Transcript Chapter 6
Chapter 6 Confidence Intervals © 2012 Pearson Education, Inc. All rights reserved. 1 of 83 Chapter Outline • 6.1 Confidence Intervals for the Mean (Large Samples) • 6.2 Confidence Intervals for the Mean (Small Samples) • 6.3 Confidence Intervals for Population Proportions • 6.4 Confidence Intervals for Variance and Standard Deviation © 2012 Pearson Education, Inc. All rights reserved. 2 of 83 Section 6.1 Confidence Intervals for the Mean (Large Samples) © 2012 Pearson Education, Inc. All rights reserved. 3 of 83 Section 6.1 Objectives • Find a point estimate and a margin of error • Construct and interpret confidence intervals for the population mean • Determine the minimum sample size required when estimating μ © 2012 Pearson Education, Inc. All rights reserved. 4 of 83 Point Estimate for Population μ Point Estimate • A single value estimate for a population parameter • Most unbiased point estimate of the population mean μ is the sample mean x Estimate Population with Sample Parameter… Statistic x Mean: μ © 2012 Pearson Education, Inc. All rights reserved. 5 of 83 Example: Point Estimate for Population μ A social networking website allows its users to add friends, send messages, and update their personal profiles. The following represents a random sample of the number of friends for 40 users of the website. Find a point estimate of the population mean, µ. (Source: Facebook) Enter this in list 1. We will use this for future examples in this section 140 105 130 97 80 165 232 110 214 201 122 98 65 88 154 133 121 82 130 211 153 114 58 77 51 247 236 109 126 132 125 149 122 74 59 218 192 90 117 105 © 2012 Pearson Education, Inc. All rights reserved. 6 of 83 Solution: Point Estimate for Population μ The sample mean of the data is x 5232 x 130.8 n 40 Your point estimate for the mean number of friends for all users of the website is 130.8 friends. The problem with a Point Estimate is that the real world probability of hitting that point is virtually zero. -In other words, who has 130.8 friends? © 2012 Pearson Education, Inc. All rights reserved. 7 of 83 Interval Estimate Therefore, we estimate µ using an Interval estimate • An interval, or range of values, used to estimate a population parameter. Left endpoint x 130.8 115.1 ( 115 Right endpoint Point estimate 120 125 130 135 146.5 ) 140 145 150 Interval estimate How confident do we want to be that the interval estimate contains the population mean μ? © 2012 Pearson Education, Inc. All rights reserved. 8 of 83 Level of Confidence Level of confidence c • The probability that the interval estimate contains the population parameter. c is the area under the c standard normal curve between the critical values. ½(1 – c) ½(1 – c) –zc z=0 Critical values z zc Use the Standard Normal Table to find the corresponding z-scores. The remaining area in the tails is 1 – c . © 2012 Pearson Education, Inc. All rights reserved. 9 of 83 Level of Confidence • If the level of confidence is 90%, this means that we are 90% confident that the interval contains the population mean μ. c = 0.90 ½(1 – c) = 0.05 ½(1 – c) = 0.05 zc –zc = –1.645 z=0 zc =zc1.645 z The corresponding z-scores are ±1.645. Previously- we have calculated this as InvNorm(.05) © 2012 Pearson Education, Inc. All rights reserved. 10 of 83 Levels of Confidence • There are 5 standard levels of confidence in statistics: KNOW THESE • 80% z = +- 1.28 (InvNorm(.10) Write them down • 90% z = +- 1.645 (InvNorm(.05)It will save you time • 95% z = +- 1.96 (InvNorm(.025) • 98% z = +- 2.33 (InvNorm(.01) • 99% z = +- 2.575 (InvNorm(.005) •NOTE: To have a 90% confidence level, we must have an area under the curve = .90 •Therefore, our left limit is -1.645, and our right limit is 1.645 •When using this Z value as a Z critical value (and not as a left limit/right limit) we only use the POSITIVE Z value Sampling Error Sampling error • The difference between the point estimate and the actual population parameter value. • For μ: the sampling error is the difference x – μ μ is generally unknown x varies from sample to sample • Therefore the sampling error is hard to calculate –if at all © 2012 Pearson Education, Inc. All rights reserved. 