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Chapter 7 Confident Intervals and Sample Size Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 Confidence Intervals and Sample Size Outline 7 7-1 Confidence Intervals for the Mean When Is Known 7-2 Confidence Intervals for the Mean When Is Unknown 7-3 Confidence Intervals and Sample Size for Proportions 7-4 Confidence Intervals for Variances and Standard Deviations Confidence Intervals and Sample Size 7 Objectives 1 2 3 4 5 Find the confidence interval for the mean when is known. Determine the minimum sample size for finding a confidence interval for the mean. Find the confidence interval for the mean when is unknown. Find the confidence interval for a proportion. Determine the minimum sample size for finding a confidence interval for a proportion. Confidence Intervals and Sample Size 7 Objectives 6 Find a confidence interval for a variance and a standard deviation. 7.1 Confidence Intervals for the Mean When Is Known • A point estimate is a specific numerical value estimate of a parameter. • The best point estimate of the population mean µ is the sample X. mean Bluman, Chapter 7 5 Three Properties of a Good Estimator 1. The estimator should be an unbiased estimator. That is, the expected value or the mean of the estimates obtained from samples of a given size is equal to the parameter being estimated. Bluman, Chapter 7 6 Three Properties of a Good Estimator 2. The estimator should be consistent. For a consistent estimator, as sample size increases, the value of the estimator approaches the value of the parameter estimated. Bluman, Chapter 7 7 Three Properties of a Good Estimator 3. The estimator should be a relatively efficient estimator; that is, of all the statistics that can be used to estimate a parameter, the relatively efficient estimator has the smallest variance. Bluman, Chapter 7 8 Confidence Intervals for the Mean When Is Known • An interval estimate of a parameter is an interval or a range of values used to estimate the parameter. • This estimate may or may not contain the value of the parameter being estimated. Bluman, Chapter 7 9 Confidence Level of an Interval Estimate • The confidence level of an interval estimate of a parameter is the probability that the interval estimate will contain the parameter, assuming that a large number of samples are selected and that the estimation process on the same parameter is repeated. Bluman, Chapter 7 10 Confidence Interval • A confidence interval is a specific interval estimate of a parameter determined by using data obtained from a sample and by using the specific confidence level of the estimate. Bluman, Chapter 7 11 Confidence Intervals for the Mean (σ Known or n ≥ 30) and Sample Size Formula for the Confidence Interval of the Mean for a Specific a X za / 2 X z a /2 n n For a 90% confidence interval: za / 2 1.65 For a 95% confidence interval: za / 2 1.96 For a 99% confidence interval: za / 2 2.58 Bluman, Chapter 7 13 95% Confidence Interval of the Mean Bluman, Chapter 7 14 Maximum Error of the Estimate The maximum error of the estimate is the maximum likely difference between the point estimate of a parameter and the actual value of the parameter. E za / 2 n Bluman, Chapter 7 16 Confidence Interval for a Mean Rounding Rule When you are computing a confidence interval for a population mean by using raw data, round off to one more decimal place than the number of decimal places in the original data. When you are computing a confidence interval for a population mean by using a sample mean and a standard deviation, round off to the same number of decimal places as given for the mean. Bluman, Chapter 7 17 Chapter 7 Confidence Intervals and Sample Size Section 7-1 Example 7-1 Page #358 Bluman, Chapter 7 18 Example 7-1: Days to Sell an Aveo A researcher wishes to estimate the number of days it takes an automobile dealer to sell a Chevrolet Aveo. A sample of 50 cars had a mean time on the dealer’s lot of 54 days. Assume the population standard deviation to be 6.0 days. Find the best point estimate of the population mean and the 95% confidence interval of the population mean. The best point estimate of the mean is 54 days. Xn 5 4 , 6 . 