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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 through 5. X ~ B(n, p) where n = 300 and p = 0.53. Because 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 a 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 a normal approximation 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 computer software that calculates binomial probabilities. Many students have access to calculators that calculate probabilities for 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 favors 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. 7.4 \| Central Limit Theorem (Pocket Change) 432 Chapter 7 \| The Central Limit Theorem 7.1 Central Limit Theorem (Pocket Change) Student Learning Outcome • 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. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 433 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 the 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 averages of their change. 434 Chapter 7 \| The Central Limit Theorem __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________
__________ 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 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. 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? 3. X ~ _____(_____, _____) 4. In the section titled Collecting Averages of Groups of Five, what was the approximate distribution of the averages? X ¯ ~ _____(_____, _____) 5. In one to two complete sentences, explain any differences in your answers to the previous two questions. 7.5 \| Central Limit Theorem (Cookie Recipes) This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 435 7.2 Central Limit Theorem (Cookie Recipes) Student Learning Outcome • 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 following table: 436 Chapter 7 \| The Central Limit The
orem 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. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 437 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 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 day. Sketch the graph using a ruler and pencil. Scale the axes. 438 Chapter 7 \| The Central Limit Theorem 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. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 439 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 given a random variable (RV) with a known mean, μ, and known standard deviation, σ, and sampling with size n, we are interested in two new RVs: the sample mean, X ¯, 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 distribution 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 a random events; for example, the length of time between emergency arrivals at a hospital, notation: X ~ Exp(m) The mean is μ = 1 m. The probability density function is f(x) = me–mx, x ≥ 0, and and the standard deviation is σ
= 1 m 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, the mean for a sample (denoted by x¯ ) is x¯ = sum of all values in the sample number of values in the sample, and the mean for a population (denoted by μ) is μ = sum of all values in the population number of values in the population. continuous random variable (RV) with probability density function (pdf) normal distribution f (x) = 1 σ 2π e a – (x – μ)2 2σ 2, 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 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) In a population whose distribution may be known or unknown, if the size (n) of the sample 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). 440 Chapter 7 \| The Central Limit Theorem 7.2 The Central Limit Theorem for Sums (Optional) 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) 7.2 The Central Limit Theorem for Sums (Optional) Central limit theorem for sample means: X ¯ ~ N ⎛ ⎝μ x, σx n ⎞ ⎠ Mean X Central ¯ : μx limit standard error of the mean: z = theorem for sample means z-score and x¯ − μ x σ x n ⎛ ⎝ ⎞ ⎠ Central limit theorem for sums: ∑X ~ N[(n)(μx),( n )(σx)] Mean for sums (∑X): (n)(μx) Central deviation limit theorem for sums z-score and standard sums: for z for the sample mean = Σx – (n)(μ X) ( n)(σ X) 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 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 441 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. 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? 442 Chapter 7 \| The Central Limit Theorem 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 (Optional) 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. 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 type of 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 (16) and (17). 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 exercises: 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? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 443 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 < 1186)? 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? 36. What is P(Σx > 9000)? 7.3 Using the Central Limit Theorem Use the following information to answer the next 10 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 are 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 of 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 of Exercise 7.39. 41. Find the 90th percentile for the mean weight for the 100 weights. 42. Draw the graph of 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 < 2450). 44. Draw the graph of Exercise 7.43. 45. Find the 90th percentile for the total weight of the 100 weights. 46. Draw the graph of Exercise 7.45. Use the following information to answer the next five exercises: The length of time a particular smartphone's battery lasts 444 Chapter 7 \| The Central Limit Theorem 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 7 and 11. 52. Find the 80th percentile for the total length of time 64 batteries last. 53. Find the interquartile range (IQR) for the mean amount of time 64 batteries last. 54. Find the middle 80 percent for the total amount of time 64 batteries last. Use the following information to answer the next six 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) 61. Previously, De Anza's 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 ~ _____(_____, _____) 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 amount of change 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. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 445 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 ¯ be 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 online album collection is uniformly distributed from 2 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 for the average song length is _______–_______. 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 = _____________. a. ¯ = _____________. b
. 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. ¯ 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 percentage of fat calories that a person in America consumes each day is normally distributed with a mean of about ¯ = average percentage of 36 and a standard deviation of about ten. Suppose that 16 individuals are randomly chosen. Let X fat calories. ¯ ~ ______(______, ______) a. X b. For the group of 16, find the probability that the average percentage 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 percentage of fat calories. 446 Chapter 7 \| The Central Limit Theorem 69. The distribution of income in some economically developing 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 salary 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 1. 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 (Optional) 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 = _____________. In words, ΣX = _______________. a. b. The distribution is _______. c. d. ΣX ~ _____(_____, _____) e. Find the probability that the total weight of the open boxes is less than 250 pounds. f. Find the 35th percentile for the total weight of open boxes of cereal. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 447 75. Salaries for entry-level managers at a restaurant chain are normally distributed with a mean of $44,000 and a standard deviation of $6,500. We randomly survey 10 managers from these restaurants. In words, X = ______________. In words, Σ
X = _____________. a. b. X ~ _____(_____, _____) c. d. ΣX ~ _____(_____, _____) e. Find the probability that the managers earn a total of over $400,000. f. Find the 90th percentile for an individual manager's salary. g. Find the 90th percentile for the sum of ten managers' salary. h. i. If we surveyed 70 managers instead of ten, graphically, how would that change the distribution in part (d)? If each of the 70 managers 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 10 minutes. ii. The probability that the average attention span for the 60 children is less than 10 minutes. f. Calculate the probabilities in part (e). ¯ g. Explain why the distribution for X is not exponential. 448 Chapter 7 \| The Central Limit Theorem 77. The closing stock prices of 35 U.S. semiconductor manufacturers are given as follows: Company Closing Stock Prices 1 2 3 4 5 6 7 8 9 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 Table 7.7 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 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 449 Company Closing Stock Prices 35 Table 7.
