Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Estimation and Confidence Intervals

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Define a point estimator, a point estimate, and desirable properties of a point estimator such as unbiasedness, efficiency, and consistency. Define an interval estimator and an interval estimate Define a confidence interval, confidence level, margin of error, and a confidence interval estimate Construct a confidence interval for the population mean when the population standard deviation is known When you have completed this chapter, you will be able to:

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Construct a confidence interval for the population mean when the population is normally distributed and the population standard deviation is unknown Construct a confidence interval for a population proportion Determine the sample size for attribute and variable sampling Construct a confidence interval for the population variance when the population is normally distributed

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved T erminology Point Estimate …is a single value (statistic) used to estimate a population value (parameter) Confidence Interval …is a range of values within which the population parameter is expected to occur Interval Estimate …states the range within which a population parameter probably lies

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved D esirable properties of a point estimator unbiased … possible values are concentrated close to the value of the parameter …unbiased when the expected value equals the value of the population parameter being estimated. Otherwise, it is biased ! …values are distributed evenly on both sides of the value of the parameter efficient consistent

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Standard error of the sample mean …is the standard deviation of the sampling distribution of the sample means It is computed by …is the symbol for the standard error of the sample mean …is the standard deviation of the population n …is the size of the sample T erminology  x  n    x

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Standard Error of the Means If  is not known and n > 30, the standard deviation of the sample( s ) is used to approximate the population standard deviation If  is not known and n > 30, the standard deviation of the sample( s ) is used to approximate the population standard deviation n  s x s Computed by…

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved The sample size, n 2.The variability in the population, usually estimated by s 3.The desired level of confidence 1.The sample size, n 2.The variability in the population, usually estimated by s 3.The desired level of confidence …that determine the width of a confidence interval are:

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved IN GENERAL, A confidence interval for a mean is computed by: Constructing Confidence Intervals Interpreting… z  x n s α/2

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Interpreting Confidence Intervals The Globe Suppose that you read that “…the average selling price of a family home in York Region is $ /- $15000 at 95% confidence!” This means…what?

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Interpreting Confidence Intervals In statistical terms, this means: …that we are 95% sure that the interval estimate obtained contains the value of the population mean. Lower confidence limit is $ Upper confidence limit is $ The Globe “…the average selling price of a family home in York Region is $ /- $ at 95% confidence!” Also… ( $ $15 000) ($ $15 000)

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Interpreting Confidence Intervals The Globe “…the mean time to sell a family home in York Region is 40 days. Your newspaper also reports that… You select a random sample of 36 homes sold during the past year, and determine a 90% confidence interval estimate for the population mean to be (31-39) days. Do your sample results support the paper’s claim?

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Interpreting Confidence Intervals You select a random sample of 36 homes sold during the past year, and determine a 90% confidence interval estimate for the population mean to be (31-39) days. There is a 10% chance (100%-90%) that the interval estimate does not contain the value of the population mean! Lower confidence limit is 31 days Upper confidence limit is 39 days O ur 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

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved i.e.  = % … 10% chance of falling outside this interval Interpreting Confidence Intervals 90% Confidence Interval …or, focus on tail areas …  is the probability of a value falling outside the confidence interval

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Find the appropriate value of z: Locate A rea on the normal curve 1 This is a 92% confidence interval 2 Look up a= 0.46 in Table to get the corresponding z-score Search in the centre of the table for the area of 0.46 Z = +/

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Constructing Confidence Intervals 95% C.I. for the mean: Common Confidence Intervals 99% C.I. for the mean: About 95% of the constructed intervals will contain the parameter being estimated. Also, 95% of the sample means for a specified sample size will lie within 1.96 standard deviations of the hypothesized population mean. z  x n s α/2

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Interval Estimates Use the z table… z  x n s α/2 Use the t -table… t  x n s α/2 If the population standard deviation is known or n > 30 If the population standard deviation is unknown and n<30 More on this later…

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Our best estimate is 24 hours. 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 mean of 24 hours with a standard deviation of 4 hours. What is the population mean? This is a point estimate.

