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“Blood Pressure of Males and Females.” StatCruch, 2013. Available online at http://www.statcrunch.com/5.0/ viewreport.php?reportid=11960 (accessed May 14, 2013). “The Use of Epidemiological Tools in Conflict-affected populations: Open-access educational resources for policy-makers: Calculation of at http://conflict.lshtm.ac.uk/page_125.htm (accessed May 14, 2013). and Tropical Medicine, 2009. Available online z-scores.” London School of Hygiene “2012 College-Bound Seniors Total Group Profile Report.” CollegeBoard, http://media.collegeboard.com/digitalServices/pdf/research/TotalGroup-2012.pdf (accessed May 14, 2013). 2012. Available online at “Digest of Education Statistics: ACT score average and standard deviations by sex and race/ethnicity and percentage of ACT test takers, by selected composite score ranges and planned fields of study: Selected years, 1995 through 2009.” National Center for Education Statistics. Available online at http://nces.ed.gov/programs/digest/d09/tables/dt09_147.asp (accessed May 14, 2013). Data from the San Jose Mercury News. Data from The World Almanac and Book of Facts. “List of stadiums by capacity.” Wikipedia. Available online at https://en.wikipedia.org/wiki/List_of_stadiums_by_capacity (accessed May 14, 2013). Data from the National Basketball Association. Available online at www.nba.com (accessed May 14, 2013). 6.2 Using the Normal Distribution “Naegele’s rule.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Naegele's_rule (accessed May 14, 2013). “403: NUMMI.” Chicago Public Media & Ira Glass, 2013. Available online at http://www.thisamericanlife.org/radioarchives/episode/403/nummi (accessed May 14, 2013). Tips.” WinAtTheLottery.com, Playing “Scratch-Off http://www.winatthelottery.com/public/department40.cfm (accessed May 14, 2013). Lottery Ticket 2013. Available |
online at “Smart Phone Users, By The Numbers.” Visual.ly, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed May 14, 2013). “Facebook Statistics.” Statistics Brain. Available online at http://www.statisticbrain.com/facebook-statistics/(accessed May 14, 2013). SOLUTIONS 1 ounces of water in a bottle 3 2 5 –4 7 –2 9 The mean becomes zero. 11 z = 2 13 z = 2.78 15 x = 20 17 x = 6.5 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 6 | THE NORMAL DISTRIBUTION 365 19 x = 1 21 x = 1.97 23 z = –1.67 25 z ≈ –0.33 27 0.67, right 29 3.14, left 31 about 68% 33 about 4% 35 between –5 and –1 37 about 50% 39 about 27% 41 The lifetime of a Sunshine CD player measured in years. 43 P(x < 1) 45 Yes, because they are the same in a continuous distribution: P(x = 1) = 0 47 1 – P(x < 3) or P(x > 3) 49 1 – 0.543 = 0.457 51 0.0013 53 56.03 55 0.1186 57 a. Check student’s solution. b. 3, 0.1979 59 a. Check student’s solution. b. 0.70, 4.78 years 61 c 63 a. Use the z-score formula. z = –0.5141. The height of 77 inches is 0.5141 standard deviations below the mean. An NBA player whose height is 77 inches is shorter than average. b. Use the z-score formula. z = 1.5424. The height 85 inches is 1.5424 standard deviations above the mean. An NBA player whose height is 85 inches is taller than average. c. Height = 79 + 3.5(3.89) = 90.67 inches, which is over 7.7 feet tall. There are very few NBA players this tall so the answer is no, not likely. 65 a. iv b. Kyle’s blood pressure is equal to 125 + (1 |
.75)(14) = 149.5. 366 CHAPTER 6 | THE NORMAL DISTRIBUTION 67 Let X = an SAT math score and Y = an ACT math score. a. X = 720 720 – 520 = 1.74 The exam score of 720 is 1.74 standard deviations above the mean of 520. 15 b. z = 1.5 The math SAT score is 520 + 1.5(115) ≈ 692.5. The exam score of 692.5 is 1.5 standard deviations above the mean of 520. c. X – µ ≈ 1.59, the z-score for the SAT. Y – µ ≈ 1.70, the z-scores for the ACT. With respect σ = 700 – 514 117 σ = 30 – 21 5.3 to the test they took, the person who took the ACT did better (has the higher z-score). 69 c 71 d 73 a. X ~ N(66, 2.5) b. 0.5404 c. No, the probability that an Asian male is over 72 inches tall is 0.0082 75 a. X ~ N(36, 10) b. The probability that a person consumes more than 40% of their calories as fat is 0.3446. c. Approximately 25% of people consume less than 29.26% of their calories as fat. 77 a. X = number of hours that a Chinese four-year-old in a rural area is unsupervised during the day. b. X ~ N(3, 1.5) c. The probability that the child spends less than one hour a day unsupervised is 0.0918. d. The probability that a child spends over ten hours a day unsupervised is less than 0.0001. e. 2.21 hours 79 a. X = the distribution of the number of days a particular type of criminal trial will take b. X ~ N(21, 7) c. The probability that a randomly selected trial will last more than 24 days is 0.3336. d. 22.77 81 a. mean = 5.51, s = 2.15 b. Check student's solution. c. Check student's solution. d. Check student's solution. e. X ~ N(5.51, 2.15) f. 0.6029 g. The cumulative frequency for less than 6.1 minutes is 0.64. |
h. The answers to part f and part g are not exactly the same, because the normal distribution is only an approximation to the real one. i. The answers to part f and part g are close, because a normal distribution is an excellent approximation when the sample size is greater than 30. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 j. The approximation would have been less accurate, because the smaller sample size means that the data does not fit CHAPTER 6 | THE NORMAL DISTRIBUTION 367 normal curve as well. 83 1. mean = 60,136 s = 10,468 2. Answers will vary. 3. Answers will vary. 4. Answers will vary. 5. X ~ N(60136, 10468) 6. 0.7440 7. The cumulative relative frequency is 43/60 = 0.717. 8. The answers for part f and part g are not the same, because the normal distribution is only an approximation. 85 n = 100; p = 0.1; q = 0.9 μ = np = (100)(0.10) = 10 σ = npq = (100)(0.1)(0.9) = 3 i. z = ±1: x1 = µ + zσ = 10 + 1(3) = 13 and x2 = µ – zσ = 10 – 1(3) = 7. 68% of the defective cars will fall between seven and 13. ii. iii. z = ±2: x1 = µ + zσ = 10 + 2(3) = 16 and x2 = µ – zσ = 10 – 2(3) = 4. 95 % of the defective cars will fall between four and 16 z = ±3: x1 = µ + zσ = 10 + 3(3) = 19 and x2 = µ – zσ = 10 – 3(3) = 1. 99.7% of the defective cars will fall between one and 19. 87 n = 190; p = 1 5 = 0.2; q = 0.8 μ = np = (190)(0.2) = 38 σ = npq = (190)(0.2)(0.8) = 5.5136 a. For this problem: P(34 < x < 54) = normalcdf(34 |
,54,48,5.5136) = 0.7641 b. For this problem: P(54 < x < 64) = normalcdf(54,64,48,5.5136) = 0.0018 c. For this problem: P(x > 64) = normalcdf(64,1099,48,5.5136) = 0.0000012 (approximately 0) 368 CHAPTER 6 | THE NORMAL DISTRIBUTION This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 369 7 | THE CENTRAL LIMIT THEOREM Figure 7.1 If you want to figure out the distribution of the change people carry in their pockets, using the central limit theorem and assuming your sample is large enough, you will find that the distribution is normal and bell-shaped. (credit: John Lodder) Introduction Chapter Objectives By the end of this chapter, the student should be able to: • Recognize central limit theorem problems. • Classify continuous word problems by their distributions. • Apply and interpret the central limit theorem for means. • Apply and interpret the central limit theorem for sums. Why are we so concerned with means? Two reasons are: they give us a middle ground for comparison, and they are easy to calculate. In this chapter, you will study means and the central limit theorem. The central limit theorem (clt for short) is one of the most powerful and useful ideas in all of statistics. There are two alternative forms of the theorem, and both alternatives are concerned with drawing finite samples size n from a population with a known mean, μ, and a known standard deviation, σ. The first alternative says that if we collect samples of size n with 370 CHAPTER 7 | THE CENTRAL LIMIT THEOREM a "large enough n," calculate each sample's mean, and create a histogram of those means, then the resulting histogram will tend to have an approximate normal bell shape. The second alternative says that if we again collect samples of size n that are "large enough," calculate the sum of each sample and create a histogram, then the resulting histogram will again tend to have a normal bell-shape. In either case, it does not matter what the distribution of the original population is, or whether you even need to |
know it. The important fact is that the distribution of sample means and the sums tend to follow the normal distribution. The size of the sample, n, that is required in order to be "large enough" depends on the original population from which the samples are drawn (the sample size should be at least 30 or the data should come from a normal distribution). If the original population is far from normal, then more observations are needed for the sample means or sums to be normal. Sampling is done with replacement. Suppose eight of you roll one fair die ten times, seven of you roll two fair dice ten times, nine of you roll five fair dice ten times, and 11 of you roll ten fair dice ten times. Each time a person rolls more than one die, he or she calculates the sample mean of the faces showing. For example, one person might roll five fair dice and get 2, 2, 3, 4, 6 on one roll. The mean is.4. The 3.4 is one mean when five fair dice are rolled. This same person would roll the five dice nine more times and calculate nine more means for a total of ten means. Your instructor will pass out the dice to several people. Roll your dice ten times. For each roll, record the faces, and find the mean. Round to the nearest 0.5. Your instructor (and possibly you) will produce one graph (it might be a histogram) for one die, one graph for two dice, one graph for five dice, and one graph for ten dice. Since the "mean" when you roll one die is just the face on the die, what distribution do these means appear to be representing? Draw the graph for the means using two dice. Do the sample means show any kind of pattern? Draw the graph for the means using five dice. Do you see any pattern emerging? Finally, draw the graph for the means using ten dice. Do you see any pattern to the graph? What can you conclude as you increase the number of dice? As the number of dice rolled increases from one to two to five to ten, the following is happening: 1. The mean of the sample means remains approximately the same. 2. The spread of the sample means (the standard deviation of the sample means) gets smaller. 3. The graph appears steeper and thinner. You have just demonstrated the central limit theorem (clt). The central limit theorem tells you that as you increase the number of dice, the sample means tend toward a normal |
distribution (the sampling distribution). 7.1 | The Central Limit Theorem for Sample Means (Averages) Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution). Using a subscript that matches the random variable, suppose: a. μX = the mean of X b. σX = the standard deviation of X If you draw random samples of size n, then as n increases, the random variable X to be normally distributed and ¯ which consists of sample means, tends This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 371 ¯ ~ N ⎛ X ⎝µ x, σx n ⎞ ⎠. The central limit theorem for sample means says that if you keep drawing larger and larger samples (such as rolling one, two, five, and finally, ten dice) and calculating their means, the sample means form their own normal distribution (the sampling distribution). The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by, the sample size. The variable n is the number of values that are averaged together, not the number of times the experiment is done. ¯, which consists To put it more formally, if you draw random samples of size n, the distribution of the random variable X of sample means, is called the sampling distribution of the mean. The sampling distribution of the mean approaches a normal distribution as n, the sample size, increases. The random variable X ¯ has a different z-score associated with it from that of the random variable X. The mean x¯ is the ¯ value of X in one sample. z = x¯ − µ x σ x n ⎞ ⎠ ⎛ ⎝ ¯. μX is the average of both X and X σ x¯ = σx n = standard deviation of X ¯ and is called the standard error of the mean. To find probabilities for means on the calculator, follow these steps. 2nd DISTR 2:normalcdf normalcd f ⎛ ⎝lower value o f the area, upper value o f the area, mean, where: • mean is the mean of the original distribution • • standard deviation is the standard deviation of the original distribution sample |
size = n standard deviation sample size ⎞ ⎠ Example 7.1 An unknown distribution has a mean of 90 and a standard deviation of 15. Samples of size n = 25 are drawn randomly from the population. a. Find the probability that the sample mean is between 85 and 92. Solution 7.1 a. Let X = one value from the original unknown population. The probability question asks you to find a probability for the sample mean. Let X ¯ = the mean of a sample of size 25. Since μX = 90, σX = 15, and n = 25, ¯ ~ N ⎛ X ⎝90, 15 25 ⎞ ⎠. Find P(85 < x¯ < 92). Draw a graph. 372 CHAPTER 7 | THE CENTRAL LIMIT THEOREM P(85 < x¯ < 92) = 0.6997 The probability that the sample mean is between 85 and 92 is 0.6997. Figure 7.2 normalcdf(lower value, upper value, mean, standard error of the mean) The parameter list is abbreviated (lower value, upper value, μ, σ n ) normalcdf(85,92,90, 15 25 ) = 0.6997 b. Find the value that is two standard deviations above the expected value, 90, of the sample mean. Solution 7.1 b. To find the value that is two standard deviations above the expected value 90, use the formula: value = μx + (#ofTSDEVs) ⎛ ⎝ σ x n ⎞ ⎠ value = 90 + 2 ⎛ ⎝ ⎞ ⎠ 15 25 = 96 The value that is two standard deviations above the expected value is 96. The standard error of the mean is σx n = 15 25 = 3. Recall that the standard error of the mean is a description of how far (on average) that the sample mean will be from the population mean in repeated simple random samples of size n. 7.1 An unknown distribution has a mean of 45 and a standard deviation of eight. Samples of size n = 30 are drawn randomly from the population. Find the probability that the sample mean is between 42 and 50. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 7 |
| THE CENTRAL LIMIT THEOREM 373 Example 7.2 The length of time, in hours, it takes an "over 40" group of people to play one soccer match is normally distributed with a mean of two hours and a standard deviation of 0.5 hours. A sample of size n = 50 is drawn randomly from the population. Find the probability that the sample mean is between 1.8 hours and 2.3 hours. Solution 7.2 Let X = the time, in hours, it takes to play one soccer match. The probability question asks you to find a probability for the sample mean time, in hours, it takes to play one soccer match. Let X ¯ = the mean time, in hours, it takes to play one soccer match. If μX = _________, σX = __________, and n = ___________, then X ~ N(______, ______) by the central limit theorem for means. μX = 2, σX = 0.5, n = 50, and X ~ N ⎛ ⎝2, 0.5 50 ⎞ ⎠ Find P(1.8 < x¯ < 2.3). Draw a graph. P(1.8 < x¯ < 2.3) = 0.9977 normalcdf ⎛ ⎝1.8,2.3,2,.5 50 ⎞ ⎠ = 0.9977 The probability that the mean time is between 1.8 hours and 2.3 hours is 0.9977. 7.2 The length of time taken on the SAT for a group of students is normally distributed with a mean of 2.5 hours and a standard deviation of 0.25 hours. A sample size of n = 60 is drawn randomly from the population. Find the probability that the sample mean is between two hours and three hours. To find percentiles for means on the calculator, follow these steps. 2nd DIStR 3:invNorm k = invNorm ⎛ ⎝area to the left of k, mean, standard deviation sample size ⎞ ⎠ where: • k = the kth percentile • mean is the mean of the original distribution • • standard deviation is the standard deviation of the original distribution sample size = n 374 CHAPTER 7 | THE CENTRAL LIMIT THEOREM Example 7.3 In a recent study reported Oct. 29, 2012 on the Flurry Blog, |
the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. Take a sample of size n = 100. a. What are the mean and standard deviation for the sample mean ages of tablet users? b. What does the distribution look like? c. Find the probability that the sample mean age is more than 30 years (the reported mean age of tablet users in this particular study). d. Find the 95th percentile for the sample mean age (to one decimal place). Solution 7.3 a. Since the sample mean tends to target the population mean, we have μχ = μ = 34. The sample standard deviation is given by σχ = σ n = 15 100 = 15 10 = 1.5 b. The central limit theorem states that for large sample sizes(n), the sampling distribution will be approximately normal. c. The probability that the sample mean age is more than 30 is given by P(Χ > 30) = normalcdf(30,E99,34,1.5) = 0.9962 d. Let k = the 95th percentile. ⎝0.95,34, 15 100 k = invNorm ⎛ ⎞ ⎠ = 36.5 7.3 In an article on Flurry Blog, a gaming marketing gap for men between the ages of 30 and 40 is identified. You are researching a startup game targeted at the 35-year-old demographic. Your idea is to develop a strategy game that can be played by men from their late 20s through their late 30s. Based on the article’s data, industry research shows that the average strategy player is 28 years old with a standard deviation of 4.8 years. You take a sample of 100 randomly selected gamers. If your target market is 29- to 35-year-olds, should you continue with your development strategy? Example 7.4 The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of 60. a. What are the mean and standard deviation for the sample mean number of app engagement by a tablet user? b. What is the standard error of the mean? c. Find the 90th percentile for the sample mean time for app engagement for a tablet user. Interpret this value in a complete sentence. d. Find the probability that the sample mean is between eight minutes and 8.5 minutes. Solution 7.4 a. µ |
x¯ = µ = 8.2 σ x¯ = σ n = 1 60 = 0.13 b. This allows us to calculate the probability of sample means of a particular distance from the mean, in repeated samples of size 60. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 375 c. Let k = the 90th percentile ⎛ ⎝0.90,8.2, 1 60 time for table users is less than 8.37 minutes. k = invNorm ⎞ ⎠ = 8.37. This values indicates that 90 percent of the average app engagement d. P(8 < x¯ < 8.5) = normalcdf ⎛ ⎝8,8.5,8.2, 1 60 ⎞ ⎠ = 0.9293 7.4 Cans of a cola beverage claim to contain 16 ounces. The amounts in a sample are measured and the statistics are n = 34, x¯ = 16.01 ounces. If the cans are filled so that μ = 16.00 ounces (as labeled) and σ = 0.143 ounces, find the probability that a sample of 34 cans will have an average amount greater than 16.01 ounces. Do the results suggest that cans are filled with an amount greater than 16 ounces? 7.2 | The Central Limit Theorem for Sums Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution) and suppose: a. μX = the mean of Χ b. σΧ = the standard deviation of X If you draw random samples of size n, then as n increases, the random variable ΣX consisting of sums tends to be normally distributed and ΣΧ ~ N((n)(μΧ), ( n )(σΧ)). The central limit theorem for sums says that if you keep drawing larger and larger samples and taking their sums, the sums form their own normal distribution (the sampling distribution), which approaches a normal distribution as the sample size increases. The normal distribution has a mean equal to the original mean multiplied by the sample size and a standard deviation equal to the original standard deviation multiplied by the square root of the sample size. The random variable ΣX has the |
following z-score associated with it: a. Σx is one sum. b. z = Σx – (n)(µ X) ( n)(σ X) i. ii. (n)(μX) = the mean of ΣX ( n)(σ X) = standard deviation of ΣX To find probabilities for sums on the calculator, follow these steps. 2nd DISTR 2:normalcdf normalcdf(lower value of the area, upper value of the area, (n)(mean), ( n )(standard deviation)) where: • mean is the mean of the original distribution • • standard deviation is the standard deviation of the original distribution sample size = n 376 CHAPTER 7 | THE CENTRAL LIMIT THEOREM Example 7.5 An unknown distribution has a mean of 90 and a standard deviation of 15. A sample of size 80 is drawn randomly from the population. a. Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7,500. b. Find the sum that is 1.5 standard deviations above the mean of the sums. Solution 7.5 Let X = one value from the original unknown population. The probability question asks you to find a probability for the sum (or total of) 80 values. ΣX = the sum or total of 80 values. Since μX = 90, σX = 15, and n = 80, ΣX ~ N((80)(90), ( 80 )(15)) • mean of the sums = (n)(μX) = (80)(90) = 7,200 • • standard deviation of the sums = ( n)(σ X ) = ( 80) (15) sum of 80 values = Σx = 7,500 a. Find P(Σx > 7,500) P(Σx > 7,500) = 0.0127 Figure 7.3 normalcdf(lower value, upper value, mean of sums, stdev of sums) The parameter list is abbreviated(lower, upper, (n)(μX, ( n) (σX)) normalcdf (7500,1E99,(80)(90), ⎛ ⎝ 80⎞ ⎠ (15)) = 0.0127 REMINDER 1E99 = 1099. Press the EE key for E. b. Find Σx where z = 1.5 |
. Σx = (n)(μX) + (z) ( n) (σΧ) = (80)(90) + (1.5)( 80 )(15) = 7,401.2 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 377 7.5 An unknown distribution has a mean of 45 and a standard deviation of eight. A sample size of 50 is drawn randomly from the population. Find the probability that the sum of the 50 values is more than 2,400. To find percentiles for sums on the calculator, follow these steps. 2nd DIStR 3:invNorm k = invNorm (area to the left of k, (n)(mean), ( n) (standard deviation)) where: • k is the kth percentile • mean is the mean of the original distribution • • standard deviation is the standard deviation of the original distribution sample size = n Example 7.6 In a recent study reported Oct. 29, 2012 on the Flurry Blog, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. The sample of size is 50. a. What are the mean and standard deviation for the sum of the ages of tablet users? What is the distribution? b. Find the probability that the sum of the ages is between 1,500 and 1,800 years. c. Find the 80th percentile for the sum of the 50 ages. Solution 7.6 a. μΣx = nμx = 50(34) = 1,700 and σΣx = n σx = ( 50 ) (15) = 106.01 The distribution is normal for sums by the central limit theorem. b. P(1500 < Σx < 1800) = normalcdf (1,500, 1,800, (50)(34), ( 50 ) (15)) = 0.7974 c. Let k = the 80th percentile. k = invNorm(0.80,(50)(34), ( 50 ) (15)) = 1,789.3 7.6 In a recent study reported Oct.29, 2012 on the Flurry Blog, the mean age of tablet users is 35 years. Suppose the standard deviation is ten years. The sample size is 39 |