7 29.25 a. b. In words, Χ = ______________. 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 10. d. e. Randomly average five stock prices together. (Use a random number generator.) Continue averaging five prices In words, describe the distribution of the stock prices. together until you have 10 averages. List those 10 averages. f. Use the 10 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 10. 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 on 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) d. N(2, 1) 79. The average wait time is: a. one hour two hours b. two and a half hours c. four hours d. 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. 65.2 450 Chapter 7 \| The Central Limit Theorem 82. Would you
be surprised, based on 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 more than $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 the 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. 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(1400 < Σx < 1550) = _____ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 451 88. Suppose that the length of research papers is uniformly distributed from 10 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 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 managers in a restaurant chain are normally distributed with a mean of $44,000 and a standard deviation of $6,500. We randomly survey 10 managers from that district. a. Find the 90th percentile for an individual manager's salary. b. Find the 90th percentile
for the average manager'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 a 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, what is the probability of getting a random sample of 30 batteries in which the sample mean life span 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 3,600 pounds. b. If 20 men have a sum weight greater than 3,500 pounds, then their total weight exceeds the safety limits for water taxis. Based on (a), is this a safety concern? Explain. 452 Chapter 7 \| The Central Limit Theorem 93. Large bags of a brand of multicolored candies 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 (g) Orange
(g) Yellow (g) Brown (g) Blue (g) Green (g) 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 0.751 0.841 0.856 0.799 0.966 0.859 0.857 0.942 0.873 0.809 0.890 0.878 0.905 0.735 0.895 0.865 0.864 0.852 0.866 0.859 0.838 0.863 0.888 0.925 0.793 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 0.925 0.914 0.881 0.865 0.865 1.015 0.876 0.809 0.865 0.848 0.940 0.833 0.845 0.852 0.778 0.814 0.791 0.810 0.881 0.881 0.863 0.775 0.854 0.810 0.858 0.818 0.868 0.803 0.932 0.842 0.832 0.807 0.841 0.932 0.833 0.881 0.818 0.864 0.825 0.855 0.942 0.825 0.869 0.912 0.887 Table 7.8 The bag contained 465 candies and the 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 of the weights. If 465 candies are randomly selected, find the probability that their weights sum to at least 396.9 g. c. Is the candy company's labeling accurate? d. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter
7 \| The Central Limit Theorem 453 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. 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 0.742 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 7.9 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 test, 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 g 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) Bar
an, D. (2010). 20 percent of Americans have never used email. WebGuild. Retrieved from http://www.webguild.org/ 20080519/20-percent-of-americans-have-never-used-email The Flurry Blog. (2013). Retrieved from http://blog.flurry.com U.S. Department of Agriculture. (n.d.). Retrieved from https://www.usda.gov/ 7.2 The Central Limit Theorem for Sums (Optional) Farago, P. (2012, Oct. 29). The truth about cats and dogs: Smartphone vs tablet usage differences. Flurry Analytics Blog. Retrieved from http://flurrymobile.tumblr.com/post/113379683050/the-truth-about-cats-and-dogs-smartphone-vs 7.3 Using the Central Limit Theorem The Wall Street Journal. (n.d.). Retrieved from https://www.wsj.com/ Centers for Disease Control and Prevention. (2017, April 16). National health and nutrition examination survey. National Center for Health Statistics. Retrieved from http://www.cdc.gov/nchs/nhanes.htm SOLUTIONS 1 mean = 4 hours, standard deviation = 1.2 hours, sample size = 16 454 Chapter 7 \| The Central Limit Theorem 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. 7 0.3345 9 7833.92 11 0.0089 13 7326.49 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 1000 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(2500, 5.7735) b. 0 45 2507.40 47 a. 10 b. 1 10 49 N ⎛ ⎝10, 10 8 ⎞ ⎠ 51 0.7799 53 1.69 55 0.0072 57 391.54 59 405.51 61 a. Χ = amount of change students
carry This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 455 b. Χ ~ E(0.88, 0.88) ¯ = average amount of change carried by a sample of 25 students. c. X ¯ ~ N(0.88, 0.176) d. X 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 ¯ ∼ N ⎛ c. X ⎝2,000, 8,000 1,000 ⎞ ⎠ 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(2,000 < X P(2,100 < X ¯ < 2,100) = 0.1537 ¯ < 2,200) = 0.1317 71 b 73 a. the total length of time for nine criminal trials b. N
(189, 21) c. 0.0432 456 Chapter 7 \| The Central Limit Theorem d. 162.09; 90 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(44000, 6500) c. ΣX ~ sum of the salaries of 10 elementary school teachers in the sample d. ΣX ~ N(44,000, 20,554.80) e. 0.9742 f. $52,330.09 g. 466,342.04 h. Sampling 70 teachers instead of 10 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. ¯ ~ N ⎛ b. X ⎝60, 9 25 ⎞ ⎠ c. 0.5000 d. 59.06 e. 0.8536 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 \| The Central Limit Theorem 457 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 life span 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, and s = 0.05. b. Σ x¯ = 85.65, Σs = 5.18 c. normalcdf(396.9,E99,(465)(0.8565),(0.05)( 465 )) ≈ 1 d. Because the probability of a sample of size of 465 having at least a mean sum of 396.9 is appproximately 1, we can conclude that the company is correctly labeling their candy packages. 95 Use normalcdf ⎛ ⎝E – 99,1.1,1, 1 70 ⎞ ⎠ = 0.7986. This means that there is an 80 percent chance that the service time will be less than 1.1 hours. It may be wise to schedule more time because there is an associated 20 percent chance that the maintenance time will be greater than 1.1 hours. 