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© 2012 McGraw-Hill Ryerson Limited © 2009 McGraw-Hill Ryerson Limited 1 Lind Marchal Wathen Waite © 2012 McGraw-Hill Ryerson Limited 2 Learning Objectives LO 1 Define a point estimate. LO 2 Define level of confidence. LO 3 Construct a confidence interval for the population mean when the population standard deviation is known. LO 4 Construct a confidence interval for a population mean when the population standard deviation is unknown. LO 5 Construct a confidence interval for a population proportion. LO 6 Determine the sample size for attribute and variable sampling. © 2012 McGraw-Hill Ryerson Limited 3 LO 1 Point Estimates © 2012 McGraw-Hill Ryerson Limited 4 Point and Interval Estimates A point estimate is the statistic (single value), computed from sample information, that is used to estimate the population parameter. A confidence interval estimate is a range of values constructed from sample data so that the population parameter is likely to occur within that range at a specified probability. The specified probability is called the level of confidence. LO 1 © 2012 McGraw-Hill Ryerson Limited 5 Types of Estimates Estimates of the population mean: The population standard deviation (σ) is known. The population standard deviation is unknown. – In this case, we substitute the sample standard deviation (s) for the population standard deviation (σ) Estimate of the population proportion. LO 1 © 2012 McGraw-Hill Ryerson Limited 6 LO 2 Level of Confidence © 2012 McGraw-Hill Ryerson Limited 7 Point Estimates and Confidence Intervals The sample mean, x is a point estimate of the population mean, μ. p ,a sample proportion, is a point estimate of p, the population proportion. s, the sample standard deviation, is a point estimate of σ, the population standard deviation. LO 2 © 2012 McGraw-Hill Ryerson Limited 8 Point Estimates and Confidence Intervals z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 : : : : : : : : : : : 1.7 : : 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 : : 1.8 : : 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 : : 1.9 : : 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 : : 2.0 : : 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 : : 2.1 : : 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 : : 2.2 : : 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 : : : : : : : : : : : : : The area between z = –1.96 and z = +1.96 is 0.95. LO 3 © 2012 McGraw-Hill Ryerson Limited 9 LO 3 Construct a confidence interval for the population mean when the population standard deviation is known. © 2012 McGraw-Hill Ryerson Limited 10 Confidence Levels and z-values LO 3 Confidence Level Nearest Probability z-Value 80 percent 0.3997 1.28 94 percent 0.4699 1.88 96 percent 0.4798 2.05 © 2012 McGraw-Hill Ryerson Limited 11 Factors Affecting Confidence Interval Estimates The factors that determine the width of a confidence interval are: 1. the sample size, n 2. the variability in the population σ, usually estimated by s 3. the desired level of confidence LO 3 © 2012 McGraw-Hill Ryerson Limited 12 Factors Affecting Confidence Interval Estimates If the population standard deviation is known or the sample is 30 or more, we use the z distribution. x z n LO 3 Sample mean z-value for a particular confidence level the population standard deviation the number of observations in the sample © 2012 McGraw-Hill Ryerson Limited 13 Example – Population Standard Deviation (σ) Known A survey company wants to determine the mean income of middle level employees in the retail industry. A random sample of 361 employees reveals a sample mean of $54 520. The standard deviation of this population is $3060. The company would like answers to the following questions: 1.What is the population mean? What is a reasonable value to use as an estimate of the population mean? 2.What is a reasonable range of values for the population mean? 3.What do these results mean? LO 3 © 2012 McGraw-Hill Ryerson Limited 14 Solution – Population Standard Deviation (σ) Known 1. In this case, we do not know the population mean. We do know the sample mean is $54 520. Hence, our best estimate of the unknown population value is the corresponding sample statistic. Thus, the sample mean of $54 520 is a point estimate of the unknown population mean. 2. Suppose the association decides to use the 