12 of 83 Margin of Error Margin of error • The greatest possible distance between the point estimate (sample mean) and the value of the parameter (Population mean) it is estimating for a given level of confidence, c. • Denoted by E. E zcσ x zc Critical Z Score σ n When n ≥ 30, the sample standard deviation, s, can be used for σ. • Sometimes called the maximum error of estimate or error tolerance. NOTE: If we had entered the long list of numbers into list 1, we could then calculate S using 13 of 83 STAT: 1 –VAR Stats Example: Finding the Margin of Error Use the social networking website data and a 95% confidence level to find the margin of error for the mean number of friends for all users of the website. Assume the sample standard deviation is about 53.0. © 2012 Pearson Education, Inc. All rights reserved. 14 of 83 Solution: Finding the Margin of Error • First find the critical values This slide is showing how they got a Z-score We have this on our cheat sheet of scores 0.95 0.025 0.025 zc –zc = –1.96 z=0 zczc= 1.96 z 95% of the area under the standard normal curve falls within 1.96 standard deviations of the mean. (You can approximate the distribution of the sample means with a normal curve by the Central Limit Theorem, because n = 40 ≥ 30.) © 2012 Pearson Education, Inc. All rights reserved. 15 of 83 Solution: Finding the Margin of Error E zc n zc 53.0 1.96 40 16.4 s n You don’t know σ, but since n ≥ 30, you can use s in place of σ. On a Calculator: Stat TestZinterval Choose Stats Enter ơ Enter “0” for x-bar (we don’t know it) Enter n Enter c-Level Calculate You are 95% confident that the margin of error for the population mean is about 16.4 friends. © 2012 Pearson Education, Inc. All rights reserved. 16 of 83 On the Calculator • The problem with calculating “The maximum error of Estimate” E is calculating the sample Standard Deviation S (they may not give you a population standard deviation to use) • This is easily overcome, however, if you make a list, then run 1var Stats • You need this to see S the sample deviation • If you make a list, you can then go to Stats TestsZinterval • Here, it asks for input • You can either type in the mean, SD etc. or choose “Data” if you made a list. • You then choose the confidence level for example .95 • It will then tell you the bottom and top value for your interval estimate. It will also give you the sample means (which is also the point estimate) and the sample standard deviation S • Nice… Confidence Intervals for the Population Mean A c-confidence interval for the population mean μ • xE xE where E zc n • The probability that the confidence interval contains μ is c. © 2012 Pearson Education, Inc. All rights reserved. 18 of 83 Example: Constructing a Confidence Interval Construct a 95% confidence interval for the mean number of friends for all users of the website. Solution: Recall x 130.8 and E ≈ 16.4 Left Endpoint: x E 130.8 16.4 114.4 Right Endpoint: xE 130.8 16.4 147.2 114.4 < μ < 147.2 © 2012 Pearson Education, Inc. All rights reserved. 21 of 83 Solution: Constructing a Confidence Interval 114.4 < μ < 147.2 • With 95% confidence, you can say that the population mean number of friends is between 114.4 and 147.2. © 2012 Pearson Education, Inc. All rights reserved. 22 of 83 Example: Constructing a Confidence Interval σ Known A college admissions director wishes to estimate the mean age of all students currently enrolled. In a random sample of 20 students, the mean age is found to be 22.9 years. From past studies, the standard deviation is known to be 1.5 years, and the population is normally distributed. Construct a 90% confidence interval of the population mean age. © 2012 Pearson Education, Inc. All rights reserved. 23 of 83 Solution: Constructing a Confidence Interval σ Known • First find the critical values c = 0.90 ½(1 – c) = 0.05 ½(1 – c) = 0.05 zc –zc = –1.645 z=0 zc =zc1.645 z zc = 1.645 © 2012 Pearson Education, Inc. All rights reserved. 24 of 83 Solution: Constructing a Confidence Interval σ Known • Margin of error: E zc n 1.645 • Confidence interval: Left Endpoint: x E 22.9 0.6 22.3 1.5 20 0.6 Sometimes this is easier than using Z interval… Right Endpoint: xE 22.9 0.6 23.5 22.3 < μ < 23.5 © 2012 Pearson Education, Inc. All rights reserved. 