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Chapter 6 Review 6-1 Preparation 6-2 The Standard Normal Distribution 6-3 Applications of Normal Distributions 6-4 Sampling Distributions and Estimators 6-5 The Central Limit Theorem 6-6 Assessing Normality 6-7 Normal as Approximation to Binomial Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-1 Example Bone mineral density test scores are normally distributed with a mean of 0 and a standard deviation of 1. . These scores can be helpful in identifying the presence of osteoporosis. The result of the test is can be measured as z scores. If a randomly selected adult undergoes a bone density test, find the probability that the result is a reading less than 1.27. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-2 Look at Table A-2 The are from the left hand side of the curve to 1.27 is .8980 note that the hundredths are on top. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-3 Here is a picture that describes this P( z 1.27) 0.8980 The probability of random adult having a bone density less than 1.27 is 0.8980. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-4 Example 2 Using the same bone density test, find the probability that a randomly selected person has a result above –1.00 (which is considered to be in the “normal” range of bone density readings. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-5 Example 2 – continued Since we are looking for the area to the right we have to subtract. The probability of a randomly selected adult having a bone density above –1 is 0.8413. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-6 Example 3 Find this probability of getting a bone density reading between –1.00 and –2.50. This indicates the subject has osteopenia. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-7 Example 3 1. Find the area to the left of z = –2.50 0.0062. 2. Find the area to the left of z = –1.00 0.1587. 3. The area between z = –2.50 and z = –1.00 is the difference between these areas. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-8 Example 4 Find the 95th Percentile. This is the z score which separates the lower 95%. 5% or 0.05 This is an example of finding a z score from an area. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-9 Solution 5% or 0.05 1.645 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-10 Example Find the value of z0.025. This means find the 97.5 percentile. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-11 Solution The notation z0.025 is used to represent the z score with an area of 0.025 to its right. Referring back to the bone density example, z0.025 = 1.96. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-12 Chapter 6 Normal Probability Distributions 6-1 Preparation 6-2 The Standard Normal Distribution 6-5 The Central Limit Theorem 6-6 Assessing Normality 6-7 Normal as Approximation to Binomial Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-13 6-3 Applications of Normal Distributions Example Tall Clubs International has a requirement that women must be at least 70 inches tall. Given that women have normally distributed heights with a mean of 63.8 inches and a standard deviation of 2.6 inches, find the percentage of women who satisfy that height requirement. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-14 Solution Convert to a z score and use Table A-2 or technology to find the shaded area. z x 70 63.8 2.38 2.6 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-15 Solution Draw the normal distribution and shade the region. The area to the right of 2.38 is 0.008656, and so about 0.87% of all women meet the requirement. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-16 Example Men’s heights are normally distributed with a mean of 69.5 inches and a standard deviation of 2.4 inches. When designing aircraft cabins, what ceiling height will allow 95% of men to stand without bumping their heads? This is an example of finding a z score from an area. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-17 Solution With z = 1.645, μ = 69.5, and σ = 2.4. we can solve for x. x z 69.5 1.645 2.4 73.448 inches Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-18 6-4 Sampling Distributions and Estimators Example 2 x Roll a die 5 times. Find the mean , variance s, and the proportion of odd numbers of the results. What will happen if you continue this process? What will the mean of all of your sample means approach? Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-19 Example – Elevators Assume male weights follow a normal distribution with a mean of 182.9 lb and a standard deviation of 40.8 lb. Suppose an elevator has a maximum capacity of 16 passengers with a total weight of 2500 lb. Assuming a worst case scenario in which the passengers are all male, what are the chances the elevator is overloaded? a. Find the probability that 1 randomly selected male has a weight greater than 156.25 lb. b. Find the probability that a sample of 16 males have a mean weight greater than 156.25 lb (which puts the total weight at 2500 lb, exceeding the maximum capacity). Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-20 Solution a. Find the probability that 1 randomly selected male has a weight greater than 156.25 lb. Use the methods presented in Section 6.3. We can convert to a z score and use Table A-2. z x 156.25 182.9 0.65 40.8 Using Table A-2, the area to the right is 0.7422. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-21 Solution b. Find the probability that a sample of 16 males have a mean weight greater than 156.25 lb. Since the distribution of male weights is assumed to be normal, the sample mean will also be normal. x x 182.9 x 40.8 x 10.2 n 16 Converting to z: 156.25 182.9 z 2.61 10.2 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-22 Solution b. Find the probability that a sample of 16 males have a mean weight greater than 156.25 lb. While there is 0.7432 probability that any given male will weigh more than 156.25 lb, there is a 0.9955 probability that the sample of 16 males will have a mean weight of 156.25 lb or greater. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-23 Review Binomial Probability Distribution 1. The procedure must have a fixed number of trials. 2. The trials must be independent. 3. Each trial must have all outcomes classified into two categories (commonly, success and failure). 4. The probability of success remains the same in all trials. Solve by binomial probability formula, Table A-1, or technology. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-24 Approximation of a Binomial Distribution with a Normal Distribution np 5 nq 5 then np , npq and the random variable has a distribution. (normal) Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-25 Example – NFL Coin Toss In 431 NFL football games that went to over time, the teams that won the coin toss went on to win 235 of those games. What is the probability of such an event occurring assuming there is a 0.5 probability of winning a game after winning the coin toss? Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-26 Solution If the coin-toss method is fair, teams winning the toss would win about 50% of the games (we’d expect 215.5 wins in 431 overtime games). The given problem involves a binomial distribution with n = 431 trials and an assumed probability of success of p = 0.5. Use the normal approximation to the binomial distribution. Step 1: The conditions check: np 431 0.5 215.5 5 nq 431 0.5 215.5 5 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-27 Example – NFL Coin Toss Step 2: Find the mean and standard deviation of the normal distribution: np 431 0.5 215.5 npq 431 0.5 0.5 10.38027 Step 3: We want the probability of at least 235 wins, so x = 235. Step 4: The vertical strip will go from 234.5 to 235.5. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-28 Example – NFL Coin Toss Step 5: We will shade the area to the right of 234.5. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-29 Example – NFL Coin Toss Step 6: Find the z score and use technology or Table A-2 to determine the probability. z x 234.5 215.5 1.83 10.380270 The probability is 0.0336 for the coin flip winning team to win at least 235 games. This probability is low enough to suggest the team winning coin flip has an unfair advantage. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-30 Don’t Forget When we use the normal distribution (which is a continuous probability distribution) as an approximation to the binomial distribution (which is discrete), a continuity correction is made to a discrete whole number x in the binomial distribution by representing the discrete whole number x by the interval from x – 0.5 to x + 0.5 (that is, adding and subtracting 0.5). Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-31 Example In the Chapter Problem we noted that a Pew Research Center poll of 1007 randomly selected adults showed that 85% of respondents know what Twitter is. The sample results are n = 1007 and pˆ 0.70. a. Find the margin of error E that corresponds to a 95% confidence level. b. Find the 95% confidence interval estimate of the population proportion p. c. Based on the results, can we safely conclude that more than 75% of adults know what Twitter is? d. Assuming that you are a newspaper reporter, write a brief statement that accurately describes the results and includes all of the relevant information. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-32 Example - Continued Requirement check: simple random sample; fixed number of trials, 1007; trials are independent; two outcomes per trial; probability remains constant. Note: number of successes and failures are both at least 5. a) Use the formula to find the margin of error. ˆˆ pq E z 2 1.96 n E 0.0220545 0.85 0.15 1007 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-33 Example - Continued b) The 95% confidence interval: pˆ E p pˆ E 0.85 0.0220545 p 0.85 0.0220545 0.828 p 0.872 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-34 Example - Continued c) Based on the confidence interval obtained in part (b), it does appear that more than 75% of adults know what Twitter is. Because the limits of 0.828 and 0.872 are likely to contain the true population proportion, it appears that the population proportion is a value greater than 0.75. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-35 Example - Continued d) Here is one statement that summarizes the results: 85% of U.S. adults know what Twitter is. That percentage is based on a Pew Research Center poll of 1007 randomly selected adults. In theory, in 95% of such polls, the percentage should differ by no more than 2.2 percentage points in either direction from the percentage that would be found by interviewing all adults in the United States. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-36 Example Many companies are interested in knowing the percentage of adults who buy clothing online. How many adults must be surveyed in order to be 95% confident that the sample percentage is in error by no more than three percentage points? a. Use a recent result from the Census Bureau: 66% of adults buy clothing online. b. Assume that we have no prior information suggesting a possible value of the proportion. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-37 Example - Continued a) Use pˆ 0.66 and qˆ 1 pˆ 0.34 0.05 so z 2 1.96 E 0.03 z n 2 ˆˆ pq 2 E2 1.96 0.66 0.34 2 0.03 2 To be 95% confident that our sample percentage is within three percentage points of the true percentage for all adults, we should obtain a simple random sample of 958 adults. 957.839 958 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-38 Example - Continued b) Use 0.05 so z 2 1.96 E 0.03 z n 2 2 0.25 E2 1.96 0.25 2 0.03 2 To be 95% confident that our sample percentage is within three percentage points of the true percentage for all adults, we should obtain a simple random sample of 1068 adults. 1067.1111 1068 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-39 Example A common claim is that garlic lowers cholesterol levels. In a test of the effectiveness of garlic, 49 subjects were treated with doses of raw garlic, and their cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol (in mg/dL) have a mean of 0.4 and a standard deviation of 21.0. Use the sample statistics of n = 49, x = 0.4, and s = 21.0 to construct a 95% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-40 Example - Continued Requirements are satisfied: simple random sample and n = 49 (i.e., n > 30). 95% implies α = 0.05. With n = 49, the df = 49 – 1 = 48 Closest df is 50, two tails, so = 2.009 Using = 2.009, s = 21.0 and n = 49 the margin of error is: E t 2 21.0 2.009 6.027 n 49 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-41 Example - Continued Construct the confidence interval: x 0.4, E 6.027 x E x E 0.4 6.027 0.4 6.027 5.6 6.4 We are 95% confident that the limits of –5.6 and 6.4 actually do contain the value of μ, the mean of the changes in LDL cholesterol for the population. Because the confidence interval limits contain the value of 0, it is very possible that the mean of the changes in LDL cholesterol is equal to 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels. It does not appear that the garlic treatment is effective in lowering LDL cholesterol. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-42 Important Properties of the Student t Distribution 1. The Student t distribution is different for different sample sizes. (See the following slide for the cases n = 3 and n = 12.) 2. The Student t distribution has the same general symmetric bell shape as the standard normal distribution but it reflects the greater variability (with wider distributions) that is expected with small samples. 3. The Student t distribution has a mean of t = 0 (just as the standard normal distribution has a mean of z = 0). 4. The standard deviation of the Student t distribution varies with the sample size and is greater than 1 (unlike the standard normal distribution, which has σ = 1). 5. As the sample size n gets larger, the Student t distribution gets closer to the normal distribution. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-43 Student t Distributions for n = 3 and n = 12 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-44 Finding the Point Estimate and E from a Confidence Interval Point estimate of μ: = (upper confidence limit) + (lower confidence limit) 2 Margin of Error: = (upper confidence limit) – (lower confidence limit) 2 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-45 Finding a Sample Size for Estimating a Population Mean = population mean = population standard deviation x = sample mean = desired margin of error = z score separating an area of the standard normal distribution in the right tail of Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-46 Round-Off Rule for Sample Size n If the computed sample size n is not a whole number, round the value of n up to the next larger whole number. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-47 Finding the Sample Size n When σ is Unknown 1. Use the range rule of thumb (see Section 3-3) to estimate the standard deviation as follows: . 2. Start the sample collection process without knowing σ and, using the first several values, calculate the sample standard deviation s and use it in place of σ. The estimated value of σ can then be improved as more sample data are obtained, and the sample size can be refined accordingly. 3. Estimate the value of σ by using the results of some other earlier study. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-48 Example Assume that we want to estimate the mean IQ score for the population of statistics students. How many statistics students must be randomly selected for IQ tests if we want 95% confidence that the sample mean is within 3 IQ points of the population mean? α α/2 zα/2 E σ = 0.05 = 0.025 = 1.96 = 3 = 15 With a simple random sample of only 97 statistics students, we will be 95% confident that the sample mean is within 3 IQ points of the true population mean μ. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-49 Part 2: Key Concept This section presents methods for estimating a population mean. In addition to knowing the values of the sample data or statistics, we must also know the value of the population standard deviation, σ. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-50 Confidence Interval for Estimating a Population Mean (with σ Known) μ = population mean x = sample mean σ = population standard deviation n = number of sample values E = margin of error z α/2 = critical z value separating an area of α/2 in the right tail of the standard normal distribution Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-51 Confidence Interval for Estimating a Population Mean (with σ Known) 1. The sample is a simple random sample. (All samples of the same size have an equal chance of being selected.) 2. The value of the population standard deviation σ is known. 3. Either or both of these conditions is satisfied: The population is normally distributed or n > 30. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-52 Confidence Interval for Estimating a Population Mean (with σ Known) x E x E where E z 2 or or n x E x E,x E Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-53 Example People have died in boat and aircraft accidents because an obsolete estimate of the mean weight of men was used. In recent decades, the mean weight of men has increased considerably, so we need to update our estimate of that mean so that boats, aircraft, elevators, and other such devices do not become dangerously overloaded. Using the weights of men from a random sample, we obtain these sample statistics for the simple random sample: n = 40 and x = 172.55 lb. Research from several other sources suggests that the population of weights of men has a standard deviation given by σ = 26 lb. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-54 Example - Continued a. Find the best point estimate of the mean weight of the population of all men. b. Construct a 95% confidence interval estimate of the mean weight of all men. c. What do the results suggest about the mean weight of 166.3 lb that was used to determine the safe passenger capacity of water vessels in 1960 (as given in the National Transportation and Safety Board safety recommendation M-04-04)? Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-55 Example - Continued a. The sample mean of 172.55 lb is the best point estimate of the mean weight of the population of all men. b. A 95% confidence interval implies that α = 0.05, so zα/2= 1.96. Calculate the margin of error. 26 E z 2 1.96 8.0574835 n 40 Construct the confidence interval. x E x E 172.55 8.0574835 172.55 8.0574835 164.49 180.61 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-56 Example - Continued c. Based on the confidence interval, it is possible that the mean weight of 166.3 lb used in 1960 could be the mean weight of men today. However, the best point estimate of 172.55 lb suggests that the mean weight of men is now considerably greater than 166.3 lb. Considering that an underestimate of the mean weight of men could result in lives lost through overloaded boats and aircraft, these results strongly suggest that additional data should be collected. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-57 Choosing the Appropriate Distribution Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-58 Choosing the Appropriate Distribution Use the normal (z) distribution σ known and normally distributed population or n > 30 Use t distribution σ not known and normally distributed population or n > 30 Use a nonparametric method or bootstrapping Population is not normally distributed and n ≤ 30 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-59 Chapter 7 Estimates and Sample Sizes 7-1 Review and Preview 7-2 Estimating a Population Proportion 7-3 Estimating a Population Mean 7-4 Estimating a Population Standard Deviation or Variance Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-60 Key Concept This section we introduce the chi-square probability distribution so that we can construct confidence interval estimates of a population standard deviation or variance. We also present a method for determining the sample size required to estimate a population standard deviation or variance. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-61 Chi-Square Distribution In a normally distributed population with variance σ2 , assume that we randomly select independent samples of size n and, for each sample, compute the sample variance s2 (which is the square of the sample standard deviation s). The sample statistic χ2 (pronounced chi-square) has a sampling distribution called the chi-square distribution. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-62 Chi-Square Distribution 2 n 1 s 2 2 where s s2 σ2 df = sample size = sample variance = population variance = n – 1 degrees of freedom Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-63 Properties of the Distribution of the Chi-Square Statistic 1. The chi-square distribution is not symmetric, unlike the normal and Student t distributions. As the number of degrees of freedom increases, the distribution becomes more symmetric. Chi-Square Distribution Chi-Square Distribution for df = 10 and df = 20 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-64 Properties of the Distribution of the Chi-Square Statistic 2. The values of chi-square can be zero or positive, but it cannot be negative. 3. The chi-square distribution is different for each number of degrees of freedom, which is df = n – 1. As the number of degrees of freedom increases, the chi-square distribution approaches a normal distribution. In Table A-4, each critical value of χ2 corresponds to an area given in the top row of the table, and that area represents the cumulative area located to the right of the critical value. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-65 Example A simple random sample of 22 IQ scores is obtained. Construction of a confidence interval for the population standard deviation σ requires the left and right critical values of χ2 corresponding to a confidence level of 95% and a sample size of n = 22. Find the critical χ2 values corresponding to a 95% level of confidence. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-66 Example - Continued Critical Values of the Chi-Square Distribution Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-67 Estimators of σ2 The sample variance s2 is the best point estimate of the population variance σ2. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-68 Estimator of σ The sample standard deviation s is a commonly used point estimate of σ (even though it is a biased estimate). Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-69 Confidence Interval for Estimating a Population Standard Deviation or Variance σ σ2 s s2 n E L2 R2 = = = = = = = = population standard deviation population variance sample standard deviation sample variance number of sample values margin of error 2 left-tailed critical value of 2 right-tailed critical value of Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-70 Confidence Interval for Estimating a Population Standard Deviation or Variance Requirements: 1. The sample is a simple random sample. 2. The population must have normally distributed values (even if the sample is large). Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-71 Confidence Interval for Estimating a Population Standard Deviation or Variance Confidence Interval for the Population Variance σ2 n 1s 2 R 2 2 n 1s Copyright © 2014, 2012, 2010 Pearson Education, Inc. 2 2 L Section 6.3-72 Confidence Interval for Estimating a Population Standard Deviation or Variance Confidence Interval for the Population Standard Deviation σ n 1s 2 R 2 n 1s Copyright © 2014, 2012, 2010 Pearson Education, Inc. 2 2 L Section 6.3-73 Procedure for Constructing a Confidence Interval for σ or σ2 1. Verify that the required assumptions are satisfied. 2. Using n – 1 degrees of freedom, refer to Table A-4 or use technology to find the critical values R2 and L2 that correspond to the desired confidence level. 3. Evaluate the upper and lower confidence interval limits using this format of the confidence interval: n 1s R2 2 2 n 1s 2 L2 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-74 Procedure for Constructing a Confidence Interval for σ or σ2 4. If a confidence interval estimate of σ is desired, take the square root of the upper and lower confidence interval limits and change σ2 to σ. 5. Round the resulting confidence level limits. If using the original set of data to construct a confidence interval, round the confidence interval limits to one more decimal place than is used for the original set of data. If using the sample standard deviation or variance, round the confidence interval limits to the same number of decimal places. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-75 Caution Confidence intervals can be used informally to compare the variation in different data sets, but the overlapping of confidence intervals should not be used for making formal and final conclusions about equality of variances or standard deviations. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-76 Example A group of 22 subjects took an IQ test during part of a study. The subjects had a standard deviation IQ score of 14.3. Construct a 95% confidence interval estimate of σ, the standard deviation of the population from which the sample was obtained. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-77 Example - Continued We must first verify the requirements are met. 1. We can treat the sample as a simple random sample. 2. The following display shows a histogram of the data and the normality assumption can be reasonably met. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-78 Example - Continued n = 22 so df = 22 – 1 = 21 Use table A-4 to find: 10.283 and 35.479 2 L 2 R n 1 s 2 R2 22 114.3 35.479 2 n 1 s 2 L2 2 2 22 114.3 2 10.283 Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-79 Example - Continued Evaluation of the preceding expression yields: 121.0 417.6 2 Finding the square root of each part (before rounding), then rounding to two decimal places, yields this 95% confidence interval estimate of the population standard deviation: 11.0 20.4 Based on this result, we have 95% confidence that the limits of 11.0 and 20.4 contain the true value of σ. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-80 Determining Sample Sizes The procedures for finding the sample size necessary to estimate σ2 are much more complex than the procedures given earlier for means and proportions. Instead of using very complicated procedures, we will use Table 7-2. STATDISK also provides sample sizes. With STATDISK, select Analysis, Sample Size Determination, and then Estimate St Dev. Minitab, Excel, and the TI-83/84 Plus calculator do not provide such sample sizes. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-81 Determining Sample Sizes Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-82 Example We want to estimate the standard deviation σ of all voltage levels in a home. We want to be 95% confident that our estimate is within 20% of the true value of σ. How large should the sample be? Assume that the population is normally distributed. From Table 7-2, we can see that 95% confidence and an error of 20% for σ correspond to a sample of size 48. We should obtain a simple random sample of 48 voltage levels form the population of voltage levels. Copyright © 2014, 2012, 2010 Pearson Education, Inc. Section 6.3-83