95 percent level of confidence: xz n $54520 1.96 $3060 $54520 316 361 The confidence limits are $54 836 and $54 204. The margin of error is ±$316. LO 3 © 2012 McGraw-Hill Ryerson Limited 15 Solution – Population Standard Deviation (σ) Known Continued 3. If we select many samples of 361 employees, and for each sample we compute the mean and then construct a 95% confidence interval, we could expect about 95% of these confidence intervals to contain the population mean. Conversely, about 5% of the intervals would not contain the population mean annual income, µ. LO 3 © 2012 McGraw-Hill Ryerson Limited 16 Confidence Intervals in Excel LO 3 © 2012 McGraw-Hill Ryerson Limited 17 You Try It Out! The mean daily sales are $3000 for a sample of 40 days at a convenience store. The standard deviation of the population is $400. a) What is the estimated mean daily sales of the population? What is this estimate called? b) What is the 99 percent confidence interval? c) Interpret your findings. LO 3 © 2012 McGraw-Hill Ryerson Limited 18 LO 4 Construct a Confidence Interval for the Population Mean When the Population Standard Deviation is Unknown. © 2012 McGraw-Hill Ryerson Limited 19 Population Standard Deviation (σ) Unknown In most sampling situations the population standard deviation (σ) is not known. We can use the sample standard deviation to estimate the population standard deviation. But in doing so, we can no longer use formula (81), and because we do not know σ, we cannot use the z distribution. However, there is a remedy: We use the sample standard deviation and replace the z distribution with the t distribution. LO 4 © 2012 McGraw-Hill Ryerson Limited 20 Characteristics of the t Distribution 1. It is, like the z distribution, a continuous distribution. 2. It is, like the z distribution, bell-shaped and symmetrical. 3. There is not one t distribution, but rather a “family” of t distributions. All t distributions have a mean of 0, but their standard deviations differ according to the sample size, n. The standard deviation for a t distribution with 5 observations is larger than for a t distribution with 20 observations. LO 4 © 2012 McGraw-Hill Ryerson Limited 21 Characteristics of the t Distribution 4. The t distribution is more spread out and flatter at the centre than is the standard normal distribution. As the sample size increases, however, the t distribution approaches the standard normal distribution, because the errors in using s to estimate σ decrease with larger samples. LO 4 © 2012 McGraw-Hill Ryerson Limited 22 Characteristics of the t Distribution The Standard Normal Distribution and Student’s t Distribution LO 4 © 2012 McGraw-Hill Ryerson Limited 23 Characteristics of the t distribution Values of z and t for the 95 Percent Level of Confidence LO 4 © 2012 McGraw-Hill Ryerson Limited 24 Confidence Interval for Population Standard Deviation (σ) Unknown To develop a confidence interval for the population mean using the t distribution, we adjust the formula as follows: LO 4 © 2012 McGraw-Hill Ryerson Limited 25 Confidence Interval for Population Standard Deviation (σ) Unknown To develop a confidence interval for the population mean with an unknown population standard deviation: 1.Assume the sampled population is either normal or approximately normal. 2.Estimate the population standard deviation (σ) with the sample standard deviation (s). 3.Use the t distribution rather than the z distribution. LO 4 © 2012 McGraw-Hill Ryerson Limited 26 Confidence Interval for Population Standard Deviation (σ) Unknown Assume the population is normal. Is the population standard deviation known? No Use the t distribution. Yes Use the z distribution. Determining When to Use the z Distribution or the t Distribution LO 4 © 2012 McGraw-Hill Ryerson Limited 27 Example – t distribution A light bulb manufacturer wishes to investigate the life of its bulbs. A sample of 10 bulbs in use since 60 days revealed a sample mean of 0.71 days of life remaining with a standard deviation of 0.13 days. Construct a 95% confidence interval for the population mean. Would it be reasonable for the manufacturer to conclude that after 60 days the population mean amount of life remaining is 0.70 days? LO 4 © 2012 McGraw-Hill Ryerson Limited 28 Solution – t distribution Given in the problem: n = 10 = x0.71 s = 0.13 Compute the confidence interval using the t