. - . 4 2700 250 ± 1 . 8 1 . 645 ( )( ) s N - n ± Formula Zα/2 - X 2700 1 250 n N - 1 N = n = x = s = /2 = S olution ± 1 . 8 0 . 04 Finite-Population Correction Factor The attendance at the college hockey game last night was 2700. A sample of 250 of those in attendance revealed that the average number of drinks consumed per person was 1.8 with a standard deviation of 0.40. Developa 90%confidence intervalestimate.… 2700 250 1.8 0.05 0.40 Since 250/2700>.05,use the correction factor 90% CL =

  37. Selecting the Sample Size

  38. Factors …that determine the sample size are: 1. The degree of confidence selected 2. The maximum allowable error 3. The variation in the population

  39. Selecting the Sample Size 2 zα/2 s æ ö ç ÷ E è ø Formula n = where E… is the allowable error Z …is the z-score for the chosen level of confidence S …is the sample deviation of the pilot survey

  40. Selecting the Sample Size Q uestion A consumer group would like to estimate the mean monthly electricity charge for a single family house in July (within $5)using a 99 percent level of confidence. Based on similar studies the standard deviation is estimated to be $20.00. How large a sample is required?

  41. Selecting the Sample Size 2 zα/2 s Formula æ ö ç ÷ E è ø 2 æ æ 2.58 · 20 = ç ç 5.00 è è S olution A minimum of 107 homes must be sampled. 90% CL = A consumer group would like to estimate the mean monthly electricity charge for a single family house in July (within $5)using a 99 percent level of confidence. Based on similar studies the standard deviation is estimated to be $20.00. = (10.32)2 = 106.5

  42. Selecting the Sample Size Q uestion The Kennel Club wants to estimate theproportion of children that have a dog as a pet. Assume a 95% level ofconfidence and that the club estimates that 30% of the children have a dog as a pet. If the club wants the estimate to bewithin 3% of the population proportion, how many children would they need to contact?

  43. Selecting the Sample Size Q uestion New Formula 2 æ ö Z = - ç ÷ n p ( 1 p ) è ø E 2 æ ö 1 . 96 = - ç ÷ . 3 ( 1 . 3 ) è . 03 ø 2 = ) ( (. 21 ) 65 . 33 The Kennel Club wants to estimate theproportion of children that have a dog as a pet. Assume a 95% level ofconfidence and that the club estimates that 30% of the children have a dog as a pet. n= 896.4 A minimum of 897 children must be sampled.

  44. www.mcgrawhill.ca/college/lind for quizzes extra content data sets searchable glossary access to Statistics Canada’s E-Stat data …and much more! Test your learning… Click on… Online Learning Centre

  45. This completes Chapter 9

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