25 of 83 Solution: Constructing a Confidence Interval σ Known 22.3 < μ < 23.5 Point estimate 22.3 ( x E 22.9 23.5 x xE • ) With 90% confidence, you can say that the mean age of all the students is between 22.3 and 23.5 years. © 2012 Pearson Education, Inc. All rights reserved. 26 of 83 Interpreting the Results • μ is a fixed number. It is either in the confidence interval or not. • Incorrect: “There is a 90% probability that the actual mean is in the interval (22.3, 23.5).” • Correct: “If a large number of samples is collected and a confidence interval is created for each sample, approximately 90% of these intervals will contain μ. © 2012 Pearson Education, Inc. All rights reserved. 27 of 83 Interpreting the Results The horizontal segments represent 90% confidence intervals for different samples of the same size. In the long run, 9 of every 10 such intervals will contain μ. © 2012 Pearson Education, Inc. All rights reserved. μ 28 of 83 Sample Size • Given a c-confidence level and a margin of error E, the minimum sample size n needed to estimate the population mean µ is zc n E • If σ is unknown, you can estimate it using s, provided you have a preliminary sample with at least 30 members. • You will need to do this manually. Make a note of this equation. 2 © 2012 Pearson Education, Inc. All rights reserved. 29 of 83 Example: Sample Size You want to estimate the mean number of friends for all users of the website. How many users must be included in the sample if you want to be 95% confident that the sample mean is within seven friends of the population mean? Assume the sample standard deviation is about 53.0. In this example, we are setting our margin of error –how close do we want to be- and still maintaining our confidence level © 2012 Pearson Education, Inc. All rights reserved. 30 of 83 Solution: Sample Size • First find the critical values 0.95 0.025 0.025 zc –zc = –1.96 z=0 zczc= 1.96 z zc = 1.96 © 2012 Pearson Education, Inc. All rights reserved. 31 of 83 Solution: Sample Size zc = 1.96 σ ≈ s ≈ 53.0 E=7 zc 1.96 53.0 n 220.23 7 E 2 2 When necessary, round up to obtain a whole number. Always round up. You should include at least 221 users in your sample. © 2012 Pearson Education, Inc. All rights reserved. 32 of 83 Section 6.1 Summary • Found a point estimate and a margin of error • Constructed and interpreted confidence intervals for the population mean • Determined the minimum sample size required when estimating μ © 2012 Pearson Education, Inc. All rights reserved. 33 of 83 Assignment • Page 311 5-32 skip 17-20 • Page 312 35-50 skip 47,48 Larson/Farber 5th ed 34 Chapter 6 Quiz 1 ( 5 points each) A college admissions director wishes to estimate the mean age of all students currently enrolled. In a random sample of 25 students, the mean age is found to be 22.9 years. From past studies, the standard deviation is known to be 1.5 years, and the population is normally distributed. 1. 2. 3. 4. What is the point estimate of the mean age? What is the Z critical value for 95%? What is the Margin of Error (E) using this data (at 95%) Construct a 95% confidence interval of the population mean age. 35 of 83 Section 6.2 Confidence Intervals for the Mean (Small Samples) © 2012 Pearson Education, Inc. All rights reserved. 36 of 83 Section 6.2 Objectives • Interpret the t-distribution and use a t-distribution table • Construct confidence intervals when n < 30, the population is normally distributed, and σ is unknown © 2012 Pearson Education, Inc. All rights reserved. 37 of 83 The t-Distribution • When the population standard deviation is unknown, the sample size is less than 30, and the random variable x is approximately normally distributed, it follows a t-distribution. How would we figure out if the variable x is approximately normally distributed? x s n • Critical values of t are denoted by tc. t Question: When do we use the T-Distribution instead of the Normal Distribution? Answer: When the population σ is unknown and the sample is less than 30. KNOW THIS 38 of 83 Properties of the t-Distribution 1. The t-distribution is bell shaped and symmetric about the mean. 2. The t-distribution is a family of curves, each determined by a parameter called the degrees of freedom. The degrees of freedom are the number of free choices left after a sample statistic such as x is calculated. When you use a t-distribution to estimate a population mean, the degrees of freedom are equal to one less than the sample size. d.f. = n – 1 Degrees of freedom the number of degrees of freedom is the number of values in the final calculation of a statistic that are free to vary. © 2012 Pearson Education, Inc. All rights reserved. 