. a. What are the mean and standard deviation for the sum of the ages of tablet users? What is the distribution? b. Find the probability that the sum of the ages is between 1,400 and 1,500 years. c. Find the 90th percentile for the sum of the 39 ages. 378 CHAPTER 7 | THE CENTRAL LIMIT THEOREM Example 7.7 The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of size 70. a. What are the mean and standard deviation for the sums? b. Find the 95th percentile for the sum of the sample. Interpret this value in a complete sentence. c. Find the probability that the sum of the sample is at least ten hours. Solution 7.7 a. μΣx = nμx = 70(8.2) = 574 minutes and σΣx = ( n)(σ x) = ( 70 ) (1) = 8.37 minutes b. Let k = the 95th percentile. k = invNorm (0.95,(70)(8.2), ( 70) (1)) = 587.76 minutes Ninety five percent of the app engagement times are at most 587.76 minutes. c. ten hours = 600 minutes P(Σx ≥ 600) = normalcdf(600,E99,(70)(8.2), ( 70) (1)) = 0.0009 7.7 The mean number of minutes for app engagement by a table use is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample size of 70. a. What is the probability that the sum of the sample is between seven hours and ten hours? What does this mean in context of the problem? b. Find the 84th and 16th percentiles for the sum of the sample. Interpret these values in context. 7.3 | Using the Central Limit Theorem It is important for you to understand when to use the central limit theorem. If you are being asked to find the probability of the mean, use the clt for the mean. If you are being asked to find the probability of a sum or total, use the clt for sums. This also applies to percentiles for means and sums. NOTE If you are being asked to find the probability of an individual value, do not use the clt. Use the distribution of its random variable |
. Examples of the Central Limit Theorem Law of Large Numbers The law of large numbers says that if you take samples of larger and larger size from any population, then the mean x¯ of the sample tends to get closer and closer to μ. From the central limit theorem, we know that as n gets larger and larger, the sample means follow a normal distribution. The larger n gets, the smaller the standard deviation gets. (Remember that the.) This means that the sample mean x¯ must be close to the population mean μ. We can ¯ standard deviation for X is σ n say that μ is the value that the sample means approach as n gets larger. The central limit theorem illustrates the law of large numbers. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 Central Limit Theorem for the Mean and Sum Examples Example 7.8 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 379 A study involving stress is conducted among the students on a college campus. The stress scores follow a uniform distribution with the lowest stress score equal to one and the highest equal to five. Using a sample of 75 students, find: a. The probability that the mean stress score for the 75 students is less than two. b. The 90th percentile for the mean stress score for the 75 students. c. The probability that the total of the 75 stress scores is less than 200. d. The 90th percentile for the total stress score for the 75 students. Let X = one stress score. Problems a and b ask you to find a probability or a percentile for a mean. Problems c and d ask you to find a probability or a percentile for a total or sum. The sample size, n, is equal to 75. Since the individual stress scores follow a uniform distribution, X ~ U(1, 5) where a = 1 and b = 5 (See Continuous Random Variables for an explanation on the uniform distribution). μX = σX = (b – a)2 12 = (5 – 1)2 12 = 1.15 For problems 1. and 2., let X ¯ = the mean stress score for the 75 students. Then, ¯ X ∼ N ⎛ ⎝3, 1.15 75 ⎞ ⎠ where n = 75. a. Find P( x¯ < 2). Draw the graph. Solution 7.8 |
a. P( x¯ < 2) = 0 The probability that the mean stress score is less than two is about zero. Figure 7.4 normalcdf ⎛ ⎝1,2,3,1.15 75 ⎞ ⎠ = 0 380 CHAPTER 7 | THE CENTRAL LIMIT THEOREM REMINDER The smallest stress score is one. b. Find the 90th percentile for the mean of 75 stress scores. Draw a graph. Solution 7.8 b. Let k = the 90th precentile. Find k, where P( x¯ < k) = 0.90. k = 3.2 Figure 7.5 The 90th percentile for the mean of 75 scores is about 3.2. This tells us that 90% of all the means of 75 stress scores are at most 3.2, and that 10% are at least 3.2. invNorm ⎛ ⎝0.90,3,1.15 75 ⎞ ⎠ = 3.2 For problems c and d, let ΣX = the sum of the 75 stress scores. Then, ΣX ~ N[(75)(3), ( 75) (1.15)] c. Find P(Σx < 200). Draw the graph. Solution 7.8 c. The mean of the sum of 75 stress scores is (75)(3) = 225 The standard deviation of the sum of 75 stress scores is ( 75) (1.15) = 9.96 P(Σx < 200) = 0 Figure 7.6 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 381 The probability that the total of 75 scores is less than 200 is about zero. normalcdf (75,200,(75)(3), ( 75) (1.15)). REMINDER The smallest total of 75 stress scores is 75, because the smallest single score is one. d. Find the 90th percentile for the total of 75 stress scores. Draw a graph. Solution 7.8 d. Let k = the 90th percentile. Find k where P(Σx < k) = 0.90. k = 237.8 Figure 7.7 The 90th percentile for the sum of 75 scores |
is about 237.8. This tells us that 90% of all the sums of 75 scores are no more than 237.8 and 10% are no less than 237.8. invNorm(0.90,(75)(3), ( 75) (1.15)) = 237.8 7.8 Use the information in Example 7.8, but use a sample size of 55 to answer the following questions. a. Find P( x¯ < 7). b. Find P(Σx > 170). c. Find the 80th percentile for the mean of 55 scores. d. Find the 85th percentile for the sum of 55 scores. Example 7.9 Suppose that a market research analyst for a cell phone company conducts a study of their customers who exceed the time allowance included on their basic cell phone contract; the analyst finds that for those people who exceed the time included in their basic contract, the excess time used follows an exponential distribution with a mean of 22 minutes. 382 CHAPTER 7 | THE CENTRAL LIMIT THEOREM Consider a random sample of 80 customers who exceed the time allowance included in their basic cell phone contract. Let X = the excess time used by one INDIVIDUAL cell phone customer who exceeds his contracted time allowance. X ∼ Exp ⎛ ⎝ ⎞ ⎠ 1 22. From previous chapters, we know that μ = 22 and σ = 22. Let X ¯ = the mean excess time used by a sample of n = 80 customers who exceed their contracted time allowance. ¯ ~ N ⎛ X ⎝22, 22 80 ⎞ ⎠ by the central limit theorem for sample means Using the clt to find probability a. Find the probability that the mean excess time used by the 80 customers in the sample is longer than 20 minutes. This is asking us to find P( x¯ > 20). Draw the graph. b. Suppose that one customer who exceeds the time limit for his cell phone contract is randomly selected. Find the probability that this individual customer's excess time is longer than 20 minutes. This is asking us to find P(x > 20). c. Explain why the probabilities in parts a and b are different. Solution 7.9 a. Find: P( x¯ > 20) P( x¯ > 20) = 0.79199 using normalcdf ⎛ ⎝20,1E99,22, 22 80 ⎞ ⎠ The probability is 0 |
.7919 that the mean excess time used is more than 20 minutes, for a sample of 80 customers who exceed their contracted time allowance. Figure 7.8 REMINDER 1E99 = 1099 and –1E99 = –1099. Press the EE key for E. Or just use 1099 instead of 1E99. b. Find P(x > 20). Remember to use the exponential distribution for an individual: X ~ Exp⎛ ⎝ ⎞ ⎠. 1 22 P(x > 20) = e ⎛ ⎝− ⎛ ⎝ 1 22 ⎞ ⎞ ⎠(20) ⎠ or e(–0.04545(20)) = 0.4029 c. 1. P(x > 20) = 0.4029 but P( x¯ > 20) = 0.7919 2. The probabilities are not equal because we use different distributions to calculate the probability for individuals and for means. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 383 3. When asked to find the probability of an individual value, use the stated distribution of its random variable; do not use the clt. Use the clt with the normal distribution when you are being asked to find the probability for a mean. Using the clt to find percentiles Find the 95th percentile for the sample mean excess time for samples of 80 customers who exceed their basic contract time allowances. Draw a graph. Solution 7.9 Let k = the 95th percentile. Find k where P( x¯ < k) = 0.95 k = 26.0 using invNorm ⎛ ⎝0.95,22, 22 80 ⎞ ⎠ = 26.0 Figure 7.9 The 95th percentile for the sample mean excess time used is about 26.0 minutes for random samples of 80 customers who exceed their contractual allowed time. Ninety five percent of such samples would have means under 26 minutes; only five percent of such samples would have means above 26 minutes. 7.9 Use the information in Example 7.9, but change the sample size to 144. a. Find P(20 < x¯ < 30). b. Find P(Σx is at least |
3,000). c. Find the 75th percentile for the sample mean excess time of 144 customers. d. Find the 85th percentile for the sum of 144 excess times used by customers. Example 7.10 In the United States, someone is sexually assaulted every two minutes, on average, according to a number of studies. Suppose the standard deviation is 0.5 minutes and the sample size is 100. a. Find the median, the first quartile, and the third quartile for the sample mean time of sexual assaults in the United States. b. Find the median, the first quartile, and the third quartile for the sum of sample times of sexual assaults in the United States. c. Find the probability that a sexual assault occurs on the average between 1.75 and 1.85 minutes. 384 CHAPTER 7 | THE CENTRAL LIMIT THEOREM d. Find the value that is two standard deviations above the sample mean. e. Find the IQR for the sum of the sample times. Solution 7.10 a. We have, μx = μ = 2 and σx = σ n = 0.5 10 = 0.05. Therefore: 1. 50th percentile = μx = μ = 2 2. 25th percentile = invNorm(0.25,2,0.05) = 1.97 3. 75th percentile = invNorm(0.75,2,0.05) = 2.03 b. We have μΣx = n(μx) = 100(2) = 200 and σμx = n (σx) = 10(0.5) = 5. Therefore 1. 50th percentile = μΣx = n(μx) = 100(2) = 200 2. 25th percentile = invNorm(0.25,200,5) = 196.63 3. 75th percentile = invNorm(0.75,200,5) = 203.37 c. P(1.75 < x¯ < 1.85) = normalcdf(1.75,1.85,2,0.05) = 0.0013 d. Using the z-score equation, z = x¯ – µ x¯ σ x¯, and solving for x, we have x = 2(0.05) + 2 = 2.1 e. The IQR is 75th percentile – 25th percentile = 203.37 – 196.63 = |
6.74 7.10 Based on data from the National Health Survey, women between the ages of 18 and 24 have an average systolic blood pressures (in mm Hg) of 114.8 with a standard deviation of 13.1. Systolic blood pressure for women between the ages of 18 to 24 follow a normal distribution. a. b. c. If one woman from this population is randomly selected, find the probability that her systolic blood pressure is greater than 120. If 40 women from this population are randomly selected, find the probability that their mean systolic blood pressure is greater than 120. If the sample were four women between the ages of 18 to 24 and we did not know the original distribution, could the central limit theorem be used? Example 7.11 A study was done about violence against prostitutes and the symptoms of the posttraumatic stress that they developed. The age range of the prostitutes was 14 to 61. The mean age was 30.9 years with a standard deviation of nine years. a. b. c. d. In a sample of 25 prostitutes, what is the probability that the mean age of the prostitutes is less than 35? Is it likely that the mean age of the sample group could be more than 50 years? Interpret the results. In a sample of 49 prostitutes, what is the probability that the sum of the ages is no less than 1,600? Is it likely that the sum of the ages of the 49 prostitutes is at most 1,595? Interpret the results. e. Find the 95th percentile for the sample mean age of 65 prostitutes. Interpret the results. f. Find the 90th percentile for the sum of the ages of 65 prostitutes. Interpret the results. Solution 7.11 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 385 a. P( x¯ < 35) = normalcdf(-E99,35,30.9,1.8) = 0.9886 b. P( x¯ > 50) = normalcdf(50, E99,30.9,1.8) ≈ 0. For this sample group, it is almost impossible for the group’s average age to be more than 50. However, it is still possible for an individual in this group to have an age greater than |
50. c. P(Σx ≥ 1,600) = normalcdf(1600,E99,1514.10,63) = 0.0864 d. P(Σx ≤ 1,595) = normalcdf(-E99,1595,1514.10,63) = 0.9005. This means that there is a 90% chance that the sum of the ages for the sample group n = 49 is at most 1595. e. The 95th percentile = invNorm(0.95,30.9,1.1) = 32.7. This indicates that 95% of the prostitutes in the sample of 65 are younger than 32.7 years, on average. f. The 90th percentile = invNorm(0.90,2008.5,72.56) = 2101.5. This indicates that 90% of the prostitutes in the sample of 65 have a sum of ages less than 2,101.5 years. 7.11 According to Boeing data, the 757 airliner carries 200 passengers and has doors with a mean height of 72 inches. Assume for a certain population of men we have a mean of 69.0 inches and a standard deviation of 2.8 inches. a. What mean doorway height would allow 95% of men to enter the aircraft without bending? b. Assume that half of the 200 passengers are men. What mean doorway height satisfies the condition that there is a 0.95 probability that this height is greater than the mean height of 100 men? c. For engineers designing the 757, which result is more relevant: the height from part a or part b? Why? HISTORICAL NOTE : Normal Approximation to the Binomial Historically, being able to compute binomial probabilities was one of the most important applications of the central limit theorem. Binomial probabilities with a small value for n(say, 20) were displayed in a table in a book. To calculate the probabilities with large values of n, you had to use the binomial formula, which could be very complicated. Using the normal approximation to the binomial distribution simplified the process. To compute the normal approximation to the binomial distribution, take a simple random sample from a population. You must meet the conditions for a binomial distribution: • • there are a certain number n of independent trials the outcomes of any trial are success or failure • each trial has the same probability of a success p Recall that if X is the bin |
omial random variable, then X ~ B(n, p). The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must both be greater than five (np > 5 and nq > 5; the approximation is better if they are both greater than or equal to 10). Then the binomial can be approximated by the normal distribution with mean μ = np and standard deviation σ = npq. Remember that q = 1 – p. In order to get the best approximation, add 0.5 to x or subtract 0.5 from x (use x + 0.5 or x – 0.5). The number 0.5 is called the continuity correction factor and is used in the following example. Example 7.12 Suppose in a local Kindergarten through 12th grade (K - 12) school district, 53 percent of the population favor a charter school for grades K through 5. A simple random sample of 300 is surveyed. a. Find the probability that at least 150 favor a charter school. b. Find the probability that at most 160 favor a charter school. 386 CHAPTER 7 | THE CENTRAL LIMIT THEOREM c. Find the probability that more than 155 favor a charter school. d. Find the probability that fewer than 147 favor a charter school. e. Find the probability that exactly 175 favor a charter school. Let X = the number that favor a charter school for grades K trough 5. X ~ B(n, p) where n = 300 and p = 0.53. Since np > 5 and nq > 5, use the normal approximation to the binomial. The formulas for the mean and standard deviation are μ = np and σ = npq. The mean is 159 and the standard deviation is 8.6447. The random variable for the normal distribution is Y. Y ~ N(159, 8.6447). See The Normal Distribution for help with calculator instructions. For part a, you include 150 so P(X ≥ 150) has normal approximation P(Y ≥ 149.5) = 0.8641. normalcdf(149.5,10^99,159,8.6447) = 0.8641. For part b, you include 160 so P(X ≤ 160) has normal appraximation P(Y ≤ 160.5) = 0.5689. normalcdf(0,160.5,159,8. |
6447) = 0.5689 For part c, you exclude 155 so P(X > 155) has normal approximation P(y > 155.5) = 0.6572. normalcdf(155.5,10^99,159,8.6447) = 0.6572. For part d, you exclude 147 so P(X < 147) has normal approximation P(Y < 146.5) = 0.0741. normalcdf(0,146.5,159,8.6447) = 0.0741 For part e,P(X = 175) has normal approximation P(174.5 < Y < 175.5) = 0.0083. normalcdf(174.5,175.5,159,8.6447) = 0.0083 Because of calculators and computer software that let you calculate binomial probabilities for large values of n easily, it is not necessary to use the the normal approximation to the binomial distribution, provided that you have access to these technology tools. Most school labs have Microsoft Excel, an example of computer software that calculates binomial probabilities. Many students have access to the TI-83 or 84 series calculators, and they easily calculate probabilities for the binomial distribution. If you type in "binomial probability distribution calculation" in an Internet browser, you can find at least one online calculator for the binomial. For Example 7.10, the probabilities are calculated using the following binomial distribution: (n = 300 and p = 0.53). Compare the binomial and normal distribution answers. See Discrete Random Variables for help with calculator instructions for the binomial. P(X ≥ 150) :1 - binomialcdf(300,0.53,149) = 0.8641 P(X ≤ 160) :binomialcdf(300,0.53,160) = 0.5684 P(X > 155) :1 - binomialcdf(300,0.53,155) = 0.6576 P(X < 147) :binomialcdf(300,0.53,146) = 0.0742 P(X = 175) :(You use the binomial pdf.)binomialpdf(300,0.53,175) = 0.0083 7.12 In a city, 46 percent of the population favor the incumbent, Dawn Morgan, for mayor. A simple random sample of 500 is taken. |
Using the continuity correction factor, find the probability that at least 250 favor Dawn Morgan for mayor. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 387 7.1 Central Limit Theorem (Pocket Change) Class Time: Names: Student Learning Outcomes • The student will demonstrate and compare properties of the central limit theorem. NOTE This lab works best when sampling from several classes and combining data. Collect the Data 1. Count the change in your pocket. (Do not include bills.) 2. Randomly survey 30 classmates. Record the values of the change in Table 7.1. __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ Table 7.1 3. Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes. Figure 7.10 4. Calculate the following (n = 1; surveying one person at a time): a. b. x¯ = _______ s = _______ 5. Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve. 388 CHAPTER 7 | THE CENTRAL LIMIT THEOREM Collecting Averages of Pairs Repeat steps one through five of the section Collect the Data. with one exception. Instead of recording the change of 30 classmates, record the average change of 30 pairs. 1. Randomly survey 30 pairs of classmates. 2. Record the values of the average of their change in Table 7.2. __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ |
__________ __________ Table 7.2 3. Construct a histogram. Scale the axes using the same scaling you used for the section titled Collect the Data. Sketch the graph using a ruler and a pencil. Figure 7.11 4. Calculate the following (n = 2; surveying two people at a time): a. b. x¯ = _______ s = _______ 5. Draw a smooth curve through tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve. Collecting Averages of Groups of Five Repeat steps one through five (of the section titled Collect the Data) with one exception. Instead of recording the change of 30 classmates, record the average change of 30 groups of five. 1. Randomly survey 30 groups of five classmates. 2. Record the values of the average of their change. __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 389 __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ Table 7.3 3. Construct a histogram. Scale the axes using the same scaling you used for the section titled Collect the Data. Sketch the graph using a ruler and a pencil. Figure 7.12 4. Calculate the following (n = 5; surveying five people at a time): a. b. x¯ = _______ s = _______ 5. Draw a smooth curve through tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve. Discussion Questions 1. Why did the shape of the distribution of the data change, as n changed? Use one to two complete sentences to explain what happened. 2. In the section titled Collect _____(_____,_____) the Data, what was the approximate distribution of the data? X ~ 3. In the section titled Collecting Averages of Groups of Five, what was the approximate distribution of the averages? X ¯ |
~ _____(_____,_____) 4. In one to two complete sentences, explain any differences in your answers to the previous two questions. 390 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 7.2 Central Limit Theorem (Cookie Recipes) Class Time: Names: Student Learning Outcomes • The student will demonstrate and compare properties of the central limit theorem. Given X = length of time (in days) that a cookie recipe lasted at the Olmstead Homestead. (Assume that each of the different recipes makes the same quantity of cookies.) Recipe # X Recipe # X Recipe # X Recipe # 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 2 2 11 Table 7.4 Calculate the following: a. μx = _______ b. σx = _______ Collect the Data Use a random number generator to randomly select four samples of size n = 5 from the given population. Record your samples in Table 7.5. Then, for each sample, calculate the mean to the nearest tenth. Record them in the spaces provided. Record the sample means for the rest of the class. 1. Complete the table: This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 391 Sample 1 Sample 2 Sample 3 Sample 4 Sample means from other groups: Means: x¯ = ____ x¯ = ____ x¯ = ____ x¯ = ____ Table 7.5 2. Calculate the following: a. b. x¯ = _______ s x¯ = _______ 3. Again, use a random number generator to randomly select four samples from the population. This time, make the samples of size n = 10. Record the samples in Table 7.6. As before, for each sample, calculate the mean to the nearest tenth. Record them in the spaces provided. Record the sample means for the rest of the class. Sample 1 Sample 2 Sample 3 Sample 4 Sample means from other groups Means: x¯ = ____ x¯ = ____ x¯ = ____ x¯ = ____ Table 7.6 4. Calculate the following: a. |
b. x¯ = ______ s x¯ = ______ 5. For the original population, construct a histogram. Make intervals with a bar width of one day. Sketch the graph using a ruler and pencil. Scale the axes. 392 CHAPTER 7 | THE CENTRAL LIMIT THEOREM Figure 7.13 6. Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve. Repeat the Procedure for n = 5 1. For the sample of n = 5 days averaged together, construct a histogram of the averages (your means together with the means of the other groups). Make intervals with bar widths of 1 2 a day. Sketch the graph using a ruler and pencil. Scale the axes. Figure 7.14 2. Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve. Repeat the Procedure for n = 10 1. For the sample of n = 10 days averaged together, construct a histogram of the averages (your means together with the means of the other groups). Make intervals with bar widths of 1 2 a day. Sketch the graph using a ruler and pencil. Scale the axes. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 393 Figure 7.15 2. Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve. Discussion Questions 1. Compare the three histograms you have made, the one for the population and the two for the sample means. In three to five sentences, describe the similarities and differences. 2. State the theoretical (according to the clt) distributions for the sample means. a. n = 5: x¯ ~ _____(_____,_____) b. n = 10: x¯ ~ _____(_____,_____) 3. Are the sample means for n = 5 and n = 10 “close” to the theoretical mean, μx? Explain why or why not. 4. Which of the two distributions of sample means has the smaller standard deviation? Why? 5. As n changed, why did the shape of the distribution of the data change? Use one to two complete |