distribution, since σ is unknown. X t LO 4 s n © 2012 McGraw-Hill Ryerson Limited 29 Solution – t distribution Df 1 2 3 4 5 6 7 8 9 10 LO 4 Confidence Intervals 80% 90% 95% 98% Level of significance for One-Tailed Test 0.100 0.050 0.025 0.010 Level of Significance for Two-Tailed Test 0.20 0.10 0.05 0.02 3.078 6.314 12.706 31.821 1.886 2.920 4.303 6.965 1.638 2.353 3.182 4.541 1.533 2.132 2.776 3.747 1.476 2.015 2.571 3.365 1.440 1.943 2.447 3.143 1.415 1.895 2.365 2.998 1.397 1.860 2.306 2.896 1.383 1.833 2.262 2.821 1.372 1.812 2.228 2.764 © 2012 McGraw-Hill Ryerson Limited Continued 99% 0.005 0.01 63.657 9.925 5.841 4.604 4.032 3.707 3.499 3.355 3.250 3.169 30 Solution – t distribution Continued To determine the confidence interval we substitute the values in formula: s 0.13 X t 0.71 2.262 0.71 0.093 n 10 The endpoints of the confidence interval are 0.613 and 0.803. It is reasonable to conclude that the population mean is in this interval. The manufacturer can be reasonably sure (95 percent confident) that the mean remaining life is between 0.613 and 0.803 days. Because the value of 0.70 is in this interval, it is possible that the mean of the population is 0.70. LO 4 © 2012 McGraw-Hill Ryerson Limited 31 t distribution In Excel © 2012 McGraw-Hill Ryerson Limited 32 Example – t distribution The manager of the Inlet Square Mall wants to estimate the mean amount spent per shopping visit by customers. A sample of 20 customers reveals the following amounts spent in dollars. $48.16 $42.22 $46.82 $51.45 $23.78 $41.86 $54.86 $37.92 $52.64 48.59 50.82 46.94 61.83 61.69 49.17 61.46 52.68 52.68 58.84 43.88 What is the best estimate of the population mean? Determine a 95 percent confidence interval. Interpret the result. Would it be reasonable to conclude that the population mean is $50? What about $60? LO 4 © 2012 McGraw-Hill Ryerson Limited 33 Solution – Confidence Interval Estimates for the Mean Using Excel LO 4 © 2012 McGraw-Hill Ryerson Limited 34 Solution – Confidence Interval Estimates for the Mean Using Excel LO 4 © 2012 McGraw-Hill Ryerson Limited 35 You Try It Out! Ms. Kleman is concerned about absenteeism among her students. The information below reports the number of days absent for a sample of 10 students during the last twoweek exam period. 3 2 1 3 3 5 4 0 1 4 a) b) c) d) e) LO 4 Determine the mean and the standard deviation of the sample. What is the population mean? What is the best estimate of that value? Develop a 95 percent confidence interval for the population mean. Explain why the t distribution is used as a part of the confidence interval. Is it reasonable to conclude that the typical student does not miss any days during an exam period? © 2012 McGraw-Hill Ryerson Limited 36 LO 5 Confidence Interval for a Proportion © 2012 McGraw-Hill Ryerson Limited 37 Confidence Interval for a Proportion A proportion is the fraction, ratio, or percent indicating the part of the sample or the population having a particular trait of interest. 1. All binomial conditions are met. 2. The values np and n(1 – p) should both be greater than or equal to 5. LO 5 © 2012 McGraw-Hill Ryerson Limited 38 Confidence Interval for a Proportion To develop a confidence interval for a population proportion, we change formula (8-1) to: For sample data: We can then construct a confidence interval for a population proportion from the following formula. LO 5 © 2012 McGraw-Hill Ryerson Limited 39 Example – Confidence Interval for a Proportion A company decided to take poll on whether employees should have dress code. Employees will have dress code if at least three fourths of employees vote in favour of it. A random sample of 3000 employees reveals 2600 plan to vote for dress code. a) What is the estimate of the population proportion? b) Develop a 95% confidence interval for the population proportion. c) Basing your decision on this sample information, can you conclude that the necessary proportion of employees favours the dress code? LO 5 © 2012 McGraw-Hill Ryerson Limited 40 Solution – Confidence Interval for a Proportion First compute the sample proportion: X 2,600 p 0.87 n 3,000 Compute the 95% C.I. C.I. p z p (1 p ) n 0.87 1.96 .87(1 .87) .87 .012 3,000 (0.858, 0.882) We conclude that the dress code proposal will pass because the interval estimate includes values greater than 75% of the employees. LO 5 © 2012 McGraw-Hill Ryerson Limited 