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Find the 95 percent confidence interval for the population mean. 95% Confidence Z = +/ Substitute values: 24 = 24 +/ The Confidence Limits range from to Commonly denoted as 1-  95 percent confidence Mean = 24 SD = 4 N = z  x n s α/2

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved % confidence level 1-  = 0.9 or  =  = 0.99 or  = % confidence level Interval Estimates

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Student’s t-distribution ….used for small sample sizes Characteristics …like z, the t-distribution is continuous …takes values between –4 and +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 closer to the standard normal distribution

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Chart 9-1 Comparison of The Standard Normal Distribution and the Student’s t Distribution Z distribution The t distribution should be flatter and more spread out than the z distribution t distribution Student’s t-distribution

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Student’s t-distribution …with df = 9 and 0.10 area in the upper tail… t = t 0.10 T - table

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Student’s t-distribution Confidence Intervals 80% 90% 95% 98% 99% Level of Significance for One-Tailed Test Level of Significance for Two-Tailed Test

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved When? …to use the z Distribution or the t Distribution NO Population standard deviation known? Use a nonparametric test (see Ch16) Use the z distribution Use the t distribution YES Population Normal? NO YES n 30 or more? NO YES Use the z distribution

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Student’s t-distribution 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 mean of 24 hours with a standard deviation of 4 hours. Find the 95 percent confidence interval for the population mean. so use the t - Distribution n is small

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved …sample of only 12 students …a mean of 24 hours …a standard deviation of 4 hours Data X = 24 n = 12 s = 4 df = 12-1 = 11  = 1 – 95% =.05 Looking up 5% level of significance for a two-tailed test with 11df, we find… Formula t  x n s α/2

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Student’s t-distribution Confidence Intervals 80% 90% 95% 98% 99% Level of Significance for One-Tailed Test Level of Significance for Two-Tailed Test

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved …sample of only 12 students …a mean of 24 hours …a standard deviation of 4 hours Data X = 24 n = 12 s = 4 df = 12-1 = 11  = 1 – 95% =.05 Looking up 5% level of significance for a two-tailed test with 11df, we find… t = t = Compare these with earlier limits of to = 24 +/  4 The confidence limits range from to Formula t  x n s α/2

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved 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. Student’s t-distribution

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved = $4.53 +/- $1.03 $4.45 $4.05 $4.95 $3.25 $4.68 $5.75 $6.01 $3.99 $5.25 $2.95 Determine the sample mean and standard deviation. Step 1 n = 10 – 1 = % =.01 Enter the key data into the appropriate formula. Step 2 = We are 99% confident that the mean amount spent per customer is between $3.50 and $5.56 Student’s t-distribution = $4.53 s = $1.00 X  = df =10  Formula t  x n s α/2

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Constructing Confidence Intervals for Population Proportions A confidence interval for a population proportion is estimated by: …is the symbol for the sample proportion p Formula

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved 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. Constructing Confidence Intervals for Population Proportions

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved 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 = Constructing Confidence Intervals for Population Proportions Formula 98% CL = 500 )35.1(   

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Finite-Population Correction Factor Used when n/N is 0.05 or more The attendance at the college hockey game last night was 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 Formula   x n N -n  1 Develop a 90% confidence interval estimate for the mean number of drinks consumed per person. Correction Factor

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved N = n = x = s =  /2 = N = n = x = s =  /2 = Finite-Population Correction Factor The attendance at the college hockey game last night was 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 Develop a 90% confidence interval estimate.… Since 250/2700 >.05, use the correction factor ) )( (     90% CL = Formula N -n 1 n s Z α/2 X 

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Selecting the Sample Size

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved The degree of confidence selected 2.The maximum allowable error 3.The variation in the population 1.The degree of confidence selected 2.The maximum allowable error 3.The variation in the population …that determine the sample size are: Factors

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved 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 Selecting the Sample Size  2      E sz α/2 n = Formula

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved 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 s tandard d eviation is estimated to be $ Selecting the Sample Size How large a sample is required?

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Selecting the Sample Size 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 $ = (10.32) 2 = A minimum of 107 homes must be sampled. 90% CL =         Formula  2      E sz α/2

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Selecting the Sample Size The Kennel Club wants to estimate the proportion of children that have a dog as a pet. Assume a 95% level of confidence and that the club estimates that 30% of the children have a dog as a pet. If the club wants the estimate to be within 3% of the population proportion, how many children would they need to contact?

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved The Kennel Club wants to estimate the proportion of children that have a dog as a pet. Assume a 95% level of confidence and that the club estimates that 30% of the children have a dog as a pet. Selecting the Sample Size New Formula npp Z E        () )3.1(3.          )21(.  n = A minimum of 897 children must be sampled.

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved Test your learning … Click on… Online Learning Centre for quizzes extra content data sets searchable glossary access to Statistics Canada’s E-Stat data …and much more!

Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved This completes Chapter 9