sentences to explain what happened. 394 CHAPTER 7 | THE CENTRAL LIMIT THEOREM KEY TERMS Average a number that describes the central tendency of the data; there are a number of specialized averages, including the arithmetic mean, weighted mean, median, mode, and geometric mean. Central Limit Theorem sampling with size n, and we are interested in two new RVs: the sample mean, X Given a random variable (RV) with known mean μ and known standard deviation, σ, we are ¯, and the sample sum, ΣΧ. If the size (n) of the sample is sufficiently large, then X ¯ ~ N(μ, σ n ) and ΣΧ ~ N(nμ, ( n )(σ)). If the size (n) of the sample is sufficiently large, then the distribution of the sample means and the distribution of the sample sums will approximate a normal distributions regardless of the shape of the population. The mean of the sample means will equal the population mean, and the mean of the sample sums will equal n times the population mean. The standard deviation of the distribution of the sample means, σ n, is called the standard error of the mean. Exponential Distribution a continuous random variable (RV) that appears when we are interested in the intervals of time between some random events, for example, the length of time between emergency arrivals at a hospital, m and the standard deviation is σ = 1 notation: X ~ Exp(m). The mean is μ = 1 m. The probability density function is f(x) = me–mx, x ≥ 0 and the cumulative distribution function is P(X ≤ x) = 1 – e–mx. Mean a number that measures the central tendency; a common name for mean is "average." The term "mean" is a shortened form of "arithmetic mean." By definition, x¯ = Sum of all values in the sample Number of values in the sample and the mean, the mean for a sample (denoted by x¯ ) for a population (denoted by μ) is is µ = Sum of all values in the population Number of values in the population. Normal Distribution a continuous random variable (RV) with pdf f (x) = 1 σ 2π – (x – µ)2 2σ 2 e, where μ is the mean of the distribution and σ is the standard deviation; notation: Χ ~ N(μ, σ). If |
μ = 0 and σ = 1, the RV is called a standard normal distribution. Normal Distribution a continuous random variable (RV) with pdf f (x) = 1 σ 2π – (x – µ)2 2σ 2 e, where μ is the mean of the distribution and σ is the standard deviation.; notation: X ~ N(μ, σ). If μ = 0 and σ = 1, the RV is called the standard normal distribution. Sampling Distribution Given simple random samples of size n from a given population with a measured characteristic such as mean, proportion, or standard deviation for each sample, the probability distribution of all the measured characteristics is called a sampling distribution. Standard Error of the Mean the standard deviation of the distribution of the sample means, or σ n. Uniform Distribution a continuous random variable (RV) that has equally likely outcomes over the domain, a < x < b; often referred as the Rectangular Distribution because the graph of the pdf has the form of a rectangle. Notation: X ~ U(a, b). The mean is µ = a + b 2 and the standard deviation is σ = (b – a)2 12. The probability density function is f (x) = 1 b – a for a < x < b or a ≤ x ≤ b. The cumulative distribution is P(X ≤ x) = x – a b – a. CHAPTER REVIEW 7.1 The Central Limit Theorem for Sample Means (Averages) This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 395 In a population whose distribution may be known or unknown, if the size (n) of samples is sufficiently large, the distribution of the sample means will be approximately normal. The mean of the sample means will equal the population mean. The standard deviation of the distribution of the sample means, called the standard error of the mean, is equal to the population standard deviation divided by the square root of the sample size (n). 7.2 The Central Limit Theorem for Sums The central limit theorem tells us that for a population with any distribution, the distribution of the sums for the sample means approaches a normal distribution as the sample size increases. In other words, if the sample size is large enough, the distribution of the sums can be |
approximated by a normal distribution even if the original population is not normally distributed. Additionally, if the original population has a mean of μX and a standard deviation of σx, the mean of the sums is nμx and the standard deviation is ( n) (σx) where n is the sample size. 7.3 Using the Central Limit Theorem The central limit theorem can be used to illustrate the law of large numbers. The law of large numbers states that the larger the sample size you take from a population, the closer the sample mean x¯ gets to μ. FORMULA REVIEW 7.1 The Central Limit Theorem for Sample Means (Averages) The Central Limit Theorem for Sample Means: X ⎛ ⎝µ x, ⎞ ⎠ σx n ¯ ~ N The Mean X ¯ : μx Central Limit Theorem for Sample Means z-score and standard error of the mean: z = x.2 The Central Limit Theorem for Sums The Central Limit Theorem for Sums: ∑X ~ N[(n)(μx),( n )(σx)] Mean for Sums (∑X): (n)(μx) The Central Limit Theorem for Sums z-score and standard Σx – (n)(µ X) ( n)(σ X) deviation for sums: zfor the sample mean = Standard deviation for Sums (∑X): ( n) (σx) Standard Error of the Mean (Standard Deviation ( X ¯ )): σ x n PRACTICE 7.1 The Central Limit Theorem for Sample Means (Averages) Use the following information to answer the next six exercises: Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours. Let Χ be the random variable representing the time ¯ be the random variable representing the mean it takes her to complete one review. Assume Χ is normally distributed. Let X time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews. 1. What is the mean, standard deviation, and sample size? 2. Complete the distributions. a. X ~ _____(_____,_____) b. X ¯ ~ _____(_____,_____) 3. Find the probability that |
one review will take Yoonie from 3.5 to 4.25 hours. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability. 396 CHAPTER 7 | THE CENTRAL LIMIT THEOREM a. Figure 7.16 b. P(________ < x < ________) = _______ 4. Find the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hrs. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability. a. Figure 7.17 b. P(________________) = _______ 5. What causes the probabilities in Exercise 7.3 and Exercise 7.4 to be different? 6. Find the 95th percentile for the mean time to complete one month's reviews. Sketch the graph. a. Figure 7.18 b. The 95th Percentile =____________ 7.2 The Central Limit Theorem for Sums Use the following information to answer the next four exercises: An unknown distribution has a mean of 80 and a standard deviation of 12. A sample size of 95 is drawn randomly from the population. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 397 7. Find the probability that the sum of the 95 values is greater than 7,650. 8. Find the probability that the sum of the 95 values is less than 7,400. 9. Find the sum that is two standard deviations above the mean of the sums. 10. Find the sum that is 1.5 standard deviations below the mean of the sums. Use the following information to answer the next five exercises: The distribution of results from a cholesterol test has a mean of 180 and a standard deviation of 20. A sample size of 40 is drawn randomly. 11. Find the probability that the sum of the 40 values is greater than 7,500. 12. Find the probability that the sum of the 40 values is less than 7,000. 13. Find the sum that is one standard deviation above the mean of the sums. 14. Find the sum that is 1.5 standard deviations below the mean of the sums. 15. Find the percentage of sums between 1.5 standard deviations below the mean of the sums and one standard deviation above the mean of the sums |
. Use the following information to answer the next six exercises: A researcher measures the amount of sugar in several cans of the same soda. The mean is 39.01 with a standard deviation of 0.5. The researcher randomly selects a sample of 100. 16. Find the probability that the sum of the 100 values is greater than 3,910. 17. Find the probability that the sum of the 100 values is less than 3,900. 18. Find the probability that the sum of the 100 values falls between the numbers you found in and. 19. Find the sum with a z–score of –2.5. 20. Find the sum with a z–score of 0.5. 21. Find the probability that the sums will fall between the z-scores –2 and 1. Use the following information to answer the next four exercise: An unknown distribution has a mean 12 and a standard deviation of one. A sample size of 25 is taken. Let X = the object of interest. 22. What is the mean of ΣX? 23. What is the standard deviation of ΣX? 24. What is P(Σx = 290)? 25. What is P(Σx > 290)? 26. True or False: only the sums of normal distributions are also normal distributions. 27. In order for the sums of a distribution to approach a normal distribution, what must be true? 28. What three things must you know about a distribution to find the probability of sums? 29. An unknown distribution has a mean of 25 and a standard deviation of six. Let X = one object from this distribution. What is the sample size if the standard deviation of ΣX is 42? 30. An unknown distribution has a mean of 19 and a standard deviation of 20. Let X = the object of interest. What is the sample size if the mean of ΣX is 15,200? Use the following information to answer the next three exercises. A market researcher analyzes how many electronics devices customers buy in a single purchase. The distribution has a mean of three with a standard deviation of 0.7. She samples 400 customers. 31. What is the z-score for Σx = 840? 32. What is the z-score for Σx = 1,186? 33. What is P(Σx < 1,186)? Use the following information to answer the next three exercises: An unkwon distribution has a mean of 100, a standard deviation of 100 |
, and a sample size of 100. Let X = one object of interest. 34. What is the mean of ΣX? 35. What is the standard deviation of ΣX? 398 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 36. What is P(Σx > 9,000)? 7.3 Using the Central Limit Theorem Use the following information to answer the next ten exercises: A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of 100 weights is taken. 37. a. What is the distribution for the weights of one 25-pound lifting weight? What is the mean and standard deivation? b. What is the distribution for the mean weight of 100 25-pound lifting weights? c. Find the probability that the mean actual weight for the 100 weights is less than 24.9. 38. Draw the graph from Exercise 7.37 39. Find the probability that the mean actual weight for the 100 weights is greater than 25.2. 40. Draw the graph from Exercise 7.39 41. Find the 90th percentile for the mean weight for the 100 weights. 42. Draw the graph from Exercise 7.41 43. a. What is the distribution for the sum of the weights of 100 25-pound lifting weights? b. Find P(Σx < 2,450). 44. Draw the graph from Exercise 7.43 45. Find the 90th percentile for the total weight of the 100 weights. 46. Draw the graph from Exercise 7.45 Use the following information to answer the next five exercises: The length of time a particular smartphone's battery lasts follows an exponential distribution with a mean of ten months. A sample of 64 of these smartphones is taken. 47. a. What is the standard deviation? b. What is the parameter m? 48. What is the distribution for the length of time one battery lasts? 49. What is the distribution for the mean length of time 64 batteries last? 50. What is the distribution for the total length of time 64 batteries last? 51. Find the probability that the sample mean is between seven and 11. 52. Find the 80th percentile for the total length of time 64 batteries last. 53. Find the IQR for the mean amount of time 64 batteries last. 54. Find the middle 80% for the total amount of time 64 batteries last. Use |
the following information to answer the next eight exercises: A uniform distribution has a minimum of six and a maximum of ten. A sample of 50 is taken. 55. Find P(Σx > 420). 56. Find the 90th percentile for the sums. 57. Find the 15th percentile for the sums. 58. Find the first quartile for the sums. 59. Find the third quartile for the sums. 60. Find the 80th percentile for the sums. HOMEWORK 7.1 The Central Limit Theorem for Sample Means (Averages) This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 61. Previously, De Anza statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.88. Suppose that we randomly pick 25 daytime statistics students. In words, Χ = ____________ a. b. Χ ~ _____(_____,_____) CHAPTER 7 | THE CENTRAL LIMIT THEOREM 399 c. ¯ = ____________ In words, X ¯ ~ ______ (______, ______) d. X e. Find the probability that an individual had between $0.80 and $1.00. Graph the situation, and shade in the area to be determined. f. Find the probability that the average of the 25 students was between $0.80 and $1.00. Graph the situation, and shade in the area to be determined. g. Explain why there is a difference in part e and part f. 62. Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls. ¯ = average distance in feet for 49 fly balls, then X a. b. What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the ¯ ~ _______(_______,_______) If X horizontal axis for X ¯. Shade the region corresponding to the probability. Find the probability. c. Find the 80th percentile of the distribution of the average of 49 fly balls. 63. According to the Internal Revenue Service, the average length of time for an individual to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040 is 10.53 |
hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is two hours. Suppose we randomly sample 36 taxpayers. a. b. In words, Χ = _____________ ¯ = _____________ In words, X ¯ ~ _____(_____,_____) c. X d. Would you be surprised if the 36 taxpayers finished their Form 1040s in an average of more than 12 hours? Explain why or why not in complete sentences. e. Would you be surprised if one taxpayer finished his or her Form 1040 in more than 12 hours? In a complete sentence, explain why. 64. Suppose that a category of world-class runners are known to run a marathon (26 miles) in an average of 145 minutes ¯ with a standard deviation of 14 minutes. Consider 49 of the races. Let X the average of the 49 races. ¯ ~ _____(_____,_____) a. X b. Find the probability that the runner will average between 142 and 146 minutes in these 49 marathons. c. Find the 80th percentile for the average of these 49 marathons. d. Find the median of the average running times. 65. The length of songs in a collector’s iTunes album collection is uniformly distributed from two to 3.5 minutes. Suppose we randomly pick five albums from the collection. There are a total of 43 songs on the five albums. In words, Χ = _________ a. b. Χ ~ _____________ c. ¯ = _____________ In words, X ¯ ~ _____(_____,_____) d. X e. Find the first quartile for the average song length. f. The IQR(interquartile range) for the average song length is from _______–_______. 66. In 1940 the average size of a U.S. farm was 174 acres. Let’s say that the standard deviation was 55 acres. Suppose we randomly survey 38 farmers from 1940. In words, Χ = _____________ ¯ = _____________ b. a. In words, X ¯ ~ _____(_____,_____) ¯ d. The IQR for X c. X is from _______ acres to _______ acres. 67. Determine which of the following are true and which are false. Then, in complete sentences, justify your answers. ¯ a. When the sample size is large, the mean of X is approximately equal to the mean of Χ. 400 CHAPTER |
7 | THE CENTRAL LIMIT THEOREM ¯ b. When the sample size is large, X is approximately normally distributed. ¯ c. When the sample size is large, the standard deviation of X is approximately the same as the standard deviation of Χ. 68. The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about ¯ = average percent of fat 36 and a standard deviation of about ten. Suppose that 16 individuals are randomly chosen. Let X calories. ¯ ~ ______(______, ______) a. X b. For the group of 16, find the probability that the average percent of fat calories consumed is more than five. Graph the situation and shade in the area to be determined. c. Find the first quartile for the average percent of fat calories. 69. The distribution of income in some Third World countries is considered wedge shaped (many very poor people, very few middle income people, and even fewer wealthy people). Suppose we pick a country with a wedge shaped distribution. Let the average salary be $2,000 per year with a standard deviation of $8,000. We randomly survey 1,000 residents of that country. a. b. In words, Χ = _____________ ¯ = _____________ In words, X ¯ ~ _____(_____,_____) c. X d. How is it possible for the standard deviation to be greater than the average? e. Why is it more likely that the average of the 1,000 residents will be from $2,000 to $2,100 than from $2,100 to $2,200? 70. Which of the following is NOT TRUE about the distribution for averages? a. The mean, median, and mode are equal. b. The area under the curve is one. c. The curve never touches the x-axis. d. The curve is skewed to the right. 71. The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations. The distribution to use for the average cost of gasoline for the 16 gas stations is: a. X b. X c. X d. X ¯ ~ N(4.59, 0.10) ¯ ~ N ⎛ ⎝4.59, 0.10 16 � |
��4.59, 16 0.10 ⎝4.59, 16 0.10 ¯ ~.2 The Central Limit Theorem for Sums 72. Which of the following is NOT TRUE about the theoretical distribution of sums? a. The mean, median and mode are equal. b. The area under the curve is one. c. The curve never touches the x-axis. d. The curve is skewed to the right. 73. Suppose that the duration of a particular type of criminal trial is known to have a mean of 21 days and a standard deviation of seven days. We randomly sample nine trials. In words, ΣX = ______________ a. b. ΣX ~ _____(_____,_____) c. Find the probability that the total length of the nine trials is at least 225 days. d. Ninety percent of the total of nine of these types of trials will last at least how long? 74. Suppose that the weight of open boxes of cereal in a home with children is uniformly distributed from two to six pounds with a mean of four pounds and standard deviation of 1.1547. We randomly survey 64 homes with children. In words, X = _____________ a. b. The distribution is _______. c. In words, ΣX = _______________ This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 d. ΣX ~ _____(_____,_____) e. Find the probability that the total weight of open boxes is less than 250 pounds. f. Find the 35th percentile for the total weight of open boxes of cereal. 75. Salaries for teachers in a particular elementary school district are normally distributed with a mean of $44,000 and a standard deviation of $6,500. We randomly survey ten teachers from that district. CHAPTER 7 | THE CENTRAL LIMIT THEOREM 401 In words, X = ______________ In words, ΣX = _____________ a. b. X ~ _____(_____,_____) c. d. ΣX ~ _____(_____,_____) e. Find the probability that the teachers earn a total of over $400,000. f. Find the 90th percentile for an individual teacher's salary. g. Find the 90th percentile for the sum of ten teachers' salary. h. i. |
If we surveyed 70 teachers instead of ten, graphically, how would that change the distribution in part d? If each of the 70 teachers received a $3,000 raise, graphically, how would that change the distribution in part b? 7.3 Using the Central Limit Theorem 76. The attention span of a two-year-old is exponentially distributed with a mean of about eight minutes. Suppose we randomly survey 60 two-year-olds. a. In words, Χ = _______ b. Χ ~ _____(_____,_____) c. ¯ = ____________ In words, X ¯ ~ _____(_____,_____) d. X e. Before doing any calculations, which do you think will be higher? Explain why. i. The probability that an individual attention span is less than ten minutes. ii. The probability that the average attention span for the 60 children is less than ten minutes? f. Calculate the probabilities in part e. ¯ g. Explain why the distribution for X is not exponential. 77. The closing stock prices of 35 U.S. semiconductor manufacturers are given as follows. 8.625; 30.25; 27.625; 46.75; 32.875; 18.25; 5; 0.125; 2.9375; 6.875; 28.25; 24.25; 21; 1.5; 30.25; 71; 43.5; 49.25; 2.5625; 31; 16.5; 9.5; 18.5; 18; 9; 10.5; 16.625; 1.25; 18; 12.87; 7; 12.875; 2.875; 60.25; 29.25 In words, Χ = ______________ a. b. x¯ = _____ i. ii. sx = _____ iii. n = _____ c. Construct a histogram of the distribution of the averages. Start at x = –0.0005. Use bar widths of ten. d. e. Randomly average five stock prices together. (Use a random number generator.) Continue averaging five pieces In words, describe the distribution of stock prices. together until you have ten averages. List those ten averages. f. Use the ten averages from part e to calculate the following. i. ii. x¯ = _____ sx = _____ g. Construct a histogram of the distribution of the averages. Start at x |
= -0.0005. Use bar widths of ten. h. Does this histogram look like the graph in part c? i. In one or two complete sentences, explain why the graphs either look the same or look different? j. Based upon the theory of the central limit theorem, X ¯ ~ _____(_____,____) Use the following information to answer the next three exercises: Richard’s Furniture Company delivers furniture from 10 A.M. to 2 P.M. continuously and uniformly. We are interested in how long (in hours) past the 10 A.M. start time that individuals wait for their delivery. 78. Χ ~ _____(_____,_____) a. U(0,4) b. U(10,2) c. Eχp(2) 402 CHAPTER 7 | THE CENTRAL LIMIT THEOREM d. N(2,1) 79. The average wait time is: a. one hour. b. two hours. c. two and a half hours. d. four hours. 80. Suppose that it is now past noon on a delivery day. The probability that a person must wait at least one and a half more hours is: a. b. c. d. 1 4 1 2 3 4 3 8 Use the following information to answer the next two exercises: The time to wait for a particular rural bus is distributed uniformly from zero to 75 minutes. One hundred riders are randomly sampled to learn how long they waited. 81. The 90th percentile sample average wait time (in minutes) for a sample of 100 riders is: a. 315.0 b. 40.3 c. 38.5 d. 65.2 82. Would you be surprised, based upon numerical calculations, if the sample average wait time (in minutes) for 100 riders was less than 30 minutes? a. yes b. no c. There is not enough information. Use the following to answer the next two exercises: The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations. 83. What's the approximate probability that the average price for 16 gas stations is over $4.69? a. almost zero b. 0.1587 c. 0.0943 d. unknown 84. |