41 You Try It Out! A survey was conducted to estimate the proportion of car drivers who use seatbelts while driving. Of the 1500 drivers sampled, 520 drivers said they always use a seatbelt while driving. a) Estimate the value of the population proportion. b) Compute the standard error of the proportion. c) Develop a 99% confidence interval for the population proportion. d) Interpret your findings. LO 5 © 2012 McGraw-Hill Ryerson Limited 42 Finite Population Correction Factor The populations we have sampled so far have been very large or infinite. What if the sampled population is not very large? We need to make some adjustments in the way we compute the standard error of the sample means and the standard error of the sample proportions. A population that has a fixed upper bound is finite. LO 5 © 2012 McGraw-Hill Ryerson Limited 43 Finite Population Correction Factor For a finite population, where the total number of objects or individuals is N and the number of objects or individuals in the sample is n, we need to adjust the standard errors in the confidence interval formulas. To find the confidence interval for the mean, we adjust the standard error of the mean. For the confidence interval for a proportion, we adjust the standard error of the proportion. LO 5 © 2012 McGraw-Hill Ryerson Limited 44 Finite Population Correction Factor This adjustment is called the finite-population correction factor (FPC). FPC N n N 1 The usual rule is if the ratio of n/N is less than 0.05, then the correction factor is ignored. LO 5 © 2012 McGraw-Hill Ryerson Limited 45 Adjusting the Standard Errors with the FPC Adjust the standard error of the mean or proportion as follows: LO 5 © 2012 McGraw-Hill Ryerson Limited 46 Example – Finite Population Correction Factor There are 350 families in one area in Brooks City. A poll of 50 families reveals the mean annual charity contribution is $550 with a standard deviation of $85. 1. Develop a 90 percent confidence interval for the population mean. 2. Interpret the confidence interval. LO 5 © 2012 McGraw-Hill Ryerson Limited 47 Solution – Finite Population Correction Factor Given in Problem: N = 350; n = 50 and s = $85 Since n/N = 50/350 = 0.14, the finite population correction factor must be used. The population standard deviation is not known, therefore use the t distribution. X t s n N n $85 $550 t N 1 50 $550 1.677 350 50 350 1 $85 50 350 50 350 1 $550 $20.16 .93 $550 $18.75 ($568.75, $531.25) LO 5 © 2012 McGraw-Hill Ryerson Limited 48 Solution – Finite Population Correction Factor Continued It is likely that the population mean is more than $531.25 but less than $568.75. The population mean can be $545 but not $525, because $545 is within the confidence interval and $525 is not. LO 5 © 2012 McGraw-Hill Ryerson Limited 49 You Try It Out! The same study of charity contributions revealed that 20 of the 50 families sampled donate to charity regularly. 1. Construct the 95 percent confidence interval for the proportion of families donating to charity regularly. 2. Should the finite-population correction factor be used? Why or why not? LO 5 © 2012 McGraw-Hill Ryerson Limited 50 LO 6 Choosing an Appropriate Sample Size © 2012 McGraw-Hill Ryerson Limited 51 Choosing An Appropriate Sample Size There are three factors that determine the size of a sample, none of which has any direct relationship to the size of the population. 1. the degree of confidence selected 2. the maximum allowable error 3. the population standard deviation i. Use a comparable study. ii. Use a range-based approach. iii. Conduct a pilot study. LO 6 © 2012 McGraw-Hill Ryerson Limited 52 Sample Size for a Population Mean We can express the interaction among these three factors and the sample size in the following formula. Ez n Solving this equation for n yields the following result. Where: n is the size of the sample. z is the standard normal value corresponding to the desired level of confidence. is the population standard deviation. E is the maximum allowable error. LO 6 © 2012 McGraw-Hill Ryerson Limited 53 Example – Sample Size for a Population Mean An NGO wants to determine the mean number of trees planted in last month near city. The error in estimating the mean is to be less than 150 with a 95 percent level of confidence. An NGO found a report by the Department of Forestry that estimated the standard deviation to be 