97 Because we have normalcdf ⎛ ⎝5.111,5.291,5.201,0.065 280 within the limits; therefore, there should be no rejected coins out of a well-selected sample size of 280. ⎞ ⎠ ≈ 1, we can conclude that practically all the coins are 458 Chapter 7 \| The Central Limit Theorem This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 459 8 \| CONFIDENCE INTERVALS Figure 8.1 Have you ever wondered what the average number of chocolate candies 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 do the following: 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 attempt. In this case, you would have obtained a point estimate for the true proportion. 460 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 those 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 an internet music store. 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 instance 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 internet music 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 percent of the samples, the sample mean, x¯, will be within two standard deviations of the population mean, μ. For our internet music example, two standard deviations would be calculated as (2)(0.1) = 0.2. The sample mean, x¯, is likely to be within 0.2 units of μ. In this example, we do not know the true population mean μ (because we do not have information from all the internet music users!), but we can compute the sample mean x¯ based on our sample of 100 individuals. Because the sample mean is likely to be within 0.2 units of the true population mean 95 percent of the times that we take a sample of 100 users, we can say with 95 percent confidence that μ is within 0.2 units of x¯. In other words, μ is somewhere between x¯ − 0.2 and x¯ + 0.2. Suppose that from the sample of 100 internet music customers, we compute a sample mean download of x¯ = 2 songs per month. Since we know that the population standard deviation is σ − 1, according to the central limit theorem, the standard deviation for the sample means is σ = 1 = 0.1. 100 We know that there is a 95 percent chance that the true population mean value μ is between two standard deviations from the sample mean. That is, with 95 percent confidence we can say that μ is between x¯ − 2× σ n and x¯ − 2× σ n. Replacing the symbols for their values in this example, we say that we are 95 percent confident that the true average number between x¯ − 2× σ n downloaded = 2 −.02 = 1.8, and of songs = 2 − 2× σ internet month music from store per an is 100 x¯ + 2× σ n = 2 + 2× σ 100 = 2 +.02 = 2.2. The 95 percent confidence interval for μ is (1.8, 2.2). The 95 percent confidence interval implies two possibilities. Either the interval (1.8, 2.2) contains the true mean, μ, or our
sample produced an x¯ all the samples (95–100 percent). that is not within 0.2 units of the true mean μ. The second possibility happens for only 5 percent of Remember that a confidence interval is created for an unknown population parameter like the population mean, μ. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 461 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, you might notice that some reports 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. Those are two ways of expressing the same concept. NOTE Although the text covers only symmetrical confidence intervals, there are non-symmetrical confidence intervals (for example, a confidence interval for the standard deviation). 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 percent 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 ⎞ ⎞ ⎠, ⎠ ⎛ − ⎝ We say we are approximately 95 percent 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 percent confidence interval (5, 15), where the margin of error = 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 is called the error bound for a population mean (EBM) is called the margin of error for a population mean (EBM). The sample
mean, x¯, is the point estimate of the unknown population mean, μ. The confidence interval (CI) 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 (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 percentage of confidence intervals that contain the true population parameter when repeated samples are taken. Most often, the person constructing the confidence interval will choose a confidence level of 90 percent or higher, because that person wants to be reasonably certain of his or her conclusions. Another probability, which is called alpha (α) is related to the confidence level, CL. Alpha is the probability that the confidence interval does not contain the unknown population parameter. Mathematically, alpha can be computed as 462 α = 1 − CL. Example 8.1 Chapter 8 \| Confidence Intervals 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¯ 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 is 95 percent, then we say, "We estimate with 95 percent confidence that the true value of the population mean is between 4.5 and 9.5." 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 percent confidence interval (5, 15) where EBM = 5. To get a 90 percent confidence interval, we must include the central 90 percent of the probability of the normal distribution. If we include the central 90 percent, we leave out a total of α = 10 percent in both tails, or 5 percent in each tail, of the normal distribution. Figure 8.2 The critical value 1
.645 is the z-score in 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. To capture the central 90 percent, we must go out 1.645 standard deviations on either side of the calculated sample mean. The critical value will change depending on the confidence level of the interval. It is important that the standard deviation used 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, the following statements apply: ¯ • X is normally distributed, that is, X ¯ ~ N ⎛ ⎝μ X, σ n ⎞ ⎠. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 463 • 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 as follows: • Calculate the sample mean, x¯, from the sample data. Remember, in this section, we already know the population standard deviation, σ. • Find the z-score that corresponds to the confidence level. • Calculate the error bound EBM. • Construct the confidence interval. • If we denote the critical z-score by z a 2, and the sample size by n, then the formula for the confidence interval with confidence level Cl = 1 − α, is given by ( x¯ + z a 2 × σ n ). • 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 E