0 , 5 0 , 9 5 % z 1 . 9 6 X za 2 X za 2 n n Bluman, Chapter 7 19 Example 7-1: Days to Sell an Aveo Xn 5 4 , 6 . 0 , 5 0 , 9 5 % z 1 . 9 6 X za 2 X za 2 n n 6.0 6.0 54 1.96 54 1.96 50 50 54 1.7 54 1.7 52.3 55.7 52 56 One can say with 95% confidence that the interval between 52 and 56 days contains the population mean, based on a sample of 50 automobiles. Bluman, Chapter 7 20 Chapter 7 Confidence Intervals and Sample Size Section 7-1 Example 7-2 Page #358 Bluman, Chapter 7 21 Example 7-2: Waiting Times A survey of 30 emergency room patients found that the average waiting time for treatment was 174.3 minutes. Assuming that the population standard deviation is 46.5 minutes, find the best point estimate of the population mean and the 99% confidence of the population mean. Bluman, Chapter 7 22 Example 7-2: Waiting Times Hence, one can be 99% confident that the mean waiting time for emergency room treatment is between 152.4 and 196.2 minutes. Bluman, Chapter 7 23 95% Confidence Interval of the Mean Bluman, Chapter 7 24 95% Confidence Interval of the Mean One can be 95% confident that an interval built around a specific sample mean would contain the population Bluman, Chapter 7 25 Finding z a 2 for 98% CL. za Bluman, Chapter 7 2 2.33 26 A sample of the reading scores of 35 fifth-graders has a mean of 82. The standard deviation of the sample is 15. a. Find the 95% confidence interval of the mean reading scores of all fifthgraders. b. Find the 99% confidence interval of the mean reading scores of all fifthgraders. c. Which interval is larger? Explain why. n = 35 s = 15 X = 82 a. Find the 95% confidence interval of mean reading scores of all fifth-graders. s s X – za 2 n < < X +za 2 n 15 15 82 – (1.96) < < 82 + (1.96) 35 35 82 – 4.97< < 82 + 4.97 77 < < 87 n = 35 s = 15 X = 82 a. Find the 95% confidence interval of mean reading scores of all fifth-graders. 77 < < 87 Approximately 95% of the sample means will fall within 1.96 standard errors of the population mean, so use za 2 = 1.96. n = 35 s = 15 X = 82 b. Find the 99% confidence interval of the mean reading scores of all fifth-graders. s s X – za 2 n < < X +za 2 n 15 15 82 – (2.58) < m < 82 + (2.58) 35 35 82 – 6.54< < 82 + 6.54 75< < 89 n = 35 s = 15 X = 82 b. Find the 99% confidence interval of the mean reading scores of all fifth-graders. 75< < 89 Approximately 99% of the sample means will fall within 2.58 standard errors of the population mean, so use za 2 = 2.58 . c. Which interval is larger? Explain why. 95% confidence level = 77< < 87 99% confidence level = 75< < 89 The 99% confidence interval is larger because the confidence level is larger. A study of 40 English composition professors showed that they spent, on average, 12.6 minutes correcting a student’s term paper. a. Find the 90% confidence interval of the mean time for all composition papers when = 2.5 minutes. b. If a professor stated that he spent, on average, 30 minutes correcting a term paper, what would be your reaction? n = 40 X = 12.6 a. Find the 90% confidence interval of the mean time for all composition papers when = 2.5 minutes. s s X – za 2 n < < X +za 2 n 2.5 2.5 12.6 – (1.65) < < 12.6 + (1.65) 40 40 12.6 – 0.652< < 12.6 + 0.652 11.9< < 13.3 b. If a professor stated that he spent, on average, 30 minutes correcting a term paper, what would be your reaction? 