39 of 83 Degrees of Freedom • In statistics, the number of degrees of freedom is the number of values in the final calculation of a statistic that are free to vary • Remember, this is calculated as n-1 • For every n-value, there is the possibility of variance, so we account for that • Why n-1 then? • What if you had 25 chairs, and 25 students. As each student walked in the door, they would have a choice (freedom) of where to sit. Except for the last student. His choice is made. So you have 25-1, or 24, choices Properties of the t-Distribution 3. The total area under a t-curve is 1 or 100%. 4. The mean, median, and mode of the t-distribution are equal to zero. 5. As the degrees of freedom increase, the t-distribution approaches the normal distribution. After 30 d.f., the tdistribution is very close to the standard normal zdistribution. The tails in the tdistribution are “thicker” than those in the standard normal distribution. d.f. = 2 d.f. = 5 Standard normal curve t 0 © 2012 Pearson Education, Inc. All rights reserved. 41 of 83 T-Distribution Curve • The tails in a t-distribution are “thicker” than those in the standard normal distribution T-Distribution Curves • William S. Gosset developed this curve • He published it under the pseudo-name “Student” thus –if you Google a T-curve, you may find “Student’s T-Distribution” Example: Critical Values of t Find the critical value tc for a 95% confidence level when the sample size is 15. Solution: d.f. = n – 1 = 15 – 1 = 14 Table 5: t-Distribution tc = 2.145 © 2012 Pearson Education, Inc. All rights reserved. 44 of 83 Solution: Critical Values of t 95% of the area under the t-distribution curve with 14 degrees of freedom lies between t = ±2.145. c = 0.95 t –tc = –2.145 © 2012 Pearson Education, Inc. All rights reserved. tc = 2.145 45 of 83 Example (using a Calculator) • Finding the Critical Values of T 2ndVarsinvT(area,D.F.) Area is almost the same as the confidence level –but you have to add one tail to get (in essence) the right limit, or critical T-value. For example, find the critical value of T for a 95% confidence when the sample size is 15 Area = .95 + one tail (1+.95)/2 = .975 D.F. = 15-1 = 14 InvT(.975,14) = 2.1447 which is the T-Critical value Therefore, 95% of the area under the t-distribution curve with 14 degrees of freedom is between -2.145 and 2.145 Chapter 6 Quiz 2 Instructions • Using data from problem 64 on page 315, do the following: 1. Enter data into list 1 2. Calculate the sample standard deviation (s), the population standard deviation (Ơ) and sample mean (xbar) calc/1Var Stats 3. Calculate a 95% confidence interval test/Zinterval (be sure to use Ơ since you have it, and input data, not “stats”) Larson/Farber 5th ed Provide the answers on a piece of paper 1. 2. 3. 4. 5. 6. n= xbar= Ơ= S=_______ Z critical score= Margin of Error (E) = Confidence interval= __________< µ < ________ • Each question worth 5 points 47 Confidence Intervals for the Population Mean A c-confidence interval for the population mean μ s Same equation • x E x E where E tc n except use T instead of Z, and S instead of Ợ • The probability that the confidence interval contains μ is c. • This should look familiar, it is the SAME EQUATION we used to calculate the margin of Error in a normal distribution, except instead of a Zcritical score, we have a Tcritical score and we use the sample standard deviation © 2012 Pearson Education, Inc. All rights reserved. 