BM 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). 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. 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 = z 0.025. 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 standard normal probability table. z α 2 Normal table (see appendices) shows that the probability for 0 to 1.96 is 0.47500, and so the probability to the right tail of the critical value 1.96 is 0.5 – 0.475 = 0.025 invNorm(0.975, 0, 1) = 1.96. In this command, the value 0.975 is the total area to the left of the critical value that we are looking to calculate. The parameters 0 and 1 are the mean value and the standard deviation of the standard normal distribution Z. 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 with mean 0 and standard deviation 1. Calculating the Margin of Error EBM The error bound formula for an unknown population mean, μ, when the population standard deviation, σ, is known is 464 Chapter 8 \| Confidence Intervals Margin of error = ⎛ ⎝z α 2 ⎞ ⎛ ⎝ ⎠ ⎞ ⎠. σ n Constructing the Confidence Interval The confidence interval estimate has the format sample mean plus or minus the margin of error. The graph gives a picture of the entire situation CL + α 2 + α 2 = CL + α = 1. Figure 8.3 Writing the Interpretation The interpretation
should clearly state the confidence level (CL), explain which population parameter is being estimated (here, a population mean), and state the confidence interval (both endpoints): "We estimate with ___percent confidence that the true population mean (include the context of the problem) is between ___ and ___ (include appropriate units)." 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 percent 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=(z α 2 )( σ n ) • The confidence level is 90 percent (CL = 0.90). σ= 3; n= 36; CL = 0.90, so α = 1 – CL = 1 – 0.90 = 0.10. α 2 = 0.05, z α 2 = z0.05 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 465 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 percent 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 3 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 We estimate with 90 percent confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82. Explanation of 90 percent 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 6 minutes. A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery time of 36 min. Find a 90 percent confidence interval estimate for the population mean delivery time. 466 Chapter 8 \| Confidence Intervals 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. For certification from the Federal Communications Commission 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 of a random cell phone company. Phone Model # SAR Phone Model # SAR Phone Model # SAR 800 900 1000 1100 1200 1300 1400 1500 1600 1700 Table 8.1 1.11 1.48 1.43 1.3 1.09 1800 1900 2000 2100 2200 0.455 2300 1.41 0.82 0.78 1.25 2400 2500 2600 2700 1.36 1.34 1.18 1.3 1.26 1.29 0.36 0.52 1.6 1.39 2800 2900 3000 3100 3200 3300 3400 3500 3600 3700 0.74 0.5 0.4 0.867 0.68 0.51 1.13 0.3 1.48 1.38 Find a 98 percent confidence interval
for the true (population) mean of the 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. This is calculated by adding the specific absorption rate for the 30 cell phones in the sample, and dividing the result by 30. Next, find the EBM. Because you are creating a 98 percent confidence interval, CL = 0.98. Figure 8.4 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 467 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 percent 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 percent 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). 468 Chapter 8 \| Confidence Intervals 8.3 Table 8.2 shows a different random sampling of 20 cell phone models. Use these data to calculate a 93 percent 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. Phone Model SAR Phone Model SAR 450 550 650 750 850 950 1050
1150 1250 1350 Table 8.2 1.48 1450 0.8 1.15 1.36 0.77 1550 1650 1750 1850 0.462 1950 1.36 1.39 1.3 0.7 2050 2150 2250 2350 1.53 0.68 1.4 1.24 0.57 0.2 0.51 0.3 0.73 0.869 Notice the difference in the confidence intervals calculated in Example 8.3 and the following Try It exercise. These intervals are different for several reasons: they are calculated from different samples, the samples are different sizes, and the intervals are 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 percent confidence level. Find a 95 percent 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=(z α 2 )( σ n ) • The confidence level is 95 percent (CL = 0.95). σ= 3; n= 36 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 z 0.025 is 0.025, and the area to the left of z 0.025 is 1 – 0.025 = 0.975. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 469 = 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.) 3 36 ⎞ ⎛ ⎠ = 0.98 EBM = (1.96) ⎝ x¯ – EBM = 68 – 0.98 = 67.02 x¯ + EBM = 68
+ 0.98 = 68.98 Notice that the EBM is larger for a 95 percent confidence level in the original problem. Interpretation We estimate with 95 percent confidence that the true population mean for all statistics exam scores is between 67.02 and 68.98. Explanation of 95 percent Confidence Level 95 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 percent confidence interval is (67.18, 68.82). The 95 percent confidence interval is (67.02, 68.98). The 95 percent 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 percent confidence interval is wider. For more certainty 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 the pizza-delivery Try It exercise. 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 percent confidence interval estimate for the true mean pizza-delivery time. 470 Chapter 8 \| Confidence Intervals 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 percent 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 percent (CL = 0.90), z α 2 = z0.05 = 1.645. Solution 8.5 Solution A If we increase the sample size n to 100, we decrease the margin of error. 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 the pizza-delivery Try It exercise. 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 percent confidence interval estimate for the population mean delivery time. Working Backward 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 backward 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 2. Finding the Sample Mean • Subtract the error bound from the upper value of the confidence interval, This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 471 • 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 gives the confidence interval and does 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 do not know the sample mean, EBM = (68.82 − 67.18) = 0.82. The margin of error is the quantity 2 that we add and subtract from the sample mean to obtain the confidence interval. Therefore, the