11.9 < < 13.3 It would be highly unlikely since this is far larger than 13.3 minutes. 47 95 54 33 64 4 8 57 9 80 8 90 3 49 4 44 79 80 48 16 68 7 15 21 52 6 78 109 40 50 29 A random sample of the number of farms (in thousands) in various states is found above. Estimate the mean number of farms per state with 90% confidence. 47 95 54 33 64 4 8 57 9 80 8 90 3 49 4 44 79 80 48 16 68 7 15 21 52 6 78 109 40 50 29 X = 43.45 s = 31.27 31.27 31.27 43.45 – 1.65 < < 43.45 + 1.65 31 31 43.45 – 9.27 < < 43.45 + 9.27 34.18 < < 52.72 34.2 < < 52.7 or Chapter 7 Confidence Intervals and Sample Size Section 7-1 Example 7-3 Page #360 Bluman, Chapter 7 38 Example 7-3: Credit Union Assets The following data represent a sample of the assets (in millions of dollars) of 30 credit unions in southwestern Pennsylvania. Find the 90% confidence interval of the mean. 12.23 16.56 4.39 2.89 1.24 2.17 13.19 9.16 1.42 73.25 1.91 14.64 11.59 6.69 1.06 8.74 3.17 18.13 7.92 4.78 16.85 Bluman, Chapter 7 39 Example 7-3: Credit Union Assets Step 1: Find the mean and standard deviation. Using technology, we find X = 11.091 and s = 14.405. Assume 14.405. Step 2: Find α/2. 90% CL α/2 = 0.05. Step 3: Find zα/2. 90% CL α/2 = 0.05 z.05 = 1.65 Table E The Standard Normal Distribution z .00 … .04 .05 0.9495 0.9505 … .09 0.0 0.1 . . . 1.6 Bluman, Chapter 7 40 Example 7-3: Credit Union Assets Step 4: Substitute in the formula. X za 2 X za 2 n n 14.405 14.405 11.091 1.65 11.091 1.65 30 30 11.091 4.339 11.091 4.339 6.752 15.430 One can be 90% confident that the population mean of the assets of all credit unions is between $6.752 million and $15.430 million, based on a sample of 30 credit unions. Bluman, Chapter 7 41 Technology Note This chapter and subsequent chapters include examples using raw data. If you are using computer or calculator programs to find the solutions, the answers you get may vary somewhat from the ones given in the textbook. This is so because computers and calculators do not round the answers in the intermediate steps and can use 12 or more decimal places for computation. Also, they use more exact values than those given in the tables in the back of this book. These discrepancies are part and parcel of statistics. Bluman, Chapter 7 42 Formula for Minimum Sample Size Needed for an Interval Estimate of the Population Mean za 2 n E 2 where E is the margin of error. If necessary, round the answer up to obtain a whole number. That is, if there is any fraction or decimal portion in the answer, use the next whole number for sample size n. Bluman, Chapter 7 43 Chapter 7 Confidence Intervals and Sample Size Section 7-1 Example 7-4 Page #362 Bluman, Chapter 7 44 Example 7-4: Depth of a River A scientist wishes to estimate the average depth of a river. He wants to be 99% confident that the estimate is accurate within 2 feet. From a previous study, the standard deviation of the depths measured was 4.33 feet. How large a sample is required? 9 9 % z 2 . 5 8 , E 2 , 4 . 3 3 2 z a2 2 .5 84 .3 3 n 3 1 .2 32 2 E 2 Therefore, to be 99% confident that the estimate is within 2 feet of the true mean depth, the scientist needs at least a sample of 32 measurements. Bluman, Chapter 7 45 An insurance company is trying to estimate the average number of sick days that full-time food service workers use per year. A pilot study found the standard deviation to be 2.5 days. How large a sample must be selected if the company wants to be 95% confident of getting an interval that contains the true mean with a maximum error of 1 day? s = 2.5 confidence level = 95% maximum error = 1 day s = 2.5 confidence level = 95% maximum error = 1 day 2 za 2 (1.96)(2.5) 2 2 = n = = (4.9) E 1 or n = 25 workers n = 24.01 7.2 Confidence Intervals for the Mean When Is Unknown The value of , when it is not known, must be estimated by using s, the standard deviation of the sample. When s is used, especially when the sample size is small (less than 30), critical values greater than the values for za 2 are used in confidence intervals in order to keep the interval at a given level, such as the 95%. These values are taken from the Student t distribution, most often called the t distribution. Bluman, Chapter 7 48 Characteristics of the t Distribution The t distribution is similar to the standard normal distribution in these ways: 1. It is bell-shaped. 2. It is symmetric about the mean. 3. The mean, median, and mode are equal to 0 and are located at the center of the distribution. 4. The curve never touches the x axis. Bluman, Chapter 7 49 Characteristics of the t Distribution The t distribution differs from the standard normal distribution in the following ways: 1. The variance is greater than 1. 2. The t distribution is actually a family of curves based on the concept of degrees of freedom, which is related to sample size. 3. As the sample size increases, the t distribution approaches the standard normal distribution. Bluman, Chapter 7 50 Degrees of Freedom • The symbol d.f. will be used for degrees of freedom. • The degrees of freedom for a confidence interval for the mean are found by subtracting 1 from the sample size. That is, d.f. = n – 1. • Note: For some statistical tests used later in this book, the degrees of freedom are not equal to n – 1. Bluman, Chapter 7 51 Formula for a Specific Confidence Interval for the Mean When Is Unknown and n < 30 s s X ta 2 X ta 2 n n The degrees of freedom are n – 1. Bluman, Chapter 7 52 Chapter 7 Confidence Intervals and Sample Size Section 7-2 Example 7-5 Page #369 Bluman, Chapter 7 53 Example 7-5: Using Table F Find the tα/2 value for a 95% confidence interval when the sample size is 22. Degrees of freedom are d.f. = 21. Bluman, Chapter 7 54 Chapter 7 Confidence Intervals and Sample Size Section 7-2 Example 7-6 Page #370 Bluman, Chapter 7 56 Example 7-6: Sleeping Time Ten randomly selected people were asked how long they slept at night. The mean time was 7.1 hours, and the standard deviation was 0.78 hour. Find the 95% confidence interval of the mean time. Assume the variable is normally distributed. Since is unknown and s must replace it, the t distribution (Table F) must be used for the confidence interval. Hence, with 9 degrees of freedom, tα/2 = 2.262. s s X ta 2 X ta 2 n n 0.78 0.78 7.1 2.262 7.1 2.262 10 10 Bluman, Chapter 7 57 Example 7-6: Sleeping Time 0.78 0.78 7.1 2.262 7.1 2.262 10 10 7.1 0.56 7.1 0.56 6.5 7.7 One can be 95% confident that the population mean is between 6.5 and 7.7 hours. Bluman, Chapter 7 58 Chapter 7 Confidence Intervals and Sample Size Section 7-2 Example 7-7 Page #370 Bluman, Chapter 7 59 Example 7-7: Home Fires by Candles The data represent a sample of the number of home fires started by candles for the past several years. Find the 99% confidence interval for the mean number of home fires started by candles each year. 5460 5900 6090 6310 7160 8440 9930 Step 1: Find the mean and standard deviation. The mean is X = 7041.4 and standard deviation s = 1610.3. Step 2: Find tα/2 in Table F. The confidence level is 99%, and the degrees of freedom d.f. = 6 t .005 = 3.707. Bluman, Chapter 7 60 Example 7-7: Home Fires by Candles Step 3: Substitute in the formula. s s X ta 2 X ta 2 n n 1 6 1 0 . 3 1 6 1 0 . 3 7 0 4 1 . 4 3 . 7 0 7 7 0 4 1 . 4 3 . 