48 of 83 On a Calculator • • • • • • • • • On a Calculator: Stat TestTinterval Choose Stats Enter sx Enter “0” for x-bar (we don’t know it) Enter n Enter c-Level Calculate Chose the positive answer Larson/Farber 5th ed 49 Confidence Intervals and t-Distributions In Words 1. Identify the sample statistics n, x , and s. 2. Identify the degrees of freedom, the level of confidence c, and the critical value tc. 3. Find the margin of error E. © 2012 Pearson Education, Inc. All rights reserved. In Symbols x (x x )2 x s n 1 n d.f. = n – 1 E tc s n 50 of 83 Confidence Intervals and t-Distributions In Words 4. Find the left and right endpoints and form the confidence interval. © 2012 Pearson Education, Inc. All rights reserved. In Symbols Left endpoint: x E Right endpoint: x E Interval: xE xE 51 of 83 Example: Constructing a Confidence Interval You randomly select 16 coffee shops and measure the temperature of the coffee sold at each. The sample mean temperature is 162.0ºF with a sample standard deviation of 10.0ºF. Find the 95% confidence interval for the population mean temperature. Assume the temperatures are approximately normally distributed. Solution: Use the t-distribution (n < 30, σ is unknown, temperatures are approximately normally distributed). © 2012 Pearson Education, Inc. All rights reserved. 52 of 83 Solution: Constructing a Confidence Interval • n =16, x = 162.0 s = 10.0 c = 0.95 • df = n – 1 = 16 – 1 = 15 InvT(.975,15) • Critical Value Table 5: t-Distribution tc = 2.131 © 2012 Pearson Education, Inc. All rights reserved. 53 of 83 Solution: Constructing a Confidence Interval • Margin of error: s 10 E tc 2.131 5.3 n 16 • Confidence interval: Left Endpoint: x E 162 5.3 156.7 Right Endpoint: xE 162 5.3 167.3 156.7 < μ < 167.3 © 2012 Pearson Education, Inc. All rights reserved. 54 of 83 Solution: Constructing a Confidence Interval • 156.7 < μ < 167.3 Point estimate 156.7 ( 162.0 x E • x 167.3 ) xE With 95% confidence, you can say that the population mean temperature of coffee sold is between 156.7ºF and 167.3ºF. © 2012 Pearson Education, Inc. All rights reserved. 55 of 83 Example • You randomly select 16 restaurants and measure the temperature of the coffee • The sample mean is 162 degrees (F) • The sample Standard Deviation (S) = 10 deg. • Find the 95% confidence interval for the mean temperature. Assume they are normally distributed • Go to: StatTestTinterval. Choose “Stats” and enter information • You should get: 156.57---167.33, with a mean of 162 Normal or t-Distribution? Is n ≥ 30? Yes No Is the population normally, or approximately normally, distributed? No Cannot use the normal distribution or the t-distribution. Yes Use the normal distribution with Yes Is σ known? E zc No Use the t-distribution with E tc Use the normal distribution with σ E zc n If σ is unknown, use s instead. σ . n s n and n – 1 degrees of freedom. © 2012 Pearson Education, Inc. All rights reserved. 57 of 83 Example: Normal or t-Distribution? You randomly select 25 newly constructed houses. The sample mean construction cost is $181,000 and the population standard deviation is $28,000. Assuming construction costs are normally distributed, should you use the normal distribution, the t-distribution, or neither to construct a 95% confidence interval for the population mean construction cost? Solution: Use the normal distribution (the population is normally distributed and the population standard deviation is known) © 2012 Pearson Education, Inc. All rights reserved. 58 of 83 Section 6.2 Summary • Interpreted the t-distribution and used a t-distribution table • Constructed confidence intervals when n < 30, the population is normally distributed, and σ is unknown © 2012 Pearson Education, Inc. All rights reserved. 59 of 83 Assignment • Page 323 1-34 Skip 13-16 Larson/Farber 5th ed 60 Chapter 6 Quiz 3 • Using data from problem 24 on page 324, and assuming a 98% confidence level, answer the following: 1. n= 2. s= 5 points each question 3. D.F.= 4. xbar= 5. Tcritical= 6. Confidence interval (based on 98% confidence level) 7. Hint: Use InvT for question 5 8. Hint: Use Tinterval for question 6 Larson/Farber 5th ed 61 Chapter 6 Quiz 4 • Solve question #35 on page 325 1. Calculate the confidence interval indicated in the problem 2. Is the company making acceptable tennis balls? 