margin of error is half of the length of the interval. Calculate the sample mean: • • If we know the error bound, x¯ = 68.82 – 0.82 = 68. If we do not 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. In this situation, we are given the desired margin of error, EBM, and we need to compute the sample size n. The formula for sample size is n = z2 σ 2 EBM 2 n to the closest integer., found by solving the error bound formula for n. Always round up the value of In this formula, z is the critical value 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 percent 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 percent. 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. 472 Chapter 8 \| Confidence Intervals Therefore, 217 Foothill College students should be surveyed in order to be 95 percent 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 percent 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's tDistribution In practice
, we rarely know the population standard deviation. In the past, when the sample size was large, this unknown number 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. William S. Gosset (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 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 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 (df), n -– 1, are the sample size minus 1. 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. Figure 8.6 shows the graphs of the student t-distribution for 1, 2 and 5 degrees of freedom: (v
), compare to the standard normal distribution (in black). This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 473 Figure 8.6 • The exact shape of the Student's t-distribution depends on the degrees of freedom. As the degrees of freedom increase, the graph of the 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 does not 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 critical t-values 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. 474 Chapter 8 \| Confidence Intervals 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. If the population standard deviation is not known, the error bound for a population mean is • EBM = ⎛ ⎝ 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, do the following: 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 percent 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). To find the confidence interval, you need the sample mean, x¯, and the EBM. x¯ = 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 = 8.2267; 2 + (9.4 − x¯ ) 2 + ⋯ + (5.5 − x¯ ) 14 2 15 + (6.9 − x¯ ) 2 = 1.6722; (8.6 − x¯ ) s = n = 15 df = 15 – 1 = 14 CL, so α = 1 – CL = 1 – 0.95 = 0.05 α 2 = 0.025; t α 2 = t
0.025 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 = ⎛ ⎝ ⎛ EBM = (2.14) ⎝ x¯ – EBM = 8.2267 – 0.9240 = 7.3 ⎞ ⎠ = 0.924 1.6722 15 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 475 x¯ + EBM= 8.2267 + 0.9240 = 9.15 The 95 percent confidence interval is (7.30, 9.15). We estimate with 95 percent confidence that the true population mean sensory rate is between 7.30 and 9.15. Solution 8.8 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 percent confidence interval is (7.3006, 9.1527). NOTE When calculating the error bound, you can also use a probability table for the Student's t-distribution 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 hours of sleep subjects get each night. You measure hours of sleep for 12 subjects with the following results. Construct a 95 percent 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 A
group of researchers 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 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, 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 8.3 476 Chapter 8 \| Confidence Intervals Use this sample data to construct a 90 percent confidence interval for the mean number of targeted industrial chemicals to be found in an infant’s blood. Solution 8.9 Solution A From the sample data, you can calculate x¯ = 79 + 145 + ⋯ + 139 + 99 20 = 127.45 2 (79 − x¯ ) + (145 − x¯ ) 2 + ⋯ + (139 − x¯ ) 2 + (99 − x¯ ) 2 19 s = infants in the sample, so n = 20, and df = 20 – 1 = 19. There are 20 = 25.965. You are asked to calculate a 90 percent 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 = t α 2 ( s n ) = 1.729 ⎛ ⎝ 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 percent 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 percent confidence interval is (117.41, 137.49). 8.9 A random sample of statistics students was 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 the following sample data to construct a 98 percent confidence interval for the mean number of hours statistics students will spend watching television in one This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 477 week. 0 5 3 1 20 9 10 1 10 4 14 2 4 4 5 Table 8.4 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 percent of the vote within 3 percentage points (if the sample is large enough). Often, election polls are calculated with 95 percent confidence, so the pollsters would be 95 percent 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 for a population (EBP), 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 data that you are collecting is categorical, consisting of two categories: Success or Failure, Yes
or No. Examples of situations where you are the following trying to estimate the true population proportion are the following: What proportion of the population smoke? What proportion of the population will vote for candidate A? What proportion of the population has a college-level education? The distribution of the sample proportions (based on samples of size n) is denoted by P′ (read “P prime”). The central limit theorem for proportions asserts that the sample proportion distribution P′ follows a normal distribution with mean value p, and standard deviation p • q n, where p is the population proportion and q = 1 -– p. The confidence interval has the form (p′ – EBP, p′ + EBP). EBP is error bound for the proportion. p′ = the estimated proportion of successes (p′ is a point estimate for p, the true proportion.) p′ = x n 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 478 Chapter 8 \| Confidence Intervals from the data: p′ is the estimated proportion of successes, and q′ is the estimated proportion of failures. The confidence interval can be used only