7 0 7 7 7 7041.4 2256.2 7041.4 2256.2 4785.2 9297.6 One can be 99% confident that the population mean number of home fires started by candles each year is between 4785.2 and 9297.6, based on a sample of home fires occurring over a period of 7 years. Bluman, Chapter 7 61 A state representative wishes to estimate the mean number of women representatives per state legislature. A random sample of 17 states is selected, and the number of women representatives is shown.Based on the sample, what is the point estimate of the mean? Find the 90% confidence interval of the mean population. (Note: The population mean is actually 31.72, or about 32.) Compare this value to the point estimate and the confidence interval. There is something unusual about the data. Describe it and state how it would affect the confidence interval. 5 31 18 58 X = 33.4 33 16 29 132 35 45 15 37 19 39 24 13 18 s = 28.7 s s X – ta 2 n < < X + ta 2 n 28.7 28.7 33.4 – 1.746 < < 33.4 + 1.746 17 17 33.4 – 12.2 < < 33.4 + 12.2 21.2< < 45.6 A recent study of 28 employees of XYZ Company showed that the mean of the distance they traveled to work was 14.3 miles. The standard deviation of the sample mean was 2 miles. Find the 95% confidence interval of the true mean. s s X – ta 2 n < < X + ta 2 n 14.3 – 2.052 2 28 < < 14.3 + 2.052 14.3 – 0.776< < 14.3 + 0.776 2 28 s s X – ta 2 n < < X + ta 2 n 14.3 – 2.052 2 < < 14.3 + 2.052 28 14.3 – 0.776< < 14.3 + 0.776 13.5< < 15.1 2 28 If a manager wanted to be sure that most of his employees would not be late, how much time would he suggest they allow for the commute if the average speed is 30 miles per hour? 14.3 mi 30 mi/hr - 1 2 hour about 30 minutes Chapter 7 Confidence Intervals and Sample Size Section 7-3 Exercise #19 The average yearly income for 28 community college instructors was $56,718. The standard deviation was $650. Find the 95% confidence interval of the true mean. s s X – ta 2 n < < X + ta 2 n s s X – ta 2 n < < X + ta 2 n $56, 718 – 2.052 < < $56, 718 + 2.052 28 650 $56, 466 < < $56, 970 28 650 If a faculty member wishes to see if he or she is being paid below average, what salary value should he or she use? Use the lower bound of the confidence interval: $56,466 7.3 Confidence Intervals and Sample Size for Proportions p = population proportion p̂ (read p “hat”) = sample proportion For a sample proportion, X n X ˆ ˆ ˆ ˆ p a n d q o r q 1 p n n where X = number of sample units that possess the characteristics of interest and n = sample size. Bluman, Chapter 7 71 Chapter 7 Confidence Intervals and Sample Size Section 7-3 Example 7-8 Page #376 Bluman, Chapter 7 72 Example 7-8: Air Conditioned Households In a recent survey of 150 households, 54 had central air conditioning. Find p̂ and q̂, where p̂ is the proportion of households that have central air conditioning. Since X = 54 and n = 150, X 54 pˆ 0.36 36% n 150 qˆ 1 pˆ 1 0.36 0.64 64% Bluman, Chapter 7 73 Formula for a Specific Confidence Interval for a Proportion pˆ za 2 ˆˆ pq p pˆ za n 2 ˆˆ pq n when np 5 and nq 5. Rounding Rule: Round off to three decimal places. Bluman, Chapter 7 74 Chapter 7 Confidence Intervals and Sample Size Section 7-3 Example 7-9 Page #377 Bluman, Chapter 7 75 Example 7-9: Covering College Costs A survey conducted by Sallie Mae and Gallup of 1404 respondents found that 323 students paid for their education by student loans. Find the 90% confidence of the true proportion of students who paid for their education by student loans. Bluman, Chapter 7 76 Example 7-9: Covering College Costs Since α = 1 – 0.90 = 0.10, zα/2 = 1.65. Bluman, Chapter 7 77 Example 7-9: Covering College Costs You can be 90% confident that the percentage of students who pay for their college education by student