Find the probability that the average price for 30 gas stations is less than $4.55. a. 0.6554 b. 0.3446 c. 0.0142 d. 0.9858 e. 0 85. Suppose in a local Kindergarten through 12th grade (K - 12) school district, 53 percent of the population favor a charter school for grades K through five. A simple random sample of 300 is surveyed. Calculate following using the normal approximation to the binomial distribtion. a. Find the probability that less than 100 favor a charter school for grades K through 5. b. Find the probability that 170 or more favor a charter school for grades K through 5. c. Find the probability that no more than 140 favor a charter school for grades K through 5. d. Find the probability that there are fewer than 130 that favor a charter school for grades K through 5. e. Find the probability that exactly 150 favor a charter school for grades K through 5. If you have access to an appropriate calculator or computer software, try calculating these probabilities using the technology. 86. Four friends, Janice, Barbara, Kathy and Roberta, decided to carpool together to get to school. Each day the driver would be chosen by randomly selecting one of the four names. They carpool to school for 96 days. Use the normal approximation to the binomial to calculate the following probabilities. Round the standard deviation to four decimal places. a. Find the probability that Janice is the driver at most 20 days. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 403 b. Find the probability that Roberta is the driver more than 16 days. c. Find the probability that Barbara drives exactly 24 of those 96 days. ¯ be the random variable of 87. X ~ N(60, 9). Suppose that you form random samples of 25 from this distribution. Let X averages. Let ΣX be the random variable of sums. For parts c through f, sketch the graph, shade the region, label and scale the horizontal axis for X ¯, and find the probability. ¯ on the same graph. a. Sketch the distributions of X and X ¯ ~ _____(_____,_____) b. X c. P( x¯ < 60) = _____ d. Find |
the 30th percentile for the mean. e. P(56 < x¯ < 62) = _____ f. P(18 < x¯ < 58) = _____ g. Σx ~ _____(_____,_____) h. Find the minimum value for the upper quartile for the sum. i. P(1,400 < Σx < 1,550) = _____ 88. Suppose that the length of research papers is uniformly distributed from ten to 25 pages. We survey a class in which 55 research papers were turned in to a professor. The 55 research papers are considered a random collection of all papers. We are interested in the average length of the research papers. In words, X = _____________ a. b. X ~ _____(_____,_____) c. μx = _____ d. σx = _____ e. ¯ = ______________ In words, X ¯ ~ _____(_____,_____) In words, ΣX = _____________ f. X g. h. ΣX ~ _____(_____,_____) i. Without doing any calculations, do you think that it’s likely that the professor will need to read a total of more than 1,050 pages? Why? j. Calculate the probability that the professor will need to read a total of more than 1,050 pages. k. Why is it so unlikely that the average length of the papers will be less than 12 pages? 89. Salaries for teachers in a particular elementary school district are normally distributed with a mean of $44,000 and a standard deviation of $6,500. We randomly survey ten teachers from that district. a. Find the 90th percentile for an individual teacher’s salary. b. Find the 90th percentile for the average teacher’s salary. 90. The average length of a maternity stay in a U.S. hospital is said to be 2.4 days with a standard deviation of 0.9 days. We randomly survey 80 women who recently bore children in a U.S. hospital. a. b. In words, X = _____________ ¯ = ___________________ In words, X ¯ ~ _____(_____,_____) In words, ΣX = _______________ c. X d. e. ΣX ~ _____(_____,_____) f. g. h. Which is more likely: Is it likely that an individual stayed more than five days in |
the hospital? Why or why not? Is it likely that the average stay for the 80 women was more than five days? Why or why not? i. An individual stayed more than five days. ii. the average stay of 80 women was more than five days. i. If we were to sum up the women’s stays, is it likely that, collectively they spent more than a year in the hospital? Why or why not? For each problem, wherever possible, provide graphs and use the calculator. 91. NeverReady batteries has engineered a newer, longer lasting AAA battery. The company claims this battery has an average life span of 17 hours with a standard deviation of 0.8 hours. Your statistics class questions this claim. As a class, you randomly select 30 batteries and find that the sample mean life span is 16.7 hours. If the process is working properly, 404 CHAPTER 7 | THE CENTRAL LIMIT THEOREM what is the probability of getting a random sample of 30 batteries in which the sample mean lifetime is 16.7 hours or less? Is the company’s claim reasonable? 92. Men have an average weight of 172 pounds with a standard deviation of 29 pounds. a. Find the probability that 20 randomly selected men will have a sum weight greater than 3600 lbs. b. If 20 men have a sum weight greater than 3500 lbs, then their total weight exceeds the safety limits for water taxis. Based on (a), is this a safety concern? Explain. 93. M&M candies large candy bags have a claimed net weight of 396.9 g. The standard deviation for the weight of the individual candies is 0.017 g. The following table is from a stats experiment conducted by a statistics class. Red Orange Yellow Brown Blue Green 0.751 0.735 0.841 0.895 0.856 0.865 0.799 0.864 0.966 0.852 0.859 0.866 0.857 0.859 0.942 0.838 0.873 0.863 0.809 0.888 0.890 0.925 0.878 0.793 0.905 0.977 0.850 0.830 0.856 0.842 0.778 0.786 0.853 0.864 0.873 0.880 0.882 0.931 Table 7.7 0.883 0.769 |
0.859 0.784 0.824 0.858 0.848 0.851 0.696 0.876 0.855 0.806 0.840 0.868 0.859 0.982 0.881 0.925 0.863 0.914 0.775 0.881 0.854 0.865 0.810 0.865 0.858 1.015 0.818 0.876 0.868 0.809 0.803 0.865 0.932 0.848 0.842 0.940 0.832 0.833 0.807 0.845 0.841 0.852 0.932 0.778 0.833 0.814 0.881 0.791 0.818 0.810 0.864 0.881 0.825 0.855 0.942 0.825 0.869 0.912 0.887 The bag contained 465 candies and he listed weights in the table came from randomly selected candies. Count the weights. a. Find the mean sample weight and the standard deviation of the sample weights of candies in the table. b. Find the sum of the sample weights in the table and the standard deviation of the sum the of the weights. c. d. If 465 M&Ms are randomly selected, find the probability that their weights sum to at least 396.9. Is the Mars Company’s M&M labeling accurate? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 94. The Screw Right Company claims their 3 4 inch screws are within ±0.23 of the claimed mean diameter of 0.750 inches with a standard deviation of 0.115 inches. The following data were recorded. CHAPTER 7 | THE CENTRAL LIMIT THEOREM 405 0.757 0.723 0.754 0.737 0.757 0.741 0.722 0.741 0.743 0.742 0.740 0.758 0.724 0.739 0.736 0.735 0.760 0.750 0.759 0.754 0.744 0.758 0.765 0.756 0.738 0.7 |
42 0.758 0.757 0.724 0.757 0.744 0.738 0.763 0.756 0.760 0.768 0.761 0.742 0.734 0.754 0.758 0.735 0.740 0.743 0.737 0.737 0.725 0.761 0.758 0.756 Table 7.8 The screws were randomly selected from the local home repair store. a. Find the mean diameter and standard deviation for the sample b. Find the probability that 50 randomly selected screws will be within the stated tolerance levels. Is the company’s diameter claim plausible? 95. Your company has a contract to perform preventive maintenance on thousands of air-conditioners in a large city. Based on service records from previous years, the time that a technician spends servicing a unit averages one hour with a standard deviation of one hour. In the coming week, your company will service a simple random sample of 70 units in the city. You plan to budget an average of 1.1 hours per technician to complete the work. Will this be enough time? 96. A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selected adults are given an IQ tesst, what is the probability that the sample mean scores will be between 85 and 125 points? 97. Certain coins have an average weight of 5.201 grams with a standard deviation of 0.065 g. If a vending machine is designed to accept coins whose weights range from 5.111 g to 5.291 g, what is the expected number of rejected coins when 280 randomly selected coins are inserted into the machine? REFERENCES 7.1 The Central Limit Theorem for Sample Means (Averages) Baran, Daya. “20 Percent of Americans Have Never Used Email.”WebGuild, 2010. Available online at http://www.webguild.org/20080519/20-percent-of-americans-have-never-used-email (accessed May 17, 2013). Data from The Flurry Blog, 2013. Available online at http://blog.flurry.com (accessed May 17, 2013). Data from the United States Department of Agriculture. 7.2 The Central Limit Theorem for Sums Farago, Peter. “The Truth About Cats and Dogs: Smartphone vs Tablet Usage Differences.” The Flurry Blog |
, 2013. Posted October 29, 2012. Available online at http://blog.flurry.com (accessed May 17, 2013). 7.3 Using the Central Limit Theorem Data from the Wall Street Journal. “National Health and Nutrition Examination Survey.” Center for Disease Control and Prevention. Available online at http://www.cdc.gov/nchs/nhanes.htm (accessed May 17, 2013). SOLUTIONS 1 mean = 4 hours; standard deviation = 1.2 hours; sample size = 16 3 a. Check student's solution. b. 3.5, 4.25, 0.2441 5 The fact that the two distributions are different accounts for the different probabilities. 406 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 7 0.3345 9 7,833.92 11 0.0089 13 7,326.49 15 77.45% 17 0.4207 19 3,888.5 21 0.8186 23 5 25 0.9772 27 The sample size, n, gets larger. 29 49 31 26.00 33 0.1587 35 1,000 37 a. U(24, 26), 25, 0.5774 b. N(25, 0.0577) c. 0.0416 39 0.0003 41 25.07 43 a. N(2,500, 5.7735) b. 0 45 2,507.40 47 a. 10 b. 1 10 49 N ⎛ ⎝10, 10 8 ⎞ ⎠ 51 0.7799 53 1.69 55 0.0072 57 391.54 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 407 59 405.51 61 a. Χ = amount of change students carry b. Χ ~ E(0.88, 0.88) c. X ¯ = average amount of change carried by a sample of 25 sstudents. d. X ¯ ~ N(0.88, 0.176) e. 0.0819 f. 0.1882 g. The distributions are different. Part a is exponential and part b is normal. 63 a. length of time for an individual to complete IRS form 1040, in hours. |
b. mean length of time for a sample of 36 taxpayers to complete IRS form 1040, in hours. c. N ⎛ ⎝10.53, 1 3 ⎞ ⎠ d. Yes. I would be surprised, because the probability is almost 0. e. No. I would not be totally surprised because the probability is 0.2312 65 a. the length of a song, in minutes, in the collection b. U(2, 3.5) c. the average length, in minutes, of the songs from a sample of five albums from the collection d. N(2.75, 0.0220) e. 2.74 minutes f. 0.03 minutes 67 a. True. The mean of a sampling distribution of the means is approximately the mean of the data distribution. b. True. According to the Central Limit Theorem, the larger the sample, the closer the sampling distribution of the means becomes normal. c. The standard deviation of the sampling distribution of the means will decrease making it approximately the same as the standard deviation of X as the sample size increases. 69 a. X = the yearly income of someone in a third world country b. the average salary from samples of 1,000 residents of a third world country ¯ c. X ∼ N ⎛ ⎝2000, 8000 1000 ⎞ ⎠ d. Very wide differences in data values can have averages smaller than standard deviations. e. The distribution of the sample mean will have higher probabilities closer to the population mean. P(2000 < X P(2100 < X ¯ < 2100) = 0.1537 ¯ < 2200) = 0.1317 71 b 73 408 CHAPTER 7 | THE CENTRAL LIMIT THEOREM a. the total length of time for nine criminal trials b. N(189, 21) c. 0.0432 d. 162.09; ninety percent of the total nine trials of this type will last 162 days or more. 75 a. X = the salary of one elementary school teacher in the district b. X ~ N(44,000, 6,500) c. ΣX ~ sum of the salaries of ten elementary school teachers in the sample d. ΣX ~ N(44000, 20554.80) e. 0.9742 f. $52,330.09 g. 466,342.04 h. Sampling 70 teachers instead of ten would cause the distribution to |
be more spread out. It would be a more symmetrical normal curve. i. If every teacher received a $3,000 raise, the distribution of X would shift to the right by $3,000. In other words, it would have a mean of $47,000. 77 a. X = the closing stock prices for U.S. semiconductor manufacturers c. b. i. $20.71; ii. $17.31; iii. 35 d. Exponential distribution, Χ ~ Exp ⎛ ⎝ 1 20.71 ⎞ ⎠ e. Answers will vary. f. i. $20.71; ii. $11.14 g. Answers will vary. h. Answers will vary. i. Answers will vary. j. N ⎛ ⎝20.71, 17.31 5 ⎞ ⎠ 79 b 81 b 83 a 85 a. 0 b. 0.1123 c. 0.0162 d. 0.0003 e. 0.0268 87 a. Check student’s solution. b. X ¯ ~ N ⎛ ⎝60, 9 25 ⎞ ⎠ This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 7 | THE CENTRAL LIMIT THEOREM 409 c. 0.5000 d. 59.06 e. 0.8536 f. 0.1333 g. N(1500, 45) h. 1530.35 i. 0.6877 89 a. $52,330 b. $46,634 91 • We have μ = 17, σ = 0.8, x¯ = 16.7, and n = 30. To calculate the probability, we use normalcdf(lower, upper, μ, σ n ) = normalcdf ⎛ ⎝E – 99,16.7,17, 0.8 30 ⎞ ⎠ = 0.0200. • If the process is working properly, then the probability that a sample of 30 batteries would have at most 16.7 lifetime hours is only 2%. Therefore, the class was justified to question the claim. 93 a. For the sample, we have n = 100, x¯ = 0.862, s = 0.05 b. |
Σ x¯ = 85.65, Σs = 5.18 c. normalcdf(396.9,E99,(465)(0.8565),(0.05)( 465 )) ≈ 1 d. Since the probability of a sample of size 465 having at least a mean sum of 396.9 is appproximately 1, we can conclude that Mars is correctly labeling their M&M packages. 95 Use normalcdf ⎛ ⎝E – 99,1.1,1, 1 70 ⎞ ⎠ = 0.7986. This means that there is an 80% chance that the service time will be less than 1.1 hours. It could be wise to schedule more time since there is an associated 20% chance that the maintenance time will be greater than 1.1 hours. 97 Since we have normalcdf ⎛ ⎝5.111,5.291,5.201,0.065 280 the limits, therefore, there should be no rejected coins out of a well selected sample of size 280. ⎞ ⎠ ≈ 1, we can conclude that practically all the coins are within 410 CHAPTER 7 | THE CENTRAL LIMIT THEOREM This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 411 8 | CONFIDENCE INTERVALS Figure 8.1 Have you ever wondered what the average number of M&Ms in a bag at the grocery store is? You can use confidence intervals to answer this question. (credit: comedy_nose/flickr) Introduction By the end of this chapter, the student should be able to: Chapter Objectives Interpret the Student's t probability distribution as the sample size changes. • Calculate and interpret confidence intervals for estimating a population mean and a population proportion. • • Discriminate between problems applying the normal and the Student's t distributions. • Calculate the sample size required to estimate a population mean and a population proportion given a desired confidence level and margin of error. Suppose you were trying to determine the mean rent of a two-bedroom apartment in your town. You might look in the classified section of the newspaper, write down several rents listed, and average them together. You would have obtained a point estimate of the true mean. If |
you are trying to determine the percentage of times you make a basket when shooting a basketball, you might count the number of shots you make and divide that by the number of shots you attempted. In this case, you would have obtained a point estimate for the true proportion. 412 CHAPTER 8 | CONFIDENCE INTERVALS We use sample data to make generalizations about an unknown population. This part of statistics is called inferential statistics. The sample data help us to make an estimate of a population parameter. We realize that the point estimate is most likely not the exact value of the population parameter, but close to it. After calculating point estimates, we construct interval estimates, called confidence intervals. In this chapter, you will learn to construct and interpret confidence intervals. You will also learn a new distribution, the Student's-t, and how it is used with these intervals. Throughout the chapter, it is important to keep in mind that the confidence interval is a random variable. It is the population parameter that is fixed. If you worked in the marketing department of an entertainment company, you might be interested in the mean number of songs a consumer downloads a month from iTunes. If so, you could conduct a survey and calculate the sample mean, x¯, and the sample standard deviation, s. You would use x¯ to estimate the population mean and s to estimate the population standard deviation. The sample mean, x¯, is the point estimate for the population mean, μ. The sample standard deviation, s, is the point estimate for the population standard deviation, σ. Each of x¯ and s is called a statistic. A confidence interval is another type of estimate but, instead of being just one number, it is an interval of numbers. The interval of numbers is a range of values calculated from a given set of sample data. The confidence interval is likely to include an unknown population parameter. Suppose, for the iTunes example, we do not know the population mean μ, but we do know that the population standard deviation is σ = 1 and our sample size is 100. Then, by the central limit theorem, the standard deviation for the sample mean is σ n = 1 100 = 0.1. The empirical rule, which applies to bell-shaped distributions, says that in approximately 95% of the samples, the sample mean, x¯, will be within two standard deviations of the population mean μ. For our iTunes example, two standard deviations is (2)(0.1) = 0.2. The sample |
mean x¯ is likely to be within 0.2 units of μ. is within 0.2 units of μ, which is unknown, then μ is likely to be within 0.2 units of x¯ Because x¯ in 95% of the samples. The population mean μ is contained in an interval whose lower number is calculated by taking the sample mean and subtracting two standard deviations (2)(0.1) and whose upper number is calculated by taking the sample mean and adding two standard deviations. In other words, μ is between x¯ − 0.2 and x¯ + 0.2 in 95% of all the samples. For the iTunes example, suppose that a sample produced a sample mean x¯ = 2. Then the unknown population mean μ is between x¯ − 0.2 = 2 − 0.2 = 1.8 and x¯ + 0.2 = 2 + 0.2 = 2.2 We say that we are 95% confident that the unknown population mean number of songs downloaded from iTunes per month is between 1.8 and 2.2. The 95% confidence interval is (1.8, 2.2). The 95% confidence interval implies two possibilities. Either the interval (1.8, 2.2) contains the true mean μ or our sample produced an x¯ (95–100%). that is not within 0.2 units of the true mean μ. The second possibility happens for only 5% of all the samples Remember that a confidence interval is created for an unknown population parameter like the population mean, μ. Confidence intervals for some parameters have the form: (point estimate – margin of error, point estimate + margin of error) The margin of error depends on the confidence level or percentage of confidence and the standard error of the mean. When you read newspapers and journals, some reports will use the phrase "margin of error." Other reports will not use that phrase, but include a confidence interval as the point estimate plus or minus the margin of error. These are two ways of expressing the same concept. NOTE Although the text only covers symmetrical confidence intervals, there are non-symmetrical confidence intervals (for example, a confidence interval for the standard deviation). This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 413 Have your instructor record the number of meals each student in |
your class eats out in a week. Assume that the standard deviation is known to be three meals. Construct an approximate 95% confidence interval for the true mean number of meals students eat out each week. 1. Calculate the sample mean. 2. Let σ = 3 and n = the number of students surveyed. 3. Construct the interval ⎛ ⎛ ⎝ x¯ − 2 ⎝ ⎛ ⎞ ⎠ ⎝ ⎞ ⎠, ⎛ ⎛ ⎝ x¯ + 2 We say we are approximately 95% confident that the true mean number of meals that students eat out in a week is between __________ and ___________. 8.1 | A Single Population Mean using the Normal Distribution A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. Suppose that our sample has a mean of x¯ = 10 and we have constructed the 90% confidence interval (5, 15) where EBM = 5. Calculating the Confidence Interval To construct a confidence interval for a single unknown population mean μ, where the population standard deviation is known, we need x¯ as an estimate for μ and we need the margin of error. Here, the margin of error (EBM) is called the error bound for a population mean (abbreviated EBM). The sample mean x¯ population mean μ. is the point estimate of the unknown The confidence interval estimate will have the form: (point estimate - error bound, point estimate + error bound) or, in symbols,( x¯ – EBM, x¯ +EBM ) The margin of error (EBM) depends on the confidence level (abbreviated CL). The confidence level is often considered the probability that the calculated confidence interval estimate will contain the true population parameter. However, it is more accurate to state that the confidence level is the percent of confidence intervals that contain the true population parameter when repeated samples are taken. Most often, it is the choice of the person constructing the confidence interval to choose a confidence level of 90% or higher because that person wants to be reasonably certain of his or her conclusions. There is another probability called alpha (α). α is related to the confidence level, CL. α is the probability that the interval does not contain the unknown population parameter. Mathematically, α + CL = 1. Example 8.1 Suppose we have collected data from |
a sample. We know the sample mean but we do not know the mean for the entire population. The sample mean is seven, and the error bound for the mean is 2.5. x¯ = 7 and EBM = 2.5 The confidence interval is (7 – 2.5, 7 + 2.5), and calculating the values gives (4.5, 9.5). If the confidence level (CL) is 95%, then we say that, "We estimate with 95% confidence that the true value of the population mean is between 4.5 and 9.5." 414 CHAPTER 8 | CONFIDENCE INTERVALS 8.1 Suppose we have data from a sample. The sample mean is 15, and the error bound for the mean is 3.2. What is the confidence interval estimate for the population mean? A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. Suppose that our sample has a mean of x¯ = 10, and we have constructed the 90% confidence interval (5, 15) where EBM = 5. To get a 90% confidence interval, we must include the central 90% of the probability of the normal distribution. If we include the central 90%, we leave out a total of α = 10% in both tails, or 5% in each tail, of the normal distribution. Figure 8.2 To capture the central 90%, we must go out 1.645 "standard deviations" on either side of the calculated sample mean. The value 1.645 is the z-score from a standard normal probability distribution that puts an area of 0.90 in the center, an area of 0.05 in the far left tail, and an area of 0.05 in the far right tail. It is important that the "standard deviation" used must be appropriate for the parameter we are estimating, so in this section we need to use the standard deviation that applies to sample means, which is σ, is commonly called the n. The fraction σ n "standard error of the mean" in order to distinguish clearly the standard deviation for a mean from the population standard deviation σ. In summary, as a result of the central limit theorem: ¯ • X is normally distributed, that is, X ¯ ~ N ⎛ ⎝µ X, σ n ⎞ ⎠. • When the population standard deviation σ is known, we use a normal |