if the number of successes np′ and the number of failures nq′ are both greater than five. That is, in order to use the formula for confidence intervals for proportions, you need to verify that both np' ≥ 5 and nq' ≥ 5. Example 8.10 Suppose that a market research firm is hired to estimate the percentage 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 percent confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones. Solution 8.10 • 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). ⎛ ⎝500, 421 Let X = the number of people in the sample who have cell phones. X is binomial. X ~ B 500 ⎞ ⎠. To calculate the confidence interval, you must find p′, q′, and EBP. n = 500 x = the number of successes = 421 p′ = 0.842 is the sample proportion; this is the point estimate of the population proportion. p′ = x n = 421 500 = 0.842 Because CL = 0.95, then α = 1 – CL = 1 – 0.95 = 0.05 ⎛ ⎝ = 0.025. ⎞ ⎠ α 2 q′ = 1 – p′ = 1 – 0.842 = 0.158 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 z0.025is 0.025, and the area to the left of z0.025is 0.975. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table. EBP = ⎞ ⎠ p′ q′ n = (1.96) ⎛ ⎝z α 2 p′ – EBP = 0.842 – 0.032 = 0.81 (0.842)(0.158) 500 = 0.032 The confidence interval for the true binomial population proportion is (p′ – EBP, p′ + EBP) = (0.810, 0.874). p′ + EBP = 0.842 + 0.032 = 0.874 Interpretation We estimate with 95 percent confidence that between 81 percent and 87.4 percent of all adult residents of this city have cell
phones. Explanation of 95 percent 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. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 479 Solution 8.10 Press STAT and arrow over to TESTS. Arrow down to A:1-PropZint. Press ENTER. Arrow down to x and enter 421. 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 whether they own tablets. Of the 250 surveyed, 98 reported owning tablets. Using a 95 percent 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 percentage of students who are registered voters. He surveys 500 students and finds that 300 are registered voters. Compute a 90 percent confidence interval for the true percentage 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 n = 300 p′ = x 500 q′ = 1 – p′ = 1 - 0.600 = 0.400 = 0.600 Because 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 = 0.636 480 Chapter 8 \| Confidence Intervals The confidence interval for the true binomial population proportion is (p′ – EBP, p′ + EBP) = (0.564, 0.636). Interpretation • We estimate with 90 percent confidence that the true percentage of all students who are registered voters is between 56.4 percent and 63.6 percent. • Alternate wording: We estimate with 90 percent confidence that between 56.4 percent and 63.6 percent of all students are registered voters. Explanation of 90 percent Confidence Level Ninety percent of all confidence intervals constructed in this way contain the true value for the population percentage of students who 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 her school to determine whether 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 percent confidence interval for the true percentage of students who are against the new legislation, and interpret the confidence interval. b. In a sample of 300 students, 68 percent said they own an iPod and a smartphone. Compute a 97 percent confidence interval for the true percentage 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 the plus-four confidence interval for p method. It should be used when the confidence level desired is at least 90 percent and the sample size is at least ten. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 481 Example 8.12 A random sample of 25 statistics students was asked: “Have you used a product in the past week?” Six students reported using the product within the past week. Use the plus-four method to find a 95 percent confidence interval for the true proportion of statistics students who use the product weekly. Solution 8.12 Six students out of 25 reported using a product 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 q′ = 1 – p′ = 1 – 0.276 = 0.724 ≈ 0.276 Because CL = 0.95, we know α = 1 – 0.95 = 0.05, and α 2 = 0.025. z0.025 = 1.96 EPB = ⎞ ⎠ p′ q′ n = (1.96) ⎛ ⎝z α 2 p′ – EPB = 0.276 – 0.163 = 0.113 p′ + EPB = 0.276 + 0.163 = 0.439 0.276(0.724) 29 ≈ 0.163 We are 95 percent confident that the true proportion of all statistics students who use the product is between 0.113 and 0.439. Solution 8.12 Press STAT and arrow over to TESTS. Arrow down to A:1-PropZint. Press ENTER. Arrow down to x and enter 8. 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). REMINDER Remember that the plus-four method assumes 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. 8.12 Out of a random sample of 65 freshmen at State University, 31 students have declared their majors. Use the plus-four method to find a 96 percent confidence interval for the true proportion of freshmen at State University who have declared their majors. 482 Chapter 8 \| Confidence Intervals Example 8.13 A group of researchers 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 a social media site. Use the plus four method to find a 90 percent confidence interval for the true proportion of teens who would report having more than 500 online friends. Solution 8.13 Using plus-four, we have x = 13 + 2 = 15, and n = 50 + 4 = 54. p' = 15 54 ≈ 0.278 q' = 1 – p' = 1 - 0.278 = 0.722 Because 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) ⎝ ⎛ ) ⎝ p′ – EPB = 0.278 – 0.100 = 0.178 p′ + EPB = 0.278 + 0.160 = 0.378 (0.278)(0.722) 54 ⎞ ⎠ ≈ 0.100 We are 90 percent confident that between 17.8 percent and 37.8 percent of all teens would report having more than 500 friends on a social media site. Solution 8.13 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 research group 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 social media site friends, with 159 saying that they have more than 500 friends. Use the plus-four method to find a 90 percent confidence interval for the true proportion of teens who would report