3. Why or why not? Larson/Farber 5th ed 62 Section 6.3 Confidence Intervals for Population Proportions © 2012 Pearson Education, Inc. All rights reserved. 63 of 83 Section 6.3 Objectives • Find a point estimate for the population proportion • Construct a confidence interval for a population proportion • Determine the minimum sample size required when estimating a population proportion © 2012 Pearson Education, Inc. All rights reserved. 64 of 83 Point Estimate for Population p Population Proportion • The probability of success in a single trial of a binomial experiment. • Denoted by p Point Estimate for p • The proportion of successes in a sample. • Denoted by x number of successes in sample pˆ n sample size read as “p hat” © 2012 Pearson Education, Inc. All rights reserved. 65 of 83 Point Estimate for Population p Estimate Population with Sample Parameter… Statistic Proportion: p p̂ Point Estimate for q, the population proportion of failures • Denoted by qˆ 1 pˆ • Read as “q hat” © 2012 Pearson Education, Inc. All rights reserved. 66 of 83 Example: Point Estimate for p In a survey of 1000 U.S. adults, 662 said that it is acceptable to check personal e-mail while at work. Find a point estimate for the population proportion of U.S. adults who say it is acceptable to check personal e-mail while at work. (Adapted from Liberty Mutual) Solution: n = 1000 and x = 662 x 662 pˆ 0.662 66.2% n 1000 © 2012 Pearson Education, Inc. All rights reserved. 67 of 83 Confidence Intervals for p A c-confidence interval for a population proportion p • pˆ E p pˆ E where E zc pq ˆˆ n •The probability that the confidence interval contains p is c. © 2012 Pearson Education, Inc. All rights reserved. 68 of 83 Constructing Confidence Intervals for p In Words In Symbols 1. Identify the sample statistics n and x. 2. Find the point estimate p̂. 3. Verify that the sampling distribution of p̂ can be approximated by a normal distribution. 4. Find the critical value zc that corresponds to the given level of confidence c. © 2012 Pearson Education, Inc. All rights reserved. pˆ x n npˆ 5, nqˆ 5 Use the Standard Normal Table or technology. 69 of 83 Constructing Confidence Intervals for p In Words 5. Find the margin of error E. 6. Find the left and right endpoints and form the confidence interval. © 2012 Pearson Education, Inc. All rights reserved. In Symbols E zc pq ˆˆ n Left endpoint: p̂ E Right endpoint: p̂ E Interval: pˆ E p pˆ E 70 of 83 Example: Confidence Interval for p In a survey of 1000 U.S. adults, 662 said that it is acceptable to check personal e-mail while at work. Construct a 95% confidence interval for the population proportion of U.S. adults who say that it is acceptable to check personal e-mail while at work. Solution: Recall pˆ 0.662 qˆ 1 pˆ 1 0.662 0.338 © 2012 Pearson Education, Inc. All rights reserved. 71 of 83 Solution: Confidence Interval for p • Verify the sampling distribution of p̂ can be approximated by the normal distribution npˆ 1000 0.662 662 5 nqˆ 1000 0.338 338 5 • Margin of error: E zc pq ˆ ˆ 1.96 (0.662) (0.338) 0.029 n 1000 © 2012 Pearson Education, Inc. All rights reserved. 72 of 83 Solution: Confidence Interval for p • Confidence interval: Left Endpoint: pˆ E 0.662 0.029 0.633 Right Endpoint: pˆ E 0.662 0.029 0.691 0.633 < p < 0.691 © 2012 Pearson Education, Inc. All rights reserved. 73 of 83 Solution: Confidence Interval for p • 0.633 < p < 0.691 Point estimate p̂ E p̂ p̂ E With 95% confidence, you can say that the population proportion of U.S. adults who say that it is acceptable to check personal e-mail while at work is between 63.3% and 69.1%. © 2012 Pearson Education, Inc. All rights reserved. 74 of 83 On a Calculator • You can calculate a p confidence interval on a calculator • Stats Tests1PropZInt • Enter X (the number of successes in the sample) • Enter n (the size of the sample) • Enter the confidence level • This will give you the confidence level, and the p-hat • It will not give you the Margin of Error (E) should you need that, you gotta go old school Larson/Farber 5th ed 75 Sample Size • Given a c-confidence level and a margin of error E, the minimum sample size n needed to estimate p is 2 zc ˆ ˆ n pq E • This formula assumes you have an estimate for p̂ and qˆ . • If not, use pˆ 0.5 and qˆ 0.5. © 2012 Pearson Education, Inc. All rights reserved. 