loans is between 21.1 and 24.9%. Bluman, Chapter 7 78 Chapter 7 Confidence Intervals and Sample Size Section 7-3 Example 7-10 Page #377 Bluman, Chapter 7 79 Example 7-10: Religious Books A survey of 1721 people found that 15.9% of individuals purchase religious books at a Christian bookstore. Find the 95% confidence interval of the true proportion of people who purchase their religious books at a Christian bookstore. ˆˆ ˆˆ pq pq pˆ za 2 p pˆ za 2 n n 0.159 1.96 0.159 0.841 1721 p 0.159 1.96 0.159 0.841 1721 0.142 p 0.176 You can say with 95% confidence that the true percentage is between 14.2% and 17.6%. Bluman, Chapter 7 80 The proportion of students in private schools is around 11%. A random sample of 450 students from a wide geographic area indicated that 55 attended private schools. Estimate the true proportion of students attending private schools with 95% confidence. How does your estimate compare to 11%? X 55 p̂ = = = 0.12 n 450 pˆ – za 2 q̂ = 1– 0.12 = 0.88 ˆˆ pq < p < pˆ + za 2 n ˆˆ pq n X 55 p̂ = = = 0.12 n 450 pˆ – za 2 q̂ = 1– 0.12 = 0.88 ˆˆ pq < p < pˆ + za 2 n ˆˆ pq n 0.12 – 1.96 (0.12)(0.88) (0.12)(0.88) < p < 0.12 + 1.96 450 450 0.12 – 0.03 < p < 0.12 + 0.03 0.09 < p < 0.15 or 11% is contained in the confidence interval. 9% < p < 15% A survey found that out of 200 workers, 168 said they were interrupted three or more times an hour by phone messages, faxes, etc. Find the 90% confidence interval of the population proportion of workers who are interrupted three or more times an hour. p̂ = 0.84 pˆ – za 2 q̂ = 0.16 ˆˆ pq < p < pˆ + za 2 n ˆˆ pq n p̂ = 0.84 pˆ – za 2 q̂ = 0.16 ˆˆ pq < p < pˆ + za 2 n ˆˆ pq n (0.84)(0.16) (0.84)(0.16) 0.84 – 1.65 < p < 0.84 + 1.65 200 200 0.84 – 0.043 < p < 0.84 + 0.043 0.797< p < 0.883 A study by the University of Michigan found that one in five 13 and 14-year-olds is a sometime smoker. To see how the smoking rate of the students at a large school district compared to the national rate, the superintendent surveyed two hundred 13 and 14-year-old students and found that 23% said they were sometime smokers. Find the 99% confidence interval of the true proportion and compare this with the University of Michigan’s study. Find the 99% confidence interval of the true proportion. n = 200 p̂ = 0.23 q̂ = 1– 0.23 = 0.77 pˆ – za 2 ˆˆ pq < p < pˆ + za 2 n ˆˆ pq n (0.23)(0.77) (0.23)(0.77) 0.23 – 2.58 < p < 0.23 + 2.58 200 200 0.23 – 0.077 < p < 0.23 + 0.077 0.153 < p < 0.307 A study by the University of Michigan found that one in five 13- and 14-year-olds is a sometime smoker. Find the 99% confidence interval of the true proportion and compare this with the University of Michigan’s study. 0.153 < p < 0.307 1 5 = 0.20 The University of Michigan study falls within this confidence interval. Chapter 7 Confidence Intervals and Sample Size Section 7-4 Exercise #15 A medical researcher wishes to determine the percentage of females who take vitamins. He wishes to be 99% confident that the estimate is within 2 percentage points of the true proportion. A recent study of 180 females showed that 25% took vitamins. a. How large should the sample size be? b. If no estimate of the sample proportion is available, how large should the sample be? a. How large should the sample size be? p̂ = 0.25 q̂ = 0.75 za /2 n = p̂q̂ E 2 2.58 n = 0.25 0.75 .02 n = 3120.187 or 2 n = 3121 b. If no estimate of the sample proportion is available, how large should the sample be? p̂ = 0.5 q̂ = 0.5 za / 2 ˆˆ n = pq E n = 4160.25 2 2.58 n = 0.5 0.5 0.02 or n = 4161 Since there is no prior knowledge of p or q, assign the values q̂ = 0.5 and p̂ = 0.5. 