having more than 500 online 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. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 483 The margin of error formula for a population proportion is • EBP = z α 2 × p′ • q′ n, where p′ is the sample proportion, q′ = 1 – p′, and n is the sample size. • Solving for n gives you an equation for the sample size. (p′ q′) 2 ⎛ ⎞ ⎝z α ⎠ 2 EBP2 • n =. This formula tells us that we can compute the sample size n required for a confidence level of Cl = 1 − α by taking the square of the critical value z a 2, multiplying by the point estimate p′, and by q′ = 1 – p′ and finally dividing the result by the square of the margin of error. Always remember to round up the value of n. Example 8.14 Suppose a mobile phone company wants to determine the current percentage of customers ages 50+ who use text messaging on their cell phones. How many customers ages 50+ should the company survey in order to be 90 percent confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion of customers ages 50+ who use text messaging on their cell phones? Assume that p′ = 0.5. Solution 8.14 From the problem, we know that EBP = 0.03 (3 percent=0.03), and z α 2 z0.05 = 1.645 because the confidence level is 90 percent. 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 ages 50+ in order to be 90 percent confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion of all customers ages 50+ who use text messaging on their cell phones. 8.14 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 percent confident that the estimated proportion is within 5 percentage points of the true population proportion of customers who click on ads on their smartphones? Assume that the sample proportion p′ is 0.50. 8.4 \| Confidence Interval (Home Costs) 484 Chapter 8 \| Confidence Intervals 8.1 Confidence Interval (Home Costs) Student Learning Outcomes • The student will calculate the 90 percent 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 following 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: _____ 2. How much area is in both tails (combined)? α = _____ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 485 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 8.7 5. Some students think that a 90 percent confidence interval contains 90 percent 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 percentage is this? Is this percentage close to 90 percent? Explain why this percentage should or should not be close to 90 percent. 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. 8.5 \| Confidence Interval (Place of Birth) 486 Chapter 8 \| Confidence Intervals 8.2 Confidence Interval (Place of Birth) Student Learning Outcomes • The student will calculate the 90 percent confidence interval of 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 whether they were born in this state. Let X = the number who 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 8.8 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. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 487 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. 8.6 \| Confidence Interval (Women's Heights) 488 Chapter 8 \| Confidence Intervals 8.3 Confidence Interval (Women's Heights) Student Learning Outcomes • The student will calculate a 90 percent 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 10 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 in. With these values, construct a 90 percent confidence interval for your sample of 10 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. __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 489 __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ Table 8.9 90 percent Confidence Intervals Discussion Questions 1. The actual population mean for the 100 heights given in 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 percentage of confidence intervals that contain the mean μ. Write that percentage here: _____________. 3. Is the percentage of confidence intervals that contain the population mean μ close to 90 percent? 4. Suppose we had generated 100 confidence intervals. What do you think would happen to the percentage of confidence intervals that contained the population mean? 5. When we construct a 90 percent confidence interval, we say that we are 90 percent 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 percent confidence interval contains 90 percent 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 percentage is this? Is this percentage close to 90 percent? 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 10 women and calculated a confidence interval from this information. Without knowing the population mean μ, would you have any way of knowing for certain whether your interval actually contained the value of μ? Explain. 490 Chapter 8 \| Confidence Intervals KEY TERMS binomial distribution 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 the standard deviation is σ = a discrete random variable (RV) that arises from Bernoulli trials; there are a fixed number, n, of npq. The probability of exactly x successes in n trials is P⎛ ⎝X = x⎞ ⎝n ⎠ = ⎛ x ⎞ ⎠p x qn − x. confidence interval (CI) an interval estimate for an unknown population parameter. This depends on the following: • • • the desired confidence level, information that is known about the distribution (for example, known standard deviation), and the sample and its size. confidence level (CL) the percentage expression for the probability that the confidence interval contains the true population parameter; for example, if the CL = 90 percent, 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 4 percent of the production is defective. normal distribution a bell-shaped continuous random variable X, with center at the mean value (μ) and distance from the center to the inflection points of the bell curve given by the standard deviation (σ). We write X ~ N(μ, σ). If the mean value is 0 and the standard deviation is 1, the random variable is called the standard normal distribution, and it is denoted with the letter Z parameter a numerical characteristic of a population plus-four confidence interval plus-four confidence interval when you add two imaginary successes and two imaginary failures (four overall) to your sample 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 as follows: • 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. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 491 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). The calculation of EBM depends on the size of the sample and the level of confidence desired. The confidence level is the percentage 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 backward 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 of 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's 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 = ⎛ s n, standard deviation, and n is the sample size. Use a table, calculator, or computer to find t α 2 is the t-score with area to the right equal to α 2 for a given α. where t α 2, s is the sample ⎝t α 2 ⎞ ⎠ 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 by 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 492 Chapter 8 \| Confidence Intervals additional trials in the study; two are successes and two are failures. Calculate p′ = x + 2 n + 4, and proceed to find the 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 ~ N ⎛ ⎝μ X, σ n ⎞ ⎠ The distribution of sample means is normally distributed with mean equal to the population mean 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 estimate + EBM) =( x¯ − EBM, x¯ + EBM) ⎞ ⎠. ⎛ ⎝ x¯ − z σ = n, x¯ + z σ n = the error bound for the mean, or the margin EBM = z σ n 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 is expected to contain the true population parameter 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's t-Distribution s = the standard deviation of sample values t = x¯ − μ s n is the formula for the t-