76 of 83 Example: Sample Size You are running a political campaign and wish to estimate, with 95% confidence, the population proportion of registered voters who will vote for your candidate. Your estimate must be accurate within 3% of the true population proportion. Find the minimum sample size needed if 1. no preliminary estimate is available. Solution: Because you do not have a preliminary estimate for pˆ , use pˆ 0.5 and qˆ 0.5. © 2012 Pearson Education, Inc. All rights reserved. 77 of 83 Solution: Sample Size • c = 0.95 zc = 1.96 2 E = 0.03 2 zc 1.96 ˆ ˆ (0.5)(0.5) n pq 1067.11 0.03 E Round up to the nearest whole number. With no preliminary estimate, the minimum sample size should be at least 1068 voters. © 2012 Pearson Education, Inc. All rights reserved. 78 of 83 Example: Sample Size You are running a political campaign and wish to estimate, with 95% confidence, the population proportion of registered voters who will vote for your candidate. Your estimate must be accurate within 3% of the true population proportion. Find the minimum sample size needed if 2. a preliminary estimate gives pˆ 0.31. Solution: Use the preliminary estimate pˆ 0.31 qˆ 1 pˆ 1 0.31 0.69 © 2012 Pearson Education, Inc. All rights reserved. 79 of 83 Solution: Sample Size • c = 0.95 zc = 1.96 2 E = 0.03 2 zc 1.96 ˆ ˆ (0.31)(0.69) n pq 913.02 0.03 E Round up to the nearest whole number. With a preliminary estimate of pˆ 0.31, the minimum sample size should be at least 914 voters. Need a larger sample size if no preliminary estimate is available. © 2012 Pearson Education, Inc. All rights reserved. 80 of 83 Section 6.3 Summary • Found a point estimate for the population proportion • Constructed a confidence interval for a population proportion • Determined the minimum sample size required when estimating a population proportion © 2012 Pearson Education, Inc. All rights reserved. 81 of 83 Assignment • Page 332 11-24 Larson/Farber 5th ed 82 Chapter 6 Quiz 5 You are running a political campaign and wish to estimate, with 98% confidence, the population proportion of registered voters who will vote for your candidate. Your estimate must be accurate within 3% of the true population proportion. Find the minimum sample size needed. (hint: solve for n) You do not have a preliminary point estimate to the population proportion. Show equation 5 points for equation, 5 points for solution © 2012 Pearson Education, Inc. All rights reserved. 83 of 83 Solution: Sample Size • c = 0.98 zc = 2.33 2 E = 0.03 2 zc 2.33 1.96 ˆ ˆ (0.31)(0.69) n pq 1508.02 .50 .50 913.02 0.03 E Round up to the nearest whole number. Using the standard value of .50 for p-hat, and .50 for q-hat, the minimum sample size should be at least 1509 voters. © 2012 Pearson Education, Inc. All rights reserved. 84 of 83 Section 6.4 Confidence Intervals for Variance and Standard Deviation © 2012 Pearson Education, Inc. All rights reserved. 85 of 83 Section 6.4 Objectives • Interpret the chi-square distribution and use a chi-square distribution table • Use the chi-square distribution to construct a confidence interval for the variance and standard deviation © 2012 Pearson Education, Inc. All rights reserved. 86 of 83 The Chi-Square Distribution • The point estimate for σ2 is s2 • The point estimate for σ is s • s2 is the most unbiased estimate for σ2 Estimate Population with Sample Parameter… Statistic Variance: σ2 s2 Standard deviation: σ s © 2012 Pearson Education, Inc. All rights reserved. 87 of 83 The Chi-Square Distribution • You can use the chi-square distribution to construct a confidence interval for the variance and standard deviation. • If the random variable x has a normal distribution, then the distribution of 2 (n 1)s 2 σ2 forms a chi-square distribution for samples of any size n > 1. © 2012 Pearson Education, Inc. All rights reserved. 88 of 83 Properties of The Chi-Square Distribution 1. All chi-square values χ2 are greater than or equal to zero. 2. The chi-square distribution is a family of curves, each determined by the degrees of freedom. To form a confidence interval for σ2, use the χ2-distribution with degrees of freedom equal to one less than the sample size. • d.f. = n – 1 Degrees of freedom 3. The area under each curve of the chi-square distribution equals one. © 2012 Pearson Education, Inc. All rights reserved. 