distribution to calculate the error bound. Calculating the Confidence Interval To construct a confidence interval estimate for an unknown population mean, we need data from a random sample. The steps to construct and interpret the confidence interval are: • Calculate the sample mean x¯ standard deviation σ. from the sample data. Remember, in this section we already know the population • Find the z-score that corresponds to the confidence level. • Calculate the error bound EBM. • Construct the confidence interval. • Write a sentence that interprets the estimate in the context of the situation in the problem. (Explain what the confidence interval means, in the words of the problem.) We will first examine each step in more detail, and then illustrate the process with some examples. Finding the z-score for the Stated Confidence Level When we know the population standard deviation σ, we use a standard normal distribution to calculate the error bound EBM and construct the confidence interval. We need to find the value of z that puts an area equal to the confidence level (in decimal form) in the middle of the standard normal distribution Z ~ N(0, 1). This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 The confidence level, CL, is the area in the middle of the standard normal distribution. CL = 1 – α, so α is the area that is split equally between the two tails. Each of the tails contains an area equal to α 2. CHAPTER 8 | CONFIDENCE INTERVALS 415 The z-score that has an area to the right of α 2 is denoted by z α 2. For example, when CL = 0.95, α = 0.05 and α 2 = 0.025; we write z α 2 = z0.025. The area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 – 0.025 = 0.975. = z0.025 = 1.96, using a calculator, computer or a standard normal probability table. z α 2 invNorm(0.975, 0, 1) = 1.96 NOTE Remember to use the area to the LEFT of z α 2 ; in this chapter the last two inputs in the invNorm command are 0, 1, because you are using a standard normal distribution Z ~ N |
(0, 1). Calculating the Error Bound (EBM) The error bound formula for an unknown population mean μ when the population standard deviation σ is known is • EBM = ⎛ ⎝z α 2 ⎞ ⎛ ⎝ ⎠ σ n ⎞ ⎠ Constructing the Confidence Interval • The confidence interval estimate has the format ( x¯ – EBM, x¯ + EBM). The graph gives a picture of the entire situation. CL + α 2 = CL + α = 1. + α 2 Figure 8.3 Writing the Interpretation The interpretation should clearly state the confidence level (CL), explain what population parameter is being estimated (here, a population mean), and state the confidence interval (both endpoints). "We estimate with ___% confidence that the true population mean (include the context of the problem) is between ___ and ___ (include appropriate units)." 416 CHAPTER 8 | CONFIDENCE INTERVALS Example 8.2 Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of three points. A random sample of 36 scores is taken and gives a sample mean (sample mean score) of 68. Find a confidence interval estimate for the population mean exam score (the mean score on all exams). Find a 90% confidence interval for the true (population) mean of statistics exam scores. Solution 8.2 • You can use technology to calculate the confidence interval directly. • The first solution is shown step-by-step (Solution A). • The second solution uses the TI-83, 83+, and 84+ calculators (Solution B). Solution A To find the confidence interval, you need the sample mean, x¯, and the EBM. x¯ = 68 EBM = ⎛ ⎝; n = 36; The confidence level is 90% (CL = 0.90) CL = 0.90 so α = 1 – CL = 1 – 0.90 = 0.10 α 2 = 0.05 z α 2 = z0.05 The area to the right of z0.05 is 0.05 and the area to the left of z0.05 is 1 – 0.05 = 0.95. = z0.05 = 1.645 z α 2 using invNorm(0.95, 0, 1) on the TI-83,83+, and 84+ calculators. This can also |
be found using appropriate commands on other calculators, using a computer, or using a probability table for the standard normal distribution. EBM = (1.645) ⎛ ⎝ ⎞ ⎠ 3 36 = 0.8225 x¯ - EBM = 68 - 0.8225 = 67.1775 x¯ + EBM = 68 + 0.8225 = 68.8225 The 90% confidence interval is (67.1775, 68.8225). Solution 8.2 Solution B Press STAT and arrow over to TESTS. Arrow down to 7:ZInterval. Press ENTER. Arrow to Stats and press ENTER. Arrow down and enter three for σ, 68 for x¯, 36 for n, and.90 for C-level. Arrow down to Calculate and press ENTER. The confidence interval is (to three decimal places)(67.178, 68.822). Interpretation This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 417 We estimate with 90% confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82. Explanation of 90% Confidence Level Ninety percent of all confidence intervals constructed in this way contain the true mean statistics exam score. For example, if we constructed 100 of these confidence intervals, we would expect 90 of them to contain the true population mean exam score. 8.2 Suppose average pizza delivery times are normally distributed with an unknown population mean and a population standard deviation of six minutes. A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery time of 36 minutes. Find a 90% confidence interval estimate for the population mean delivery time. Example 8.3 The Specific Absorption Rate (SAR) for a cell phone measures the amount of radio frequency (RF) energy absorbed by the user’s body when using the handset. Every cell phone emits RF energy. Different phone models have different SAR measures. To receive certification from the Federal Communications Commission (FCC) for sale in the United States, the SAR level for a cell phone must be no more than 1.6 watts per kilogram. Table 8.1 shows the highest SAR level for a random selection of cell phone models as measured by the FCC. Phone Model SAR Phone Model SAR |
Phone Model Apple iPhone 4S 1.11 LG Ally BlackBerry Pearl 8120 1.48 LG AX275 BlackBerry Tour 9630 1.43 LG Cosmos Cricket TXTM8 1.3 LG CU515 HP/Palm Centro 1.09 LG Trax CU575 HTC One V 0.455 Motorola Q9h 1.36 1.34 1.18 1.3 1.26 1.29 Pantech Laser Samsung Character Samsung Epic 4G Touch Samsung M240 Samsung Messager III SCH-R750 0.68 Samsung Nexus S HTC Touch Pro 2 1.41 Motorola Razr2 V8 0.36 Samsung SGH-A227 Huawei M835 Ideos 0.82 Motorola Razr2 V9 0.52 SGH-a107 GoPhone Kyocera DuraPlus 0.78 Motorola V195s 1.6 Sony W350a Kyocera K127 Marbl 1.25 Nokia 1680 1.39 T-Mobile Concord Table 8.1 SAR 0.74 0.5 0.4 0.867 0.51 1.13 0.3 1.48 1.38 Find a 98% confidence interval for the true (population) mean of the Specific Absorption Rates (SARs) for cell phones. Assume that the population standard deviation is σ = 0.337. Solution 8.3 Solution A To find the confidence interval, start by finding the point estimate: the sample mean. x¯ = 1.024 Next, find the EBM. Because you are creating a 98% confidence interval, CL = 0.98. 418 CHAPTER 8 | CONFIDENCE INTERVALS Figure 8.4 You need to find z0.01 having the property that the area under the normal density curve to the right of z0.01 is 0.01 and the area to the left is 0.99. Use your calculator, a computer, or a probability table for the standard normal distribution to find z0.01 = 2.326. EBM = (z0.01) σ n = (2.326)0.337 30 = 0.1431 To find the 98% confidence interval, find x¯ ± EBM. x¯ – EBM = 1.024 – 0.1431 = 0.8809 x¯ – EBM = 1.024 – 0.1431 = 1.1671 We estimate with 98% confidence that the true SAR mean for the population of cell phones in the United States |
is between 0.8809 and 1.1671 watts per kilogram. Solution 8.3 Solution B Press STAT and arrow over to TESTS. Arrow down to 7:ZInterval. Press ENTER. Arrow to Stats and press ENTER. Arrow down and enter the following values: σ: 0.337 x¯ : 1.024 n: 30 C-level: 0.98 Arrow down to Calculate and press ENTER. The confidence interval is (to three decimal places) (0.881, 1.167). 8.3 Table 8.2 shows a different random sampling of 20 cell phone models. Use this data to calculate a 93% confidence interval for the true mean SAR for cell phones certified for use in the United States. As previously, assume that the population standard deviation is σ = 0.337. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 419 Phone Model SAR Phone Model Blackberry Pearl 8120 1.48 Nokia E71x HTC Evo Design 4G 0.8 Nokia N75 HTC Freestyle LG Ally LG Fathom 1.15 1.36 0.77 Nokia N79 Sagem Puma Samsung Fascinate SAR 1.53 0.68 1.4 1.24 0.57 LG Optimus Vu 0.462 Samsung Infuse 4G 0.2 Motorola Cliq XT Motorola Droid Pro 1.36 1.39 Samsung Nexus S 0.51 Samsung Replenish 0.3 Motorola Droid Razr M 1.3 Sony W518a Walkman 0.73 Nokia 7705 Twist 0.7 ZTE C79 0.869 Table 8.2 Notice the difference in the confidence intervals calculated in Example 8.3 and the following Try It 1.3 exercise. These intervals are different for several reasons: they were calculated from different samples, the samples were different sizes, and the intervals were calculated for different levels of confidence. Even though the intervals are different, they do not yield conflicting information. The effects of these kinds of changes are the subject of the next section in this chapter. Changing the Confidence Level or Sample Size Example 8.4 Suppose we change the original problem in Example 8.2 by using a 95% confidence level. Find a 95% confidence interval for the true (population) mean statistics exam score. Solution 8.4 To |
find the confidence interval, you need the sample mean, x¯, and the EBM. x¯ = 68 EBM = ⎛ ⎝; n = 36; The confidence level is 95% (CL = 0.95). CL = 0.95 so α = 1 – CL = 1 – 0.95 = 0.05 α 2 = 0.025 z α 2 = z0.025 The area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 – 0.025 = 0.975. = z0.025 = 1.96 z α 2 when using invnorm(0.975,0,1) on the TI-83, 83+, or 84+ calculators. (This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the standard normal distribution.) EBM = (1.96) ⎛ ⎝ ⎞ ⎠ 3 36 = 0.98 x¯ – EBM = 68 – 0.98 = 67.02 420 CHAPTER 8 | CONFIDENCE INTERVALS x¯ + EBM = 68 + 0.98 = 68.98 Notice that the EBM is larger for a 95% confidence level in the original problem. Interpretation We estimate with 95% confidence that the true population mean for all statistics exam scores is between 67.02 and 68.98. Explanation of 95% Confidence Level Ninety-five percent of all confidence intervals constructed in this way contain the true value of the population mean statistics exam score. Comparing the results The 90% confidence interval is (67.18, 68.82). The 95% confidence interval is (67.02, 68.98). The 95% confidence interval is wider. If you look at the graphs, because the area 0.95 is larger than the area 0.90, it makes sense that the 95% confidence interval is wider. To be more confident that the confidence interval actually does contain the true value of the population mean for all statistics exam scores, the confidence interval necessarily needs to be wider. Figure 8.5 Summary: Effect of Changing the Confidence Level • Increasing the confidence level increases the error bound, making the confidence interval wider. • Decreasing the confidence level decreases the error bound, making the confidence interval narrower. 8.4 Refer back to Try It. The population standard deviation is six minutes and the sample mean |
deliver time is 36 minutes. Use a sample size of 20. Find a 95% confidence interval estimate for the true mean pizza delivery time. Example 8.5 Suppose we change the original problem in Example 8.2 to see what happens to the error bound if the sample size is changed. Leave everything the same except the sample size. Use the original 90% confidence level. What happens to the error bound and the confidence interval if we increase the sample size and use n = 100 instead of n = 36? What happens if we decrease the sample size to n = 25 instead of n = 36? • x¯ = 68 • EBM = ⎛ ⎝; The confidence level is 90% (CL=0.90); z α 2 = z0.05 = 1.645. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 421 Solution 8.5 Solution A If we increase the sample size n to 100, we decrease the error bound. When n = 100: EBM = ⎛ ⎝1.645) ⎛ ⎝ ⎞ ⎠ 3 100 = 0.4935. Solution 8.5 Solution B If we decrease the sample size n to 25, we increase the error bound. When n = 25: EBM = ⎛ ⎝1.645) ⎛ ⎝ ⎞ ⎠ 3 25 = 0.987. Summary: Effect of Changing the Sample Size • Increasing the sample size causes the error bound to decrease, making the confidence interval narrower. • Decreasing the sample size causes the error bound to increase, making the confidence interval wider. 8.5 Refer back to Try It. The mean delivery time is 36 minutes and the population standard deviation is six minutes. Assume the sample size is changed to 50 restaurants with the same sample mean. Find a 90% confidence interval estimate for the population mean delivery time. Working Backwards to Find the Error Bound or Sample Mean When we calculate a confidence interval, we find the sample mean, calculate the error bound, and use them to calculate the confidence interval. However, sometimes when we read statistical studies, the study may state the confidence interval only. If we know the confidence interval, we can work backwards to find both the error bound and the sample mean |
. Finding the Error Bound • From the upper value for the interval, subtract the sample mean, • OR, from the upper value for the interval, subtract the lower value. Then divide the difference by two. Finding the Sample Mean • Subtract the error bound from the upper value of the confidence interval, • OR, average the upper and lower endpoints of the confidence interval. Notice that there are two methods to perform each calculation. You can choose the method that is easier to use with the information you know. Example 8.6 Suppose we know that a confidence interval is (67.18, 68.82) and we want to find the error bound. We may know that the sample mean is 68, or perhaps our source only gave the confidence interval and did not tell us the value of the sample mean. Calculate the Error Bound: • • If we know that the sample mean is 68: EBM = 68.82 – 68 = 0.82. If we don't know the sample mean: EBM = (68.82 − 67.18) 2 = 0.82. Calculate the Sample Mean: • If we know the error bound: x¯ = 68.82 – 0.82 = 68 422 CHAPTER 8 | CONFIDENCE INTERVALS • If we don't know the error bound: x¯ = (67.18 + 68.82) 2 = 68. 8.6 Suppose we know that a confidence interval is (42.12, 47.88). Find the error bound and the sample mean. Calculating the Sample Size n If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size. The error bound formula for a population mean when the population standard deviation is known is EBM = ⎛ ⎞ ⎛ ⎝ ⎠ σ n ⎞ ⎠. ⎝z α 2 The formula for sample size is n = z2 σ 2 EBM 2, found by solving the error bound formula for n. In this formula, z is z α 2, corresponding to the desired confidence level. A researcher planning a study who wants a specified confidence level and error bound can use this formula to calculate the size of the sample needed for the study. Example 8.7 The population standard deviation for the age of Foothill College students is 15 years. If we want to be 95% confident that the sample mean age is within two years of the true |
population mean age of Foothill College students, how many randomly selected Foothill College students must be surveyed? From the problem, we know that σ = 15 and EBM = 2. z = z0.025 = 1.96, because the confidence level is 95%. n = z2 σ 2 EBM 2 = (1.96)2 (15)2 22 = 216.09 using the sample size equation. Use n = 217: Always round the answer UP to the next higher integer to ensure that the sample size is large enough. Therefore, 217 Foothill College students should be surveyed in order to be 95% confident that we are within two years of the true population mean age of Foothill College students. 8.7 The population standard deviation for the height of high school basketball players is three inches. If we want to be 95% confident that the sample mean height is within one inch of the true population mean height, how many randomly selected students must be surveyed? 8.2 | A Single Population Mean using the Student t Distribution In practice, we rarely know the population standard deviation. In the past, when the sample size was large, this did not present a problem to statisticians. They used the sample standard deviation s as an estimate for σ and proceeded as before to calculate a confidence interval with close enough results. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in the confidence interval. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 423 William S. Goset (1876–1937) of the Guinness brewery in Dublin, Ireland ran into this problem. His experiments with hops and barley produced very few samples. Just replacing σ with s did not produce accurate results when he tried to calculate a confidence interval. He realized that he could not use a normal distribution for the calculation; he found that the actual distribution depends on the sample size. This problem led him to "discover" what is called the Student's t-distribution. The name comes from the fact that Gosset wrote under the pen name "Student." Up until the mid-1970s, some statisticians used the normal distribution approximation for large sample sizes and only used the Student's t-distribution only for sample sizes of at most |
30. With graphing calculators and computers, the practice now is to use the Student's t-distribution whenever s is used as an estimate for σ. If you draw a simple random sample of size n from a population that has an approximately a normal distribution with mean μ and unknown population standard deviation σ and calculate the t-score t = x¯ – µ, then the t-scores follow a Student's ⎛ ⎝ s n ⎞ ⎠ t-distribution with n – 1 degrees of freedom. The t-score has the same interpretation as the z-score. It measures how far x¯ is from its mean μ. For each sample size n, there is a different Student's t-distribution. The degrees of freedom, n – 1, come from the calculation of the sample standard deviation s. In Appendix H, we used n deviations (x – x¯ values) to calculate s. Because the sum of the deviations is zero, we can find the last deviation once we know the other n – 1 deviations. The other n – 1 deviations can change or vary freely. We call the number n – 1 the degrees of freedom (df). Properties of the Student's t-Distribution • The graph for the Student's t-distribution is similar to the standard normal curve. • The mean for the Student's t-distribution is zero and the distribution is symmetric about zero. • The Student's t-distribution has more probability in its tails than the standard normal distribution because the spread of the t-distribution is greater than the spread of the standard normal. So the graph of the Student's t-distribution will be thicker in the tails and shorter in the center than the graph of the standard normal distribution. • The exact shape of the Student's t-distribution depends on the degrees of freedom. As the degrees of freedom increases, the graph of Student's t-distribution becomes more like the graph of the standard normal distribution. • The underlying population of individual observations is assumed to be normally distributed with unknown population mean μ and unknown population standard deviation σ. The size of the underlying population is generally not relevant unless it is very small. If it is bell shaped (normal) then the assumption is met and doesn't need discussion. Random sampling is assumed, but that is a completely separate assumption from normality. Calculators and computers can easily calculate any Student's t-probabilities. The TI-83,83+, |
and 84+ have a tcdf function to find the probability for given values of t. The grammar for the tcdf command is tcdf(lower bound, upper bound, degrees of freedom). However for confidence intervals, we need to use inverse probability to find the value of t when we know the probability. For the TI-84+ you can use the invT command on the DISTRibution menu. The invT command works similarly to the invnorm. The invT command requires two inputs: invT(area to the left, degrees of freedom) The output is the t-score that corresponds to the area we specified. The TI-83 and 83+ do not have the invT command. (The TI-89 has an inverse T command.) A probability table for the Student's t-distribution can also be used. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row). (The TI-86 does not have an invT program or command, so if you are using that calculator, you need to use a probability table for the Student's t-Distribution.) When using a t-table, note that some tables are formatted to show the confidence level in the column headings, while the column headings in some tables may show only corresponding area in one or both tails. A Student's t table (See Appendix H) gives t-scores given the degrees of freedom and the right-tailed probability. The table is very limited. Calculators and computers can easily calculate any Student's t-probabilities. The notation for the Student's t-distribution (using T as the random variable) is: • T ~ tdf where df = n – 1. • For example, if we have a sample of size n = 20 items, then we calculate the degrees of freedom as df = n - 1 = 20 - 1 = 19 and we write the distribution as T ~ t19. If the population standard deviation is not known, the error bound for a population mean is: • EBM = ⎛ ⎝t α 2 ⎞ ⎛ ⎝ ⎠ s n ⎞ ⎠, 424 CHAPTER 8 | CONFIDENCE INTERVALS • t σ 2 is the t-score with area to the right equal to α 2, • use df = n – 1 degrees of freedom, and • s = sample standard deviation. The format for the confidence interval |
is: ( x¯ − EBM, x¯ + EBM). To calculate the confidence interval directly: Press STAT. Arrow over to TESTS. Arrow down to 8:TInterval and press ENTER (or just press 8). Example 8.8 Suppose you do a study of acupuncture to determine how effective it is in relieving pain. You measure sensory rates for 15 subjects with the results given. Use the sample data to construct a 95% confidence interval for the mean sensory rate for the population (assumed normal) from which you took the data. The solution is shown step-by-step and by using the TI-83, 83+, or 84+ calculators. 8.6; 9.4; 7.9; 6.8; 8.3; 7.3; 9.2; 9.6; 8.7; 11.4; 10.3; 5.4; 8.1; 5.5; 6.9 Solution 8.8 • The first solution is step-by-step (Solution A). • The second solution uses the TI-83+ and TI-84 calculators (Solution B). Solution A To find the confidence interval, you need the sample mean, x¯, and the EBM. x¯ = 8.2267 s = 1.6722 n = 15 df = 15 – 1 = 14 CL so α = 1 – CL = 1 – 0.95 = 0.05 α 2 = 0.025 t α 2 = t0.025 The area to the right of t0.025 is 0.025, and the area to the left of t0.025 is 1 – 0.025 = 0.975 = t0.025 = 2.14 using invT(.975,14) on the TI-84+ calculator. t α 2 EBM = ⎛ ⎝t α 2 ⎞ ⎛ ⎝ ⎠ s n ⎞ ⎠ ⎛ EBM = (2.14) ⎝ 1.6722 15 ⎞ ⎠ = 0.924 x¯ – EBM = 8.2267 – 0.9240 = 7.3 x¯ + EBM = 8.2267 + 0.9240 = 9.15 The 95% confidence interval is (7.30, 9.15). We estimate with 95% confidence that the true population mean sensory rate is between 7. |