2 Formula for Minimum Sample Size Needed for Interval Estimate of a Population Proportion za 2 ˆ ˆ n pq E 2 If necessary, round up to the next whole number. Bluman, Chapter 7 94 Chapter 7 Confidence Intervals and Sample Size Section 7-3 Example 7-11 Page #378 Bluman, Chapter 7 95 Example 7-11: Home Computers A researcher wishes to estimate, with 95% confidence, the proportion of people who own a home computer. A previous study shows that 40% of those interviewed had a computer at home. The researcher wishes to be accurate within 2% of the true proportion. Find the minimum sample size necessary. 2 2 za 2 1.96 ˆ ˆ n pq 0.40 0.60 0.02 2304.96 E The researcher should interview a sample of at least 2305 people. Bluman, Chapter 7 96 Chapter 7 Confidence Intervals and Sample Size Section 7-3 Example 7-12 Page #378 Bluman, Chapter 7 97 Example 7-12: M&M Colors A researcher wishes to estimate the percentage of M&M’s that are brown. He wants to be 95% confident and be accurate within 3% of the true proportion. How large a sample size would be necessary? Since no prior knowledge of p̂ is known, assign a value of 0.5 and then q̂ = 1 – 0.5 = 0.5. Substitute in the formula, using E = 0.03. Bluman, Chapter 7 98 Example 7-12: M&M Colors Bluman, Chapter 7 99 7-4 Confidence Intervals for Variances and Standard Deviations • When products that fit together (such as pipes) are manufactured, it is important to keep the variations of the diameters of the products as small as possible; otherwise, they will not fit together properly and will have to be scrapped. • In the manufacture of medicines, the variance and standard deviation of the medication in the pills play an important role in making sure patients receive the proper dosage. • For these reasons, confidence intervals for variances and standard deviations are necessary. Bluman, Chapter 7 100 Chi-Square Distributions • The chi-square distribution must be used to calculate confidence intervals for variances and standard deviations. • The chi-square variable is similar to the t variable in that its distribution is a family of curves based on the number of degrees of freedom. • The symbol for chi-square is 2(Greek letter chi, pronounced “ki”). • A chi-square variable cannot be negative, and the distributions are skewed to the right. Bluman, Chapter 7 101 Chi-Square Distributions At about 100 degrees of freedom, the chi-square distribution becomes somewhat symmetric. The area under each chi-square distribution is equal to 1.00, or 100%. Bluman, Chapter 7 102 Formula for the Confidence Interval for a Variance n 1 s 2 2 n 1 s 2 , 2 2 right d.f. = n 1 left Formula for the Confidence Interval for a Standard Deviation n 1 s 2 right 2 n 1 s Bluman, Chapter 7 2 left 2 , d.f. = n 1 103 Chapter 7 Confidence Intervals and Sample Size Section 7-4 Example 7-13 Page #385 Bluman, Chapter 7 104 Example 7-13: Using Table G Find the values for right and interval when n = 25. 2 2 for a 90% confidence left 2 To find right , subtract 1 – 0.90 = 0.10. 2 left To find Divide by 2 to get 0.05. , subtract 1 – 0.05 to get 0.95. Bluman, Chapter 7 105 Example 7-13: Using Table G Use the 0.95 and 0.05 columns and the row corresponding to 24 d.f. in Table G. 2 2 The right value is 36.415; the left value is 13.848. Bluman, Chapter 7 106 Confidence Interval for a Variance or Standard Deviation Rounding Rule When you are computing a confidence interval for a population variance or standard deviation by using raw data, round off to one more decimal places than the number of decimal places in the original data. When you are computing a confidence interval for a population variance or standard deviation by using a sample variance or standard deviation, round off to the same number of decimal places as given for the sample variance or standard