1500. What is the required sample size? LO 6 © 2012 McGraw-Hill Ryerson Limited 54 Solutions – Sample Size for a Population Mean z n E 2 (1.96)(1, 500) 150 (19.6) 2 384.16 2 384 LO 6 © 2012 McGraw-Hill Ryerson Limited 55 Sample Size for a Population Mean In Excel LO 6 © 2012 McGraw-Hill Ryerson Limited 56 Sample Size for a Population Proportion Again, three items need to be specified: 1. the desired level of confidence 2. the margin of error in the population proportion 3. an estimate of the population proportion The formula to determine the sample size of a proportion is: LO 6 © 2012 McGraw-Hill Ryerson Limited 57 Example – Sample Size for a Population Proportion The study in the previous example also estimates the proportion of tress planted. An NGO wants the estimate to be within 0.15 of the population proportion, the desired level of confidence is 90 percent, and no estimate is available for the population proportion. What is the required sample size? LO 6 © 2012 McGraw-Hill Ryerson Limited 58 Solution – Sample Size for a Population Proportion 1.645 n (0.5)(1 0.5) 0.15 30.06 n 30 trees LO 6 2 © 2012 McGraw-Hill Ryerson Limited 59 Sample Size for a Population Proportion In Excel LO 6 © 2012 McGraw-Hill Ryerson Limited 60 You Try It Out! The arithmetic mean of grade point average (GPA) of all graduating seniors during the past 10 years is to be estimated. GPAs range between 3.0 and 5.0. The mean GPA is to be estimated within plus or minus 0.05 of the population mean. The standard deviation is estimated to be 0.389. Use the 99 percent level of confidence. LO 5 © 2012 McGraw-Hill Ryerson Limited 61 Chapter Summary I. A point estimate is a single value (statistic) used to estimate a population value (parameter). II. A confidence interval is a range of values within which the population parameter is expected to occur. A. The factors that determine the width of a confidence interval for a mean are: 1. the number of observations in the sample, n 2. the variability in the population, usually estimated by the sample standard deviation, s 3. the level of confidence © 2012 McGraw-Hill Ryerson Limited 62 Chapter Summary a. To determine the confidence limits when the population standard deviation is known, we use the z distribution. The formula is: X z n 8 1 b. To determine the confidence limits when the population standard deviation is unknown, we use the t distribution. The formula is: X t s n 8 2 © 2012 McGraw-Hill Ryerson Limited 63 Chapter Summary III. The major characteristics of the t distribution are: A. It is a continuous distribution. B. It is mound-shaped and symmetrical. C. It is flatter, or more spread out, than the standard normal distribution. D. There is a family of t distributions, depending on the number of degrees of freedom. © 2012 McGraw-Hill Ryerson Limited 64 Chapter Summary IV. A proportion is a ratio, fraction, or percent that indicates the part of the sample or population that has the particular characteristic. A. A sample proportion is found by X, the number of successes, divided by n, the number of observations. B. The standard error of the sample proportion reports the variability in the distribution of sample proportions. It is found by sp p 1 p n 8 4 © 2012 McGraw-Hill Ryerson Limited 65 Chapter Summary C. We construct a confidence interval for a population proportion from the following formula. V. For the finite population, the standard error is adjusted by the factor NN 1n . . © 2012 McGraw-Hill Ryerson Limited 66 Chapter Summary VI. We can determine an appropriate sample size for estimating both means and proportions. A. There are three factors that determine the sample size when we wish to estimate the population mean. 1. the desired level of confidence, usually expressed by z 2. the maximum allowable error, E 3. the variation in the population, expressed by σ The formula to determine the sample size for the mean is: z n E 2 © 2012 McGraw-Hill Ryerson Limited 8 9 67 Chapter Summary B. There are three factors that determine the sample size when we wish to estimate a population proportion. 1. the desired level of confidence, which is usually expressed by z. 2. the maximum allowable error, E 3. an estimate of the population proportion. If no estimate is available, use .50. The formula to determine the sample size for a proportion is: z n p(1 p) E 2 8 10 © 2012 McGraw-Hill Ryerson Limited 68