30 and 9.15. Solution 8.8 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 425 Press STAT and arrow over to TESTS. Arrow down to 8:TInterval and press ENTER (or you can just press 8). Arrow to Data and press ENTER. Arrow down to List and enter the list name where you put the data. There should be a 1 after Freq. Arrow down to C-level and enter 0.95 Arrow down to Calculate and press ENTER. The 95% confidence interval is (7.3006, 9.1527) NOTE When calculating the error bound, a probability table for the Student's t-distribution can also be used to find the value of t. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row); the t-score is found where the row and column intersect in the table. 8.8 You do a study of hypnotherapy to determine how effective it is in increasing the number of hourse of sleep subjects get each night. You measure hours of sleep for 12 subjects with the following results. Construct a 95% confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data. 8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5 Example 8.9 The Human Toxome Project (HTP) is working to understand the scope of industrial pollution in the human body. Industrial chemicals may enter the body through pollution or as ingredients in consumer products. In October 2008, the scientists at HTP tested cord blood samples for 20 newborn infants in the United States. The cord blood of the "In utero/newborn" group was tested for 430 industrial compounds, pollutants, and other chemicals, including chemicals linked to brain and nervous system toxicity, immune system toxicity, and reproductive toxicity, and fertility problems. There are health concerns about the effects of some chemicals on the brain and nervous system. Table 8.2 shows how many of the targeted chemicals were found in each infant’s cord blood. 79 145 147 160 116 100 159 151 156 126 137 83 |
156 94 121 144 123 114 139 99 Table 8.3 Use this sample data to construct a 90% confidence interval for the mean number of targeted industrial chemicals to be found in an in infant’s blood. Solution 8.9 Solution A 426 CHAPTER 8 | CONFIDENCE INTERVALS From the sample, you can calculate x¯ = 127.45 and s = 25.965. There are 20 infants in the sample, so n = 20, and df = 20 – 1 = 19. You are asked to calculate a 90% confidence interval: CL = 0.90, so α = 1 – CL = 1 – 0.90 = 0.10 α 2 = t0.05 = 0.05,t α 2 By definition, the area to the right of t0.05 is 0.05 and so the area to the left of t0.05 is 1 – 0.05 = 0.95. Use a table, calculator, or computer to find that t0.05 = 1.729. EBM =.729 ⎝ 25.965 20 ⎞ ⎠ ≈ 10.038 x¯ – EBM = 127.45 – 10.038 = 117.412 x¯ + EBM = 127.45 + 10.038 = 137.488 We estimate with 90% confidence that the mean number of all targeted industrial chemicals found in cord blood in the United States is between 117.412 and 137.488. Solution 8.9 Solution B Enter the data as a list. Press STAT and arrow over to TESTS. Arrow down to 8:TInterval and press ENTER (or you can just press 8). Arrow to Data and press ENTER. Arrow down to List and enter the list name where you put the data. Arrow down to Freq and enter 1. Arrow down to C-level and enter 0.90 Arrow down to Calculate and press ENTER. The 90% confidence interval is (117.41, 137.49). 8.9 A random sample of statistics students were asked to estimate the total number of hours they spend watching television in an average week. The responses are recorded in Table 8.4. Use this sample data to construct a 98% confidence interval for the mean number of hours statistics students will spend watching television in one week. 0 5 3 1 20 9 10 1 10 4 14 2 4 4 5 Table 8.4 This content is available for free at http://textbookequity |
.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 427 8.3 | A Population Proportion During an election year, we see articles in the newspaper that state confidence intervals in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show that the candidate has 40% of the vote within three percentage points (if the sample is large enough). Often, election polls are calculated with 95% confidence, so, the pollsters would be 95% confident that the true proportion of voters who favored the candidate would be between 0.37 and 0.43: (0.40 – 0.03,0.40 + 0.03). Investors in the stock market are interested in the true proportion of stocks that go up and down each week. Businesses that sell personal computers are interested in the proportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers. The procedure to find the confidence interval, the sample size, the error bound, and the confidence level for a proportion is similar to that for the population mean, but the formulas are different. How do you know you are dealing with a proportion problem? First, the underlying distribution is a binomial distribution. (There is no mention of a mean or average.) If X is a binomial random variable, then X ~ B(n, p) where n is the number of trials and p is the probability of a success. To form a proportion, take X, the random variable for the number of successes and divide it by n, the number of trials (or the sample size). The random variable P′ (read "P prime") is that proportion, P′ = X n (Sometimes the random variable is denoted as P ^, read "P hat".) When n is large and p is not close to zero or one, we can use the normal distribution to approximate the binomial. X ~ N(np, npq) If we divide the random variable, the mean, and the standard deviation by n, we get a normal distribution of proportions with P′, called the estimated proportion, as the random variable. (Recall that a proportion as the number of successes divided by n.) n = P′ ~ N� |
�� X ⎝ npq n np n, ⎞ ⎠ Using algebra to simplify : npq n = pq n P′ follows a normal distribution for proportions: X n = P′ ~ N⎛ ⎝ np n, npq n ⎞ ⎠ The confidence interval has the form (p′ – EBP, p′ + EBP). EBP is error bound for the proportion. p′ = x n p′ = the estimated proportion of successes (p′ is a point estimate for p, the true proportion.) x = the number of successes n = the size of the sample The error bound for a proportion is EBP = ⎛ ⎝z α 2 ⎛ ⎞ ⎠ ⎝ p′ q′ n ⎞ ⎠ where q′ = 1 – p′ This formula is similar to the error bound formula for a mean, except that the "appropriate standard deviation" is different. For a mean, when the population standard deviation is known, the appropriate standard deviation that we use is σ. For a n proportion, the appropriate standard deviation is pq n. However, in the error bound formula, we use p′ q′ n as the standard deviation, instead of pq n. In the error bound formula, the sample proportions p′ and q′ are estimates of the unknown population proportions p and q. The estimated proportions p′ and q′ are used because p and q are not known. The sample proportions p′ and q′ are calculated from the data: p′ is the estimated proportion of successes, and q′ is the estimated proportion of failures. 428 CHAPTER 8 | CONFIDENCE INTERVALS The confidence interval can be used only if the number of successes np′ and the number of failures nq′ are both greater than five. NOTE For the normal distribution of proportions, the z-score formula is as follows. If P′ ~N⎛ ⎝p, pq n ⎞ ⎠ then the z-score formula is z = p′ − p pq n This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 429 Example 8.10 Suppose that a market research firm is hired to estimate the percent of adults living |
in a large city who have cell phones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes - they own cell phones. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones. Solution 8.10 Solution A • The first solution is step-by-step (Solution A). • The second solution uses a function of the TI-83, 83+ or 84 calculators (Solution B). Let X = the number of people in the sample who have cell phones. X is binomial. X ~ B⎛ ⎝500, 421 500 ⎞ ⎠. To calculate the confidence interval, you must find p′, q′, and EBP. n = 500 x = the number of successes = 421 p′ = x n = 421 500 = 0.842 p′ = 0.842 is the sample proportion; this is the point estimate of the population proportion. q′ = 1 – p′ = 1 – 0.842 = 0.158 Since CL = 0.95, then α = 1 – CL = 1 – 0.95 = 0.05 ⎛ ⎝ = 0.025. ⎞ ⎠ α 2 Then z α 2 = z0.025 = 1.96 Use the TI-83, 83+, or 84+ calculator command invNorm(0.975,0,1) to find z0.025. Remember that the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 0.975. This can also be found using appropriate commands on other calculators, using a computer, or using a Standard Normal probability table. EBP = ⎛ ⎝z α 2 ⎞ ⎠ p′ q′ n = (1.96) (0.842)(0.158) 500 = 0.032 p'– EBP = 0.842 – 0.032 = 0.81 p′ + EBP = 0.842 + 0.032 = 0.874 The confidence interval for the true binomial population proportion is (p′ – EBP, p′ + EBP) = (0.810, 0.874). Interpretation We estimate with 95% confidence that between 81% and 87.4 |
% of all adult residents of this city have cell phones. Explanation of 95% Confidence Level Ninety-five percent of the confidence intervals constructed in this way would contain the true value for the population proportion of all adult residents of this city who have cell phones. Solution 8.10 Solution B Press STAT and arrow over to TESTS. Arrow down to A:1-PropZint. Press ENTER. Arrow down to x and enter 421. 430 CHAPTER 8 | CONFIDENCE INTERVALS Arrow down to n and enter 500. Arrow down to C-Level and enter.95. Arrow down to Calculate and press ENTER. The confidence interval is (0.81003, 0.87397). 8.10 Suppose 250 randomly selected people are surveyed to determine if they own a tablet. Of the 250 surveyed, 98 reported owning a tablet. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of people who own tablets. Example 8.11 For a class project, a political science student at a large university wants to estimate the percent of students who are registered voters. He surveys 500 students and finds that 300 are registered voters. Compute a 90% confidence interval for the true percent of students who are registered voters, and interpret the confidence interval. Solution 8.11 • The first solution is step-by-step (Solution A). • The second solution uses a function of the TI-83, 83+, or 84 calculators (Solution B). Solution A x = 300 and n = 500 p′ = x n = 300 500 = 0.600 q′ = 1 - p′ = 1 - 0.600 = 0.400 Since CL = 0.90, then α = 1 – CL = 1 – 0.90 = 0.10 ⎛ ⎝ = 0.05 ⎞ ⎠ α 2 = z0.05 = 1.645 z α 2 Use the TI-83, 83+, or 84+ calculator command invNorm(0.95,0,1) to find z0.05. Remember that the area to the right of z0.05 is 0.05 and the area to the left of z0.05 is 0.95. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table. EBP = ⎛ ⎝z α 2 ⎞ ⎠ p′ q′ n = (1. |
645) (0.60)(0.40) 500 = 0.036 p′ – EBP = 0.60 − 0.036 = 0.564 p′ + EBP = 0.60 + 0.036 = 0.636 The confidence interval for the true binomial population proportion is (p′ – EBP, p′ + EBP) = (0.564,0.636). Interpretation • We estimate with 90% confidence that the true percent of all students that are registered voters is between 56.4% and 63.6%. • Alternate Wording: We estimate with 90% confidence that between 56.4% and 63.6% of ALL students are registered voters. Explanation of 90% Confidence Level This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 431 Ninety percent of all confidence intervals constructed in this way contain the true value for the population percent of students that are registered voters. Solution 8.11 Solution B Press STAT and arrow over to TESTS. Arrow down to A:1-PropZint. Press ENTER. Arrow down to x and enter 300. Arrow down to n and enter 500. Arrow down to C-Level and enter 0.90. Arrow down to Calculate and press ENTER. The confidence interval is (0.564, 0.636). 8.11 A student polls his school to see if students in the school district are for or against the new legislation regarding school uniforms. She surveys 600 students and finds that 480 are against the new legislation. a. Compute a 90% confidence interval for the true percent of students who are against the new legislation, and interpret the confidence interval. b. In a sample of 300 students, 68% said they own an iPod and a smart phone. Compute a 97% confidence interval for the true percent of students who own an iPod and a smartphone. “Plus Four” Confidence Interval for p There is a certain amount of error introduced into the process of calculating a confidence interval for a proportion. Because we do not know the true proportion for the population, we are forced to use point estimates to calculate the appropriate standard deviation of the sampling distribution. Studies have shown that the resulting estimation of the standard deviation can be flawed. Fortunately, there is a |
simple adjustment that allows us to produce more accurate confidence intervals. We simply pretend that we have four additional observations. Two of these observations are successes and two are failures. The new sample size, then, is n + 4, and the new count of successes is x + 2. Computer studies have demonstrated the effectiveness of this method. It should be used when the confidence level desired is at least 90% and the sample size is at least ten. Example 8.12 A random sample of 25 statistics students was asked: “Have you smoked a cigarette in the past week?” Six students reported smoking within the past week. Use the “plus-four” method to find a 95% confidence interval for the true proportion of statistics students who smoke. Solution 8.12 Solution A 432 CHAPTER 8 | CONFIDENCE INTERVALS Six students out of 25 reported smoking within the past week, so x = 6 and n = 25. Because we are using the “plus-four” method, we will use x = 6 + 2 = 8 and n = 25 + 4 = 29. p′ = x n = 8 29 ≈ 0.276 q′ = 1 – p′ = 1 – 0.276 = 0.724 Since CL = 0.95, we know α = 1 – 0.95 = 0.05 and α 2 = 0.025. z0.025 = 1.96 EPB = ⎛ ⎝z α 2 ⎞ ⎠ p′ q′ n = (1.96) 0.276(0.724) 29 ≈ 0.163 p′ – EPB = 0.276 – 0.163 = 0.113 p′ + EPB = 0.276 + 0.163 = 0.439 We are 95% confident that the true proportion of all statistics students who smoke cigarettes is between 0.113 and 0.439. Solution 8.12 Solution B Press STAT and arrow over to TESTS. Arrow down to A:1-PropZint. Press ENTER. REMINDER Remember that the plus-four method assume an additional four trials: two successes and two failures. You do not need to change the process for calculating the confidence interval; simply update the values of x and n to reflect these additional trials. Arrow down to x and enter eight. Arrow down to n and enter 29. Arrow down to C-Level and enter 0.95. Arrow down to Calculate and press |
ENTER. The confidence interval is (0.113, 0.439). 8.12 Out of a random sample of 65 freshmen at State University, 31 students have declared a major. Use the “plusfour” method to find a 96% confidence interval for the true proportion of freshmen at State University who have declared a major. Example 8.13 The Berkman Center for Internet & Society at Harvard recently conducted a study analyzing the privacy management habits of teen internet users. In a group of 50 teens, 13 reported having more than 500 friends on This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 Facebook. Use the “plus four” method to find a 90% confidence interval for the true proportion of teens who would report having more than 500 Facebook friends. CHAPTER 8 | CONFIDENCE INTERVALS 433 Solution 8.13 Solution A Using “plus-four,” we have x = 13 + 2 = 15 and n = 50 + 4 = 54. p' = 15 54 ≈ 0.278 q' = 1 – p' = 1 - 0.241 = 0.722 Since CL = 0.90, we know α = 1 – 0.90 = 0.10 and α 2 = 0.05. z0.05 = 1.645 EPB = (z α 2 ⎛ ) ⎝ p′ q′ n ⎞ ⎛ ⎠ = (1.645) ⎝ (0.278)(0.722) 54 ⎞ ⎠ ≈ 0.100 p′ – EPB = 0.278 – 0.100 = 0.178 p′ + EPB = 0.278 + 0.100 = 0.378 We are 90% confident that between 17.8% and 37.8% of all teens would report having more than 500 friends on Facebook. Solution 8.13 Solution B Press STAT and arrow over to TESTS. Arrow down to A:1-PropZint. Press ENTER. Arrow down to x and enter 15. Arrow down to n and enter 54. Arrow down to C-Level and enter 0.90. Arrow down to Calculate and press ENTER. The confidence interval is (0.178, 0.378). 8.13 The Berkman Center Study referenced in Example 8.13 talked |
to teens in smaller focus groups, but also interviewed additional teens over the phone. When the study was complete, 588 teens had answered the question about their Facebook friends with 159 saying that they have more than 500 friends. Use the “plus-four” method to find a 90% confidence interval for the true proportion of teens that would report having more than 500 Facebook friends based on this larger sample. Compare the results to those in Example 8.13. Calculating the Sample Size n If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size. The error bound formula for a population proportion is 434 CHAPTER 8 | CONFIDENCE INTERVALS • EBP = ⎛ ⎝z α 2 ⎛ ⎞ ⎠ ⎝ p′ q′ n ⎞ ⎠ • Solving for n gives you an equation for the sample size. (p′ q′) 2 ⎞ ⎛ ⎝z α ⎠ 2 EBP2 • n = Example 8.14 Suppose a mobile phone company wants to determine the current percentage of customers aged 50+ who use text messaging on their cell phones. How many customers aged 50+ should the company survey in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of customers aged 50+ who use text messaging on their cell phones. Solution 8.14 From the problem, we know that EBP = 0.03 (3%=0.03) and z α 2 z0.05 = 1.645 because the confidence level is 90%. However, in order to find n, we need to know the estimated (sample) proportion p′. Remember that q′ = 1 – p′. But, we do not know p′ yet. Since we multiply p′ and q′ together, we make them both equal to 0.5 because p′q′ = (0.5)(0.5) = 0.25 results in the largest possible product. (Try other products: (0.6)(0.4) = 0.24; (0.3)(0.7) = 0.21; (0.2)(0.8) = 0.16 and so on). The largest possible product gives us the largest n. This gives us a large enough sample so that we can be 90% confident that we are within three percentage |
points of the true population proportion. To calculate the sample size n, use the formula and make the substitutions. n = z2 p′ q′ EBP2 gives n = 1.6452(0.5)(0.5) 0.032 = 751.7 Round the answer to the next higher value. The sample size should be 752 cell phone customers aged 50+ in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of all customers aged 50+ who use text messaging on their cell phones. 8.14 Suppose an internet marketing company wants to determine the current percentage of customers who click on ads on their smartphones. How many customers should the company survey in order to be 90% confident that the estimated proportion is within five percentage points of the true population proportion of customers who click on ads on their smartphones? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 435 8.1 Confidence Interval (Home Costs) Class Time: Names: Student Learning Outcomes • The student will calculate the 90% confidence interval for the mean cost of a home in the area in which this school is located. • The student will interpret confidence intervals. • The student will determine the effects of changing conditions on the confidence interval. Collect the Data Check the Real Estate section in your local newspaper. Record the sale prices for 35 randomly selected homes recently listed in the county. NOTE Many newspapers list them only one day per week. Also, we will assume that homes come up for sale randomly. 1. Complete the table: __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ Table 8.5 Describe the Data 1. Compute the following: a. b. x¯ = _____ s x = _____ c. n = _____ 2. ¯. In words, define the random |
variable X 3. State the estimated distribution to use. Use both words and symbols. Find the Confidence Interval 1. Calculate the confidence interval and the error bound. a. Confidence Interval: _____ b. Error Bound: _____ 436 CHAPTER 8 | CONFIDENCE INTERVALS 2. How much area is in both tails (combined)? α = _____ 3. How much area is in each tail? α 2 = _____ 4. Fill in the blanks on the graph with the area in each section. Then, fill in the number line with the upper and lower limits of the confidence interval and the sample mean. Figure 8.6 5. Some students think that a 90% confidence interval contains 90% of the data. Use the list of data on the first page and count how many of the data values lie within the confidence interval. What percent is this? Is this percent close to 90%? Explain why this percent should or should not be close to 90%. Describe the Confidence Interval 1. In two to three complete sentences, explain what a confidence interval means (in general), as if you were talking to someone who has not taken statistics. 2. In one to two complete sentences, explain what this confidence interval means for this particular study. Use the Data to Construct Confidence Intervals 1. Using the given information, construct a confidence interval for each confidence level given. Confidence level EBM/Error Bound Confidence Interval 50% 80% 95% 99% Table 8.6 2. What happens to the EBM as the confidence level increases? Does the width of the confidence interval increase or decrease? Explain why this happens. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 437 8.2 Confidence Interval (Place of Birth) Class Time: Names: Student Learning Outcomes • The student will calculate the 90% confidence interval the proportion of students in this school who were born in this state. • The student will interpret confidence intervals. • The student will determine the effects of changing conditions on the confidence interval. Collect the Data 1. Survey the students in your class, asking them if they were born in this state. Let X = the number that were born in this state. a. n = ____________ b. x = |