deviation. Bluman, Chapter 7 107 Chapter 7 Confidence Intervals and Sample Size Section 7-4 Example 7-14 Page #387 Bluman, Chapter 7 108 Example 7-14: Nicotine Content Find the 95% confidence interval for the variance and standard deviation of the nicotine content of cigarettes manufactured if a sample of 20 cigarettes has a standard deviation of 1.6 milligrams. To find 2 right 2 left To find , subtract 1 – 0.95 = 0.05. Divide by 2 to get 0.025. , subtract 1 – 0.025 to get 0.975. In Table G, the 0.025 and 0.975 columns with the d.f. 19 row yield values of 32.852 and 8.907, respectively. Bluman, Chapter 7 109 Example 7-14: Nicotine Content n 1 s 2 2 n 1 s 2 2 2 right left 2 2 19 1.6 2 19 1.6 32.852 8.907 1.5 2 5.5 You can be 95% confident that the true variance for the nicotine content is between 1.5 and 5.5 milligrams. 1.5 5.5 1.2 2.3 You can be 95% confident that the true standard deviation is between 1.2 and 2.3 milligrams. Bluman, Chapter 7 110 Chapter 7 Confidence Intervals and Sample Size Section 7-4 Example 7-15 Page #387 Bluman, Chapter 7 111 Example 7-15: Cost of Ski Lift Tickets Find the 90% confidence interval for the variance and standard deviation for the price in dollars of an adult singleday ski lift ticket. The data represent a selected sample of nationwide ski resorts. Assume the variable is normally distributed. 59 54 53 52 51 39 49 46 49 48 Using technology, we find the variance of the data is s2 = 28.2. In Table G, the 0.05 and 0.95 columns with the d.f. 9 row yield values of 16.919 and 3.325, respectively. Bluman, Chapter 7 112 Example 7-15: Cost of Ski Lift Tickets n 1 s 2 2 n 1 s 2 2 2 right left 9 28.2 2 9 28.2 16.919 3.325 15.0 2 76.3 You can be 95% confident that the true variance for the cost of ski lift tickets is between 15.0 and 76.3. 15.0 76.3 3.87 8.73 You can be 95% confident that the true standard deviation is between $3.87 and $8.73. Bluman, Chapter 7 113 Find the 90% confidence interval for the variance and standard deviation for the time it takes a customer to place a telephone order with a large catalog company is a sample of 23 telephone orders has a standard deviation of 3.8 minutes. Assume the variable is normally distributed. 2 2 2 2 (n – 1)s 2 (n – 1)s < < 2 2 right left 22 3.8 33.924 2 < < 22 3.8 12.338 2 9.36 < < 25.75 3.06 < < 5.07 Chapter 7 Confidence Intervals and Sample Size Section 7-5 Exercise #6 Find the 99% confidence interval for the variance and standard deviation of the weights of 25 one-gallon containers of motor oil if a sample of 14 containers has a variance of 3.2. Assume the variable is normally distributed. Find the 99% confidence interval for the variance and standard deviation. n = 14 s 2 = 3.2 (n – 1)s 2 (n – 1)s 2 2 < < 2 2 right left 13(3.2) 13(3.2) 2 < < 29.819 3.565 1.4 < 2 < 11.7 1.2 < < 3.4 Chapter 7 Confidence Intervals and Sample Size Section 7-5 Exercise #9 The number of calories in a 1-ounce serving of various kinds of regular cheese is shown. Estimate the population variance and standard deviation with 90% confidence. 110 45 100 95 110 110 100 110 95 120 130 100 80 105 105 90 110 70 125 108 (n – 1)s 2 2 (n – 1)s 2 2 < < 2 right left 2 19(19.1913) 2 19(19.1913) < 2 < 30.144 10.117 2 232.1 < < 691.7 15.2 < < 26.3 Chapter 7 Confidence Intervals and Sample Size Section 7-5 Exercise #11 A service station advertises that customers will have to wait no more than 30 minutes for an oil change. A sample of 28 oil changes has a standard deviation of 5.2 minutes. Find the 95% confidence interval of the population standard deviation of the time spent waiting for an oil change. (n – 1)s 2 2 (n – 1)s 2 2 < < 2 right left 2 27(5.2) 27(5.2)2 < 2 < 43.194 14.573 16.9 < 2 < 50.1 4.1 < < 7.1