89 of 83 Properties of The Chi-Square Distribution 4. Chi-square distributions are positively skewed. Chi-square Distributions © 2012 Pearson Education, Inc. All rights reserved. 90 of 83 Critical Values for χ2 • There are two critical values for each level of confidence. • The value χ2R represents the right-tail critical value • The value χ2L represents the left-tail critical value. 1 c 2 c The area between the left and right critical values is c. 1 c 2 L2 R2 © 2012 Pearson Education, Inc. All rights reserved. χ2 91 of 83 Example: Finding Critical Values for χ2 Find the critical values R2 and L2 for a 95% confidence interval when the sample size is 18. Use Table 6 on A-19 Solution: • d.f. = n – 1 = 18 – 1 = 17 d.f. • Each area in the table represents the region under the chi-square curve to the right of the critical value. • Area to the right of 1 c 2 χ R= 2 • Area to the right of 1 c 2 χ L= 2 © 2012 Pearson Education, Inc. All rights reserved. 1 0.95 0.025 2 1 0.95 0.975 2 92 of 83 Solution: Finding Critical Values for χ2 Table 6: χ2-Distribution L2 7.564 R2 30.191 95% of the area under the curve lies between 7.564 and 30.191. © 2012 Pearson Education, Inc. All rights reserved. 93 of 83 Confidence Intervals for σ2 and σ Confidence Interval for σ2: 2 2 ( n 1) s ( n 1) s 2 • σ 2 R L2 Confidence Interval for σ: • (n 1)s 2 R2 σ (n 1)s 2 L2 • The probability that the confidence intervals contain σ2 or σ is c. © 2012 Pearson Education, Inc. All rights reserved. 94 of 83 Confidence Intervals for σ2 and σ In Words In Symbols 1. Verify that the population has a normal distribution. 2. Identify the sample statistic n and the degrees of freedom. 3. Find the point estimate s2. 4. Find the critical values χ2R and χ2L that correspond to the given level of confidence c. © 2012 Pearson Education, Inc. All rights reserved. d.f. = n – 1 2 ( x x ) s2 n 1 Use Table 6 in Appendix B. 95 of 83 Confidence Intervals for 2 and In Words 5. Find the left and right endpoints and form the confidence interval for the population variance. 6. Find the confidence interval for the population standard deviation by taking the square root of each endpoint. © 2012 Pearson Education, Inc. All rights reserved. In Symbols (n 1)s 2 R2 (n 1)s 2 R2 σ2 σ (n 1)s 2 L2 (n 1)s 2 L2 96 of 83 Example: Constructing a Confidence Interval You randomly select and weigh 30 samples of an allergy medicine. The sample standard deviation is 1.20 milligrams. Assuming the weights are normally distributed, construct 99% confidence intervals for the population variance and standard deviation. Solution: • d.f. = n – 1 = 30 – 1 = 29 d.f. © 2012 Pearson Education, Inc. All rights reserved. 97 of 83 Solution: Constructing a Confidence Interval 1 c 1 0.99 0.005 R= 2 2 • Area to the right of χ2 • Area to the right of 1 c 2 χ L= 2 1 0.99 0.995 2 • The critical values are χ2R = 52.336 and χ2L = 13.121 © 2012 Pearson Education, Inc. All rights reserved. 98 of 83 Solution: Constructing a Confidence Interval Confidence Interval for σ2: Left endpoint: (n 1)s 2 R2 (30 1)(1.20)2 0.80 52.336 2 2 ( n 1) s (30 1)(1.20) Right endpoint: 3.18 2 13.121 L 0.80 < σ2 < 3.18 With 99% confidence, you can say that the population variance is between 0.80 and 3.18. © 2012 Pearson Education, Inc. All rights reserved. 99 of 83 Solution: Constructing a Confidence Interval Confidence Interval for σ : (n 1)s 2 2 R σ (n 1)s 2 2 L (30 1)(1.20) 2 (30 1)(1.20) 2 52.336 13.121 0.89 < σ < 1.78 With 99% confidence, you can say that the population standard deviation is between 0.89 and 1.78 milligrams. © 2012 Pearson Education, Inc. All rights reserved. 100 of 83 Section 6.4 Summary • Interpreted the chi-square distribution and used a chi-square distribution table • Used the chi-square distribution to construct a confidence interval for the variance and standard deviation © 2012 Pearson Education, Inc. All rights reserved. 101 of 83 Assignment • Page 341 3-12 Larson/Farber 5th ed 102