____________ 2. In words, define the random variable P′. 3. State the estimated distribution to use. Find the Confidence Interval and Error Bound 1. Calculate the confidence interval and the error bound. a. Confidence Interval: _____ b. Error Bound: _____ 2. How much area is in both tails (combined)? α = _____ 3. How much area is in each tail? α 2 = _____ 4. Fill in the blanks on the graph with the area in each section. Then, fill in the number line with the upper and lower limits of the confidence interval and the sample proportion. Figure 8.7 Describe the Confidence Interval 1. In two to three complete sentences, explain what a confidence interval means (in general), as though you were talking to someone who has not taken statistics. 2. In one to two complete sentences, explain what this confidence interval means for this particular study. 3. Construct a confidence interval for each confidence level given. 438 CHAPTER 8 | CONFIDENCE INTERVALS Confidence level EBP/Error Bound Confidence Interval 50% 80% 95% 99% Table 8.7 4. What happens to the EBP as the confidence level increases? Does the width of the confidence interval increase or decrease? Explain why this happens. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 439 8.3 Confidence Interval (Women's Heights) Class Time: Names: Student Learning Outcomes • The student will calculate a 90% confidence interval using the given data. • The student will determine the relationship between the confidence level and the percentage of constructed intervals that contain the population mean. Given: 59.4 71.6 69.3 65.0 62.9 66.5 61.7 55.2 67.5 67.2 63.8 62.9 63.0 63.9 68.7 65.5 61.9 69.6 58.7 63.4 61.8 60.6 69.8 60.0 64.9 66.1 66.8 60.6 65.6 63.8 61.3 59.2 64.1 59.3 64.9 62.4 63.5 60.9 63.3 66.3 61.5 64 |
.3 62.9 60.6 63.8 58.8 64.9 65.7 62.5 70.9 62.9 63.1 62.2 58.7 64.7 66.0 60.5 64.7 65.4 60.2 65.0 64.1 61.1 65.3 64.6 59.2 61.4 62.0 63.5 61.4 65.5 62.3 65.5 64.7 58.8 66.1 64.9 66.9 57.9 69.8 58.5 63.4 69.2 65.9 62.2 60.0 58.1 62.5 62.4 59.1 66.4 61.2 60.4 58.7 66.7 67.5 63.2 56.6 67.7 62.5 Table 8.8 Heights of 100 Women (in Inches) 1. Table 8.8 lists the heights of 100 women. Use a random number generator to select ten data values randomly. 2. Calculate the sample mean and the sample standard deviation. Assume that the population standard deviation is known to be 3.3 inches. With these values, construct a 90% confidence interval for your sample of ten values. Write the confidence interval you obtained in the first space of Table 8.9. 3. Now write your confidence interval on the board. As others in the class write their confidence intervals on the board, copy them into Table 8.9. __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ 440 CHAPTER 8 | CONFIDENCE INTERVALS __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ Table 8.9 90% Confidence Intervals Discussion Questions 1. The actual population mean for the 100 heights given Table 8.8 is μ = 63.4. Using the class listing of confidence intervals, count how many of them contain the population mean μ; i.e., for how many intervals does the value of μ lie between the endpoints |
of the confidence interval? 2. Divide this number by the total number of confidence intervals generated by the class to determine the percent of confidence intervals that contains the mean μ. Write this percent here: _____________. 3. Is the percent of confidence intervals that contain the population mean μ close to 90%? 4. Suppose we had generated 100 confidence intervals. What do you think would happen to the percent of confidence intervals that contained the population mean? 5. When we construct a 90% confidence interval, we say that we are 90% confident that the true population mean lies within the confidence interval. Using complete sentences, explain what we mean by this phrase. 6. Some students think that a 90% confidence interval contains 90% of the data. Use the list of data given (the heights of women) and count how many of the data values lie within the confidence interval that you generated based on that data. How many of the 100 data values lie within your confidence interval? What percent is this? Is this percent close to 90%? 7. Explain why it does not make sense to count data values that lie in a confidence interval. Think about the random variable that is being used in the problem. 8. Suppose you obtained the heights of ten women and calculated a confidence interval from this information. Without knowing the population mean μ, would you have any way of knowing for certain if your interval actually contained the value of μ? Explain. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 KEY TERMS CHAPTER 8 | CONFIDENCE INTERVALS 441 Binomial Distribution a discrete random variable (RV) which arises from Bernoulli trials; there are a fixed number, n, of independent trials. “Independent” means that the result of any trial (for example, trial 1) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV X is defined as the number of successes in n trials. The notation is: X~B(n,p). The mean is μ = np and ⎠p x qn − x. the standard deviation is σ = npq. The probability of exactly x successes in n trials is P⎛ ⎝X = x⎞ ⎠ = ⎛ ⎞ ⎝ n x |
Confidence Interval (CI) an interval estimate for an unknown population parameter. This depends on: • • • the desired confidence level, information that is known about the distribution (for example, known standard deviation), the sample and its size. Confidence Level (CL) the percent expression for the probability that the confidence interval contains the true population parameter; for example, if the CL = 90%, then in 90 out of 100 samples the interval estimate will enclose the true population parameter. Degrees of Freedom (df) the number of objects in a sample that are free to vary Error Bound for a Population Mean (EBM) the margin of error; depends on the confidence level, sample size, and known or estimated population standard deviation. Error Bound for a Population Proportion (EBP) the margin of error; depends on the confidence level, the sample size, and the estimated (from the sample) proportion of successes. Inferential Statistics also called statistical inference or inductive statistics; this facet of statistics deals with estimating a population parameter based on a sample statistic. For example, if four out of the 100 calculators sampled are defective we might infer that four percent of the production is defective. Normal Distribution a continuous random variable (RV) with pdf f (x) = 1 σ 2π – (x – µ)2/ 2σ 2 e, where μ is the mean of the distribution and σ is the standard deviation, notation: X ~ N(μ,σ). If μ = 0 and σ = 1, the RV is called the standard normal distribution. Parameter a numerical characteristic of a population Point Estimate a single number computed from a sample and used to estimate a population parameter Standard Deviation a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation Student's t-Distribution investigated and reported by William S. Gossett in 1908 and published under the pseudonym Student; the major characteristics of the random variable (RV) are: • It is continuous and assumes any real values. • The pdf is symmetrical about its mean of zero. However, it is more spread out and flatter at the apex than the normal distribution. • It approaches the standard normal distribution as n get larger. • There is a "family of t–distributions: each representative of the family is completely defined by the number of degrees of freedom, which is one less |
than the number of data. CHAPTER REVIEW 8.1 A Single Population Mean using the Normal Distribution In this module, we learned how to calculate the confidence interval for a single population mean where the population standard deviation is known. When estimating a population mean, the margin of error is called the error bound for a population mean (EBM). A confidence interval has the general form: (lower bound, upper bound) = (point estimate – EBM, point estimate + EBM) 442 CHAPTER 8 | CONFIDENCE INTERVALS The calculation of EBM depends on the size of the sample and the level of confidence desired. The confidence level is the percent of all possible samples that can be expected to include the true population parameter. As the confidence level increases, the corresponding EBM increases as well. As the sample size increases, the EBM decreases. By the central limit theorem, EBM = z σ n Given a confidence interval, you can work backwards to find the error bound (EBM) or the sample mean. To find the error bound, find the difference of the upper bound of the interval and the mean. If you do not know the sample mean, you can find the error bound by calculating half the difference of the upper and lower bounds. To find the sample mean given a confidence interval, find the difference of the upper bound and the error bound. If the error bound is unknown, then average the upper and lower bounds of the confidence interval to find the sample mean. Sometimes researchers know in advance that they want to estimate a population mean within a specific margin of error for a given level of confidence. In that case, solve the EBM formula for n to discover the size of the sample that is needed to achieve this goal: n = z2 σ 2 EBM 2 8.2 A Single Population Mean using the Student t Distribution In many cases, the researcher does not know the population standard deviation, σ, of the measure being studied. In these cases, it is common to use the sample standard deviation, s, as an estimate of σ. The normal distribution creates accurate confidence intervals when σ is known, but it is not as accurate when s is used as an estimate. In this case, the Student’s t-distribution is much better. Define a t-score using the following formula: t = x¯ − µ s n The t-score follows the Student’s t-distribution with n – 1 degrees of freedom. The |
confidence interval under this distribution is calculated with EBM = ⎛ ⎝t α 2 where t α 2, s is the sample is the t-score with area to the right equal to α 2 for a given α. standard deviation, and n is the sample size. Use a table, calculator, or computer to find t α 2 s n ⎞ ⎠ 8.3 A Population Proportion Some statistical measures, like many survey questions, measure qualitative rather than quantitative data. In this case, the population parameter being estimated is a proportion. It is possible to create a confidence interval for the true population proportion following procedures similar to those used in creating confidence intervals for population means. The formulas are slightly different, but they follow the same reasoning. Let p′ represent the sample proportion, x/n, where x represents the number of successes and n represents the sample size. Let q′ = 1 – p′. Then the confidence interval for a population proportion is given by the following formula: (lower bound, upper bound) = (p′ – EBP, p′ + EBP) = ⎛ ⎝p′ – z p′ q′ n, p′ + z p′ q′ n ⎞ ⎠ The “plus four” method for calculating confidence intervals is an attempt to balance the error introduced by using estimates of the population proportion when calculating the standard deviation of the sampling distribution. Simply imagine four additional trials in the study; two are successes and two are failures. Calculate p′ =, and proceed to find the x + 2 n + 4 confidence interval. When sample sizes are small, this method has been demonstrated to provide more accurate confidence intervals than the standard formula used for larger samples. FORMULA REVIEW 8.1 A Single Population Mean using the Normal Distribution ¯ X ⎝µ X, ~ N⎛ ⎞ ⎠ The distribution of sample means is normally distributed with mean equal to the population mean σ n and standard deviation given by the population standard deviation divided by the square root of the sample size. The general form for a confidence interval for a single population mean, known standard deviation, normal distribution is given by (lower bound, upper bound) = (point estimate – EBM, point This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 estimate + |
EBM) = ( x¯ − EBM, x¯ + EBM) = ⎛ n, x¯ + z σ n ⎞ ⎠ ⎝ x¯ − z σ EBM = z σ n = the error bound for the mean, or the margin of error for a single population mean; this formula is used when the population standard deviation is known. CL = confidence level, or the proportion of confidence intervals created that are expected to contain the true population parameter α = 1 – CL = the proportion of confidence intervals that will not contain the population parameter = the z-score with the property that the area to the z α 2 CHAPTER 8 | CONFIDENCE INTERVALS 443 T~tdf the random variable, T, has a Student’s t-distribution with df degrees of freedom s n = the error bound for the population mean EBM = t α 2 when the population standard deviation is unknown is the t-score in the Student’s t-distribution with area to t α 2 the right equal to α 2 The general form for a confidence interval for a single mean, population standard deviation unknown, Student's t is given by (lower bound, upper bound) = (point estimate – EBM, point estimate + EBM) = ⎛ ⎝ x¯ – ts n, x¯ + ts n ⎞ ⎠ right of the z-score is ∝ 2 this is the z-score used in the 8.3 A Population Proportion calculation of "EBM where α = 1 – CL. n = z2 σ 2 EBM 2 = the formula used to determine the sample size p′ = x / n where x represents the number of successes and n represents the sample size. The variable p′ is the sample proportion and serves as the point estimate for the true population proportion. (n) needed to achieve a desired margin of error at a given level of confidence General form of a confidence interval q′ = 1 – p′ p′ ~ N⎛ ⎝p, pq n ⎞ ⎠ The variable p′ has a binomial distribution (lower value, upper value) = (point estimate−error bound, point estimate + error bound) that can be approximated with the normal distribution shown here. To find the error bound when you know the confidence interval error bound = upper value−point estimate OR error bound = upper value − lower value |
2 Single Population Mean, Known Standard Deviation, Normal Distribution Use the Normal Distribution for Means, Population Standard Deviation is Known EBM = z α 2 σ n ⋅ The confidence interval has the format ( x¯ − EBM, x¯ + EBM). 8.2 A Single Population Mean using the Student t Distribution s = the standard deviation of sample values. EBP = the error bound for a proportion = z α 2 p′ q′ n Confidence interval for a proportion: (lower bound, upper bound) = (p′ – EBP, p′ + EBP) = ⎛ ⎝p′ – z p′ q′ n, p′ + z p′ q′ n ⎞ ⎠ 2 p′ q′ n = z α 2 EBP2 needed to estimate confidence 1 - α and margin of error EBP. provides the number of participants the population proportion with Use the normal distribution for a single population proportion p′ = x n EBP = ⎛ ⎝z α 2 ⎞ ⎠ p′q′ n p′ + q′ = 1 The confidence interval has the format (p′ – EBP, p′ + EBP). t = x¯ − µ s n is the formula for the t-score which measures x¯ is a point estimate for μ how far away a measure is from the population mean in the Student’s t-distribution df = n - 1; distribution where n represents the size of the sample the degrees of freedom for a Student’s t- PRACTICE p′ is a point estimate for ρ s is a point estimate for σ 8.1 A Single Population Mean using the Normal Distribution 444 CHAPTER 8 | CONFIDENCE INTERVALS Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 pounds. We wish to construct a 95% confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 pounds. The sample standard deviation is 11 pounds. 1. Identify the following: x¯ = _____ a. b. σ = _____ c. n = _____ ¯. 2. In words, define the random variables X and X 3. Which distribution should you use for this problem? 4. Construct a 95% confidence interval for the population mean weight of newborn elephants. State the confidence interval, |
sketch the graph, and calculate the error bound. 5. What will happen to the confidence interval obtained, if 500 newborn elephants are weighed instead of 50? Why? Use the following information to answer the next seven exercises: The U.S. Census Bureau conducts a study to determine the time needed to complete the short form. The Bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes. The population distribution is assumed to be normal. 6. Identify the following: x¯ = _____ a. b. σ = _____ c. n = _____ ¯. 7. In words, define the random variables X and X 8. Which distribution should you use for this problem? 9. Construct a 90% confidence interval for the population mean time to complete the forms. State the confidence interval, sketch the graph, and calculate the error bound. 10. If the Census wants to increase its level of confidence and keep the error bound the same by taking another survey, what changes should it make? 11. If the Census did another survey, kept the error bound the same, and surveyed only 50 people instead of 200, what would happen to the level of confidence? Why? 12. Suppose the Census needed to be 98% confident of the population mean length of time. Would the Census have to survey more people? Why or why not? Use the following information to answer the next ten exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 pounds with a standard deviation of 0.1 pounds. The population standard deviation is known to be 0.2 pounds. 13. Identify the following: x¯ = ______ a. b. σ = ______ c. n = ______ 14. In words, define the random variable X. ¯. 15. In words, define the random variable X 16. Which distribution should you use for this problem? 17. Construct a 90% confidence interval for the population mean weight of the heads of lettuce. State the confidence interval, sketch the graph, and calculate the error bound. 18. Construct a 95% confidence interval for the population mean weight of the heads of lettuce. State the confidence interval, sketch the graph, and calculate the error bound. 19. In complete sentences, explain why the confidence interval in Exercise 8.17 is larger than in Exercise 8.18. |
20. In complete sentences, give an interpretation of what the interval in Exercise 8.18 means. 21. What would happen if 40 heads of lettuce were sampled instead of 20, and the error bound remained the same? 22. What would happen if 40 heads of lettuce were sampled instead of 20, and the confidence level remained the same? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 Use the following information to answer the next 14 exercises: The mean age for all Foothill College students for a recent Fall term was 33.2. The population standard deviation has been pretty consistent at 15. Suppose that twenty-five Winter students were randomly selected. The mean age for the sample was 30.4. We are interested in the true mean age for Winter Foothill College students. Let X = the age of a Winter Foothill College student. CHAPTER 8 | CONFIDENCE INTERVALS 445 23. x¯ = _____ 24. n = _____ 25. ________ = 15 ¯. 26. In words, define the random variable X 27. What is x¯ estimating? 28. Is σ x known? 29. As a result of your answer to Exercise 8.26, state the exact distribution to use when calculating the confidence interval. Construct a 95% Confidence Interval for the true mean age of Winter Foothill College students by working out then answering the next seven exercises. 30. How much area is in both tails (combined)? α =________ 31. How much area is in each tail? α 2 =________ 32. Identify the following specifications: a. lower limit b. upper limit c. error bound 33. The 95% confidence interval is:__________________. 34. Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample mean. Figure 8.8 35. In one complete sentence, explain what the interval means. 36. Using the same mean, standard deviation, and level of confidence, suppose that n were 69 instead of 25. Would the error bound become larger or smaller? How do you know? 37. Using the same mean, standard deviation, and sample size, how would the error bound change if the confidence level were reduced to 90%? Why? 8.2 A Single Population Mean using the Student t Distribution Use the following information to |
answer the next five exercises. A hospital is trying to cut down on emergency room wait times. It is interested in the amount of time patients must wait before being called back to be examined. An investigation committee randomly surveyed 70 patients. The sample mean was 1.5 hours with a sample standard deviation of 0.5 hours. 38. Identify the following: a. b. x¯ =_______ s x =_______ 446 CHAPTER 8 | CONFIDENCE INTERVALS c. n =_______ d. n – 1 =_______ ¯ 39. Define the random variables X and X in words. 40. Which distribution should you use for this problem? 41. Construct a 95% confidence interval for the population mean time spent waiting. State the confidence interval, sketch the graph, and calculate the error bound. 42. Explain in complete sentences what the confidence interval means. Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal. 43. Identify the following: a. b. x¯ =_______ s x =_______ c. n =_______ d. n – 1 =_______ 44. Define the random variable X in words. ¯ 45. Define the random variable X in words. 46. Which distribution should you use for this problem? 47. Construct a 99% confidence interval for the population mean hours spent watching television per month. (a) State the confidence interval, (b) sketch the graph, and (c) calculate the error bound. 48. Why would the error bound change if the confidence level were lowered to 95%? Use the following information to answer the next 13 exercises: The data in Table 8.10 are the result of a random survey of 39 national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true mean number of colors on a national flag. Let X = the number of colors on a national flag. X Freq. 1 2 3 4 5 1 7 18 7 6 Table 8.10 49. Calculate the following: a. b. x¯ =______ s x =______ c. n =______ ¯ 50. Define the random variable X in words. 51. What is x¯ estimating? 52. Is σ x known? 53. As a result of |
your answer to Exercise 8.52, state the exact distribution to use when calculating the confidence interval. Construct a 95% confidence interval for the true mean number of colors on national flags. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 447 54. How much area is in both tails (combined)? 55. How much area is in each tail? 56. Calculate the following: a. lower limit b. upper limit c. error bound 57. The 95% confidence interval is_____. 58. Fill in the blanks on the graph with the areas, the upper and lower limits of the Confidence Interval and the sample mean. Figure 8.9 59. In one complete sentence, explain what the interval means. 60. Using the same x¯, s x, and level of confidence, suppose that n were 69 instead of 39. Would the error bound become larger or smaller? How do you know? 61. Using the same x¯, s x, and n = 39, how would the error bound change if the confidence level were reduced to 90%? Why? 8.3 A Population Proportion Use the following information to answer the next two exercises: Marketing companies are interested in knowing the population percent of women who make the majority of household purchasing decisions. 62. When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 90% confident that the population proportion is estimated to within 0.05? 63. If it were later determined that it was important to be more than 90% confident and a new survey were commissioned, how would it affect the minimum number you need to survey? Why? Use the following information to answer the next five exercises: Suppose the marketing company did do a survey. They randomly surveyed 200 households and found that in 120 of them, the woman made the majority of the purchasing decisions. We are interested in the population proportion of households where women make the majority of the purchasing decisions. 64. Identify the following: a. x = ______ b. n = ______ c. p′ = ______ 65. Define the random variables X and P′ in words. 66. Which distribution should you use for this problem? 67. Construct a 95% confidence interval for the population proportion of households where the women make the majority of the purchasing |
decisions. State the confidence interval, sketch the graph, and calculate the error bound. 68. List two difficulties the company might have in obtaining random results, if this survey were done by email. Use the following information to answer the next five exercises: Of 1,050 randomly selected adults, 360 identified 448 CHAPTER 8 | CONFIDENCE INTERVALS themselves as manual laborers, 280 identified themselves as non-manual wage earners, 250 identified themselves as midlevel managers, and 160 identified themselves as executives. In the survey, 82% of manual laborers preferred trucks, 62% of non-manual wage earners preferred trucks, 54% of mid-level managers preferred trucks, and 26% of executives preferred trucks. 69. We are interested in finding the 95% confidence interval for the percent of executives who prefer trucks. Define random variables X and P′ in words. 70. Which distribution should you use for this problem? 71. Construct a 95% confidence interval. State the confidence interval, sketch the graph, and calculate the error bound. 72. Suppose we want to lower the sampling error. What is one way to accomplish that? 73. The sampling error given in the survey is ±2%. Explain what the ±2% means. Use the following information to answer the next five exercises: A poll of 1,200 voters asked what the most significant issue was in the upcoming election. Sixty-five percent answered the economy. We are interested in the population proportion of voters who feel the economy is the most important. 74. Define the random variable X in words. 75. Define the random variable P′ in words. 76. Which distribution should you use for this problem? 77. Construct a 90% confidence interval, and state the confidence interval and the error bound. 78. What would happen to the confidence interval if the level of confidence were 95%? Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning iceskating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population. 79. What is being counted? 80. In words |
, define the random variable X. 81. Calculate the following: a. x = _______ b. n = _______ c. p′ = _______ 82. State the estimated distribution of X. X~________ 83. Define a new random variable P′. What is p′ estimating? 84. In words, define the random variable P′. 85. State the estimated distribution of P′. Construct a 92% Confidence Interval for the true proportion of girls in the ages 8 to 12 beginning ice-skating classes at the Ice Chalet. 86. How much area is in both tails (combined)? 87. How much area is in each tail? 88. Calculate the following: a. lower limit b. upper limit c. error bound 89. The 92% confidence interval is _______. 90. Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample proportion. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 8 | CONFIDENCE INTERVALS 449 Figure 8.10 91. In one complete sentence, explain what the interval means. 92. Using the same p′ and level of confidence, suppose that n were increased to 100. Would the error bound become larger or smaller? How do you know? 93. Using the same p′ and n = 80, how would the error bound change if the confidence level were increased to 98%? Why? 94. If you decreased the allowable error bound, why would the minimum sample size increase (keeping the same level of confidence)? HOMEWORK 8.1 A Single Population Mean using the Normal Distribution 95. Among various ethnic groups, the standard deviation of heights is known to be approximately three inches. We wish to construct a 95% confidence interval for the mean height of male Swedes. Forty-eight male Swedes are surveyed. The sample mean is 71 inches. The sample standard deviation is 2.8 inches. a. x¯ =________ i. ii. σ =________ iii. n =________ ¯. In words, define the random variables X and X b. c. Which distribution should you use for this problem? Explain your choice. d. Construct a 95% confidence interval for the population mean height of male Swedes. i. State the confidence interval. ii. Sketch |
the graph. iii. Calculate the error bound. e. What will happen to the level of confidence obtained if 1,000 male Swedes are surveyed instead of 48? Why? 96. Announcements for 84 upcoming engineering conferences were randomly picked from a stack of IEEE Spectrum magazines. The mean length of the conferences was 3.94 days, with a standard deviation of 1.28 days. Assume the underlying population is normal. ¯. In words, define the random variables X and X a. b. Which distribution should you use for this problem? Explain your choice. c. Construct a 95% confidence interval for the population mean length of engineering conferences. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. 97. Suppose that an accounting firm does a study to determine the time needed to complete one person’s tax forms. It randomly surveys 100 people. The sample mean is 23.6 hours. There is a known standard deviation of 7.0 hours. The population distribution is assumed to be normal. a. x¯ =________ i. ii. σ =________ iii. n =________ 450 CHAPTER 8 | CONFIDENCE INTERVALS ¯. In words, define the random variables X and X b. c. Which distribution should you use for this problem? Explain your choice. d. Construct a 90% confidence interval for the population mean time to complete the tax forms. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. f. If the firm wished to increase its level of confidence and keep the error bound the same by taking another survey, what changes should it make? If the firm did another survey, kept the error bound the same, and only surveyed 49 people, what would happen to the level of confidence? Why? g. Suppose that the firm decided that it needed to be at least 96% confident of the population mean length of time to within one hour. How would the number of people the firm surveys change? Why? 98. A sample of 16 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is normal. The weight of each bag was then recorded. The mean weight was two ounces with a standard deviation of 0.12 ounces. The population standard deviation is known to be 0.1 ounce. a. x¯ =________ i. ii. σ =________ iii. sx =________ b. In |
words, define the random variable X. ¯. In words, define the random variable X c. d. Which distribution should you use for this problem? Explain your choice. e. Construct a 90% confidence interval for the population mean weight of the candies. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. f. Construct a 98% confidence interval for the population mean weight of the candies. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. g. h. In complete sentences, explain why the confidence interval in part f is larger than the confidence interval in part e. In complete sentences, give an interpretation of what the interval in part f means. 99. A camp director is interested in the mean number of letters each child sends during his or her camp session. The population standard deviation is known to be 2.5. A survey of 20 campers is taken. The mean from the sample is 7.9 with a sample standard deviation of 2.8. a. x¯ =________ i. ii. σ =________ iii. n =________ ¯ b. Define the random variables X and X c. Which distribution should you use for this problem? Explain your choice. d. Construct a 90% confidence interval for the population mean number of letters campers send home. in words. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. What will happen to the error bound and confidence interval if 500 campers are surveyed? Why? 100. What is meant by the term “90% confident” when constructing a confidence interval for a mean? a. b. c. d. If we took repeated samples, approximately 90% of the samples would produce the same confidence interval. If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the sample mean. If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the true value of the population mean. If we took repeated samples, the sample mean would equal the population mean in approximately 90% of the samples. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 101. The Federal Election Commission collects information about campaign contributions and disbursements for candidates and political committees |
each election cycle. During the 2012 campaign season, there were 1,619 candidates for the House of Representatives across the United States who received contributions from individuals. Table 8.11 shows the total receipts from individuals for a random selection of 40 House candidates rounded to the nearest $100. The standard deviation for this data to the nearest hundred is σ = $909,200. CHAPTER 8 | CONFIDENCE INTERVALS 451 $3,600 $1,243,900 $10,900 $385,200 $581,500 $7,400 $2,900 $400 $3,714,500 $632,500 $391,000 $467,400 $56,800 $5,800 $405,200 $733,200 $8,000 $468,700 $75,200 $41,000 $13,300 $9,500 $953,800 $1,113,500 $1,109,300 $353,900 $986,100 $88,600 $378,200 $13,200 $3,800 $745,100 $5,800 $3,072,100 $1,626,700 $512,900 $2,309,200 $6,600 $202,400 $15,800 Table 8.11 a. Find the point estimate for the population mean. b. Using 95% confidence, calculate the error bound. c. Create a 95% confidence interval for the mean total individual contributions. d. Interpret the confidence interval in the context of the problem. 102. The American Community Survey (ACS), part of the United States Census Bureau, conducts a yearly census similar to the one taken every ten years, but with a smaller percentage of participants. The most recent survey estimates with 90% confidence that the mean household income in the U.S. falls between $69,720 and $69,922. Find the point estimate for mean U.S. household income and the error bound for mean U.S. household income. 103. The average height of young adult males has a normal distribution with standard deviation of 2.5 inches. You want to estimate the mean height of students at your college or university to within one inch with 93% confidence. How many male students must you measure? 8.2 A Single Population Mean using the Student t Distribution 104. In six packages of “The Flintstones® Real Fruit Snacks” there were five Bam-Bam snack pieces. The total |
number of snack pieces in the six bags was 68. We wish to calculate a 96% confidence interval for the population proportion of Bam-Bam snack pieces. a. Define the random variables X and P′ in words. b. Which distribution should you use for this problem? Explain your choice c. Calculate p′. d. Construct a 96% confidence interval for the population proportion of Bam-Bam snack pieces per bag. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. Do you think that six packages of fruit snacks yield enough data to give accurate results? Why or why not? 105. A random survey of enrollment at 35 community colleges across the United States yielded the following figures: 6,414; 1,550; 2,109; 9,350; 21,828; 4,300; 5,944; 5,722; 2,825; 2,044; 5,481; 5,200; 5,853; 2,750; 10,012; 6,357; 27,000; 9,414; 7,681; 3,200; 17,500; 9,200; 7,380; 18,314; 6,557; 13,713; 17,768; 7,493; 2,771; 2,861; 1,263; 7,285; 28,165; 5,080; 11,622. Assume the underlying population is normal. a. i. ii. x¯ = __________ s x = __________ iii. n = __________ iv. n – 1 = __________ ¯ b. Define the random variables X and X c. Which distribution should you use for this problem? Explain your choice. in words. 452 CHAPTER 8 | CONFIDENCE INTERVALS d. Construct a 95% confidence interval for the population mean enrollment at community colleges in the United States. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. What will happen to the error bound and confidence interval if 500 community colleges were surveyed? Why? 106. Suppose that a committee is studying whether or not there is waste of time in our judicial system. It is interested in the mean amount of time individuals waste at the courthouse waiting to be called for jury duty. The committee randomly surveyed 81 people who recently served as jurors. The sample mean wait time was eight hours with |
a sample standard deviation of four hours. a. i. ii. x¯ = __________ s x = __________ iii. n = __________ iv. n – 1 = __________ ¯ b. Define the random variables X and X c. Which distribution should you use for this problem? Explain your choice. d. Construct a 95% confidence interval for the population mean time wasted. in words. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. Explain in a complete sentence what the confidence interval means. 107. A pharmaceutical company makes tranquilizers. It is assumed that the distribution for the length of time they last is approximately normal. Researchers in a hospital used the drug on a random sample of nine patients. The effective period of the tranquilizer for each patient (in hours) was as follows: 2.7; 2.8; 3.0; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4. a. i. ii. x¯ = __________ s x = __________ iii. n = __________ iv. n – 1 = __________ b. Define the random variable X in words. ¯ c. Define the random variable X d. Which distribution should you use for this problem? Explain your choice. e. Construct a 95% confidence interval for the population mean length of time. in words. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. f. What does it mean to be “95% confident” in this problem? 108. Suppose that 14 children, who were learning to ride two-wheel bikes, were surveyed to determine how long they had to use training wheels. It was revealed that they used them an average of six months with a sample standard deviation of three months. Assume that the underlying population distribution is normal. a. i. ii. x¯ = __________ s x = __________ iii. n = __________ iv. n – 1 = __________ b. Define the random variable X in words. ¯ c. Define the random variable X d. Which distribution should you use for this problem? Explain your choice. e. Construct a 99% confidence interval for the population mean length of time using training wheels. in words. i. State the confidence interval. ii. Sketch the graph. iii. |
Calculate the error bound. f. Why would the error bound change if the confidence level were lowered to 90%? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 109. The Federal Election Commission (FEC) collects information about campaign contributions and disbursements for candidates and political committees each election cycle. A political action committee (PAC) is a committee formed to raise money for candidates and campaigns. A Leadership PAC is a PAC formed by a federal politician (senator or representative) to raise money to help other candidates’ campaigns. The FEC has reported financial information for 556 Leadership PACs that operating during the 2011–2012 election cycle. The following table shows the total receipts during this cycle for a random selection of 20 Leadership PACs. CHAPTER 8 | CONFIDENCE INTERVALS 453 $46,500.00 $0 $40,966.50 $105,887.20 $5,175.00 $29,050.00 $19,500.00 $181,557.20 $31,500.00 $149,970.80 $2,555,363.20 $12,025.00 $409,000.00 $60,521.70 $18,000.00 $61,810.20 $76,530.80 $119,459.20 $0 $63,520.00 $6,500.00 $502,578.00 $705,061.10 $708,258.90 $135,810.00 $2,000.00 $2,000.00 $0 $1,287,933.80 $219,148.30 Table 8.12 x¯ = $251, 854.23 s = $521, 130.41 Use this sample data to construct a 96% confidence interval for the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle. Use the Student's t-distribution. 110. Forbes magazine published data on the best small firms in 2012. These were firms that had been publicly traded for at least a year, have a stock price of at least $5 per share, and have reported annual revenue between $5 million and $1 billion. The Table 8.13 shows the ages of the corporate CEOs for a random sample of these firms. 48 58 51 61 56 59 74 |
63 53 50 59 60 60 57 46 55 63 57 47 55 57 43 61 62 49 67 67 55 55 49 Table 8.13 Use this sample data to construct a 90% confidence interval for the mean age of CEO’s for these top small firms. Use the Student's t-distribution. 111. Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.1 seats. a. i. ii. x¯ = __________ s x = __________ iii. n = __________ iv. n-1 = __________ ¯ b. Define the random variables X and X c. Which distribution should you use for this problem? Explain your choice. d. Construct a 92% confidence interval for the population mean number of unoccupied seats per flight. in words. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. 454 CHAPTER 8 | CONFIDENCE INTERVALS 112. In a recent sample of 84 used car sales costs, the sample mean was $6,425 with a standard deviation of $3,156. Assume the underlying distribution is approximately normal. a. Which distribution should you use for this problem? Explain your choice. ¯ b. Define the random variable X in words. c. Construct a 95% confidence interval for the population mean cost of a used car. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. d. Explain what a “95% confidence interval” means for this study. 113. Six different national brands of chocolate chip cookies were randomly selected at the supermarket. The grams of fat per serving are as follows: 8; 8; 10; 7; 9; 9. Assume the underlying distribution is approximately normal. a. Construct a 90% confidence interval for the population mean grams of fat per serving of chocolate chip cookies sold in supermarkets. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. b. If you wanted a smaller error bound while keeping the same level of confidence, what should have been changed in the study before it was done? c. Go |
to the store and record the grams of fat per serving of six brands of chocolate chip cookies. d. Calculate the mean. e. Is the mean within the interval you calculated in part a? Did you expect it to be? Why or why not? 114. A survey of the mean number of cents off that coupons give was conducted by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News. The following data were collected: 20¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢; $1.50; 40¢; 65¢; 40¢. Assume the underlying distribution is approximately normal. a. i. ii. x¯ = __________ s x = __________ iii. n = __________ iv. n-1 = __________ ¯ b. Define the random variables X and X c. Which distribution should you use for this problem? Explain your choice. d. Construct a 95% confidence interval for the population mean worth of coupons. in words. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. If many random samples were taken of size 14, what percent of the confidence intervals constructed should contain the population mean worth of coupons? Explain why. Use the following information to answer the next two exercises: A quality control specialist for a restaurant chain takes a random sample of size 12 to check the amount of soda served in the 16 oz. serving size. The sample mean is 13.30 with a sample standard deviation of 1.55. Assume the underlying population is normally distributed. 115. Find the 95% Confidence Interval for the true population mean for the amount of soda served. a. b. c. d. (12.42, 14.18) (12.32, 14.29) (12.50, 14.10) Impossible to determine 116. What is the error bound? a. 0.87 b. 1.98 c. 0.99 d. 1.74 8.3 A Population Proportion 117. Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 a. When designing a study to |
determine this population proportion, what is the minimum number you would need b. to survey to be 95% confident that the population proportion is estimated to within 0.03? If it were later determined that it was important to be more than 95% confident and a new survey was commissioned, how would that affect the minimum number you would need to survey? Why? 118. Suppose that the insurance companies did do a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up. CHAPTER 8 | CONFIDENCE INTERVALS 455 a. i. x = __________ ii. n = __________ iii. p′ = __________ b. Define the random variables X and P′, in words. c. Which distribution should you use for this problem? Explain your choice. d. Construct a 95% confidence interval for the population proportion who claim they always buckle up. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. If this survey were done by telephone, list three difficulties the companies might have in obtaining random results. 119. According to a recent survey of 1,200 people, 61% feel that the president is doing an acceptable job. We are interested in the population proportion of people who feel the president is doing an acceptable job. a. Define the random variables X and P′ in words. b. Which distribution should you use for this problem? Explain your choice. c. Construct a 90% confidence interval for the population proportion of people who feel the president is doing an acceptable job. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. 120. An article regarding interracial dating and marriage recently appeared in the Washington Post. Of the 1,709 randomly selected adults, 315 identified themselves as Latinos, 323 identified themselves as blacks, 254 identified themselves as Asians, and 779 identified themselves as whites. In this survey, 86% of blacks said that they would welcome a white person into their families. Among Asians, 77% would welcome a white person into their families, 71% would welcome a Latino, and 66% would welcome a black person. a. We are interested in finding the 95% confidence interval for the percent of all black adults who would welcome a white person into their families. Define the random variables X and P′, in words. b. Which distribution should you |
use for this problem? Explain your choice. c. Construct a 95% confidence interval. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. 121. Refer to the information in Exercise 8.120. a. Construct three 95% confidence intervals. i. percent of all Asians who would welcome a white person into their families. ii. percent of all Asians who would welcome a Latino into their families. iii. percent of all Asians who would welcome a black person into their families. b. Even though the three point estimates are different, do any of the confidence intervals overlap? Which? c. For any intervals that do overlap, in words, what does this imply about the significance of the differences in the true proportions? d. For any intervals that do not overlap, in words, what does this imply about the significance of the differences in the true proportions? 122. Stanford University conducted a study of whether running is healthy for men and women over age 50. During the first eight years of the study, 1.5% of the 451 members of the 50-Plus Fitness Association died. We are interested in the proportion of people over 50 who ran and died in the same eight-year period. a. Define the random variables X and P′ in words. b. Which distribution should you use for this problem? Explain your choice. c. Construct a 97% confidence interval for the population proportion of people over 50 who ran and died in the same eight–year period. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. 456 CHAPTER 8 | CONFIDENCE INTERVALS d. Explain what a “97% confidence interval” means for this study. 123. A telephone poll of 1,000 adult Americans was reported in an issue of Time Magazine. One of the questions asked was “What is the main problem facing the country?” Twenty percent answered “crime.” We are interested in the population proportion of adult Americans who feel that crime is the main problem. a. Define the random variables X and P′ in words. b. Which distribution should you use for this problem? Explain your choice. c. Construct a 95% confidence interval for the population proportion of adult Americans who feel that crime is the main problem. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. d. Suppose we want to lower the sampling error |
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