score, which measures how far away a measure is from the population mean in the Student’s t-distribution. df = n – 1; t-distribution, where n represents the size of the sample the degrees of freedom for a Student’s T~tdf the random variable, T, has a Student’s t-distribution with df degrees of freedom EBM = t α 2 s n = the error bound for the population mean when the population standard deviation is unknown α = 1 – CL = the proportion of confidence intervals that will not contain the population parameter t α 2 is the t-score in the Student’s t-distribution with area to = the z-score with the property that the area to the z α 2 right of the z-score is ∝ 2 ; this is the z-score, used in the calculation of EBM, where α = 1 – CL. n = z2 σ 2 EBM 2 = the formula used to determine the sample size (n) needed to achieve a desired margin of error at a given level of confidence 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 ⎞ ⎠. General form of a confidence interval 8.3 A Population Proportion (lower value, upper value) = (point estimate error bound, point estimate + error bound) 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. 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. q′ = 1 – p′ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 493 p′ ~ N ⎛ ⎝p, pq n ⎞ ⎠ The variable p′ has a binomial distribution that can be approximated with the normal distribution shown here, 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 ⎞ ⎠ n = 2 p′ q′ z α 2 EBP2 provides the number of participants needed to estimate the population proportion with confidence 1 – α and margin of error EBP. PRACTICE 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). is a point estimate for μ. x¯ p′ is a point estimate for ρ. s is a point estimate for σ. 8.1 A Single Population Mean Using the Normal Distribution Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 lb. We wish to construct a 95 percent confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 lb. The sample standard deviation is 11 lb. 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 percent 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 percent 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? 494 Chapter 8 \| Confidence Intervals 12. Suppose the Census needed to be 98 percent 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 10 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 lb, with a standard deviation of 0.1 lb. The population standard deviation is known to be 0.2 lb. 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 percent 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 percent 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? 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 25 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. 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 percent confidence interval for the true mean age of winter Foothill College students by working out and then answering the next eight 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 percent confidence interval is __________________. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 495 34. Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample mean. Figure 8.9 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 percent? Why? 8.2 A Single Population Mean Using the Student's 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 hr, with a sample standard deviation of 0.5 hr. 38. Identify the following: a. b. x¯ =_______ s x =_______ 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 percent 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 watch 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? 496 Chapter 8 \| Confidence Intervals 47. Construct a 99 percent confidence interval for the population mean hours spent watching television per month. State the confidence interval, sketch the graph, and calculate the error bound. 48. Why would the error bound change if the confidence level were lowered to 95 percent? 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 percent confidence interval for the true mean number of colors on national flags. 54. How much area is in both tails (combined)? 55. How much area is in each tail? 56. Calculate the following: lower limit a. b. upper limit c. error bound 57. The 95 percent confidence interval is_____. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 497 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 8.10 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 percent? Why? 8.3 A Population Proportion Use the following information to answer the next two exercises: Marketing companies are interested in knowing the population percentage 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 percent 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 percent 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 a marketing company conducted a survey. It randomly surveyed 200 households and found that in 120 of them, the women 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 percent 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 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 percent of manual laborers preferred trucks, 62 percent of non-manual wage earners preferred trucks, 54 percent of mid-level managers preferred trucks, and 26 percent of executives preferred trucks. 498 Chapter 8 \| Confidence Intervals 69. We are interested in finding the 95 percent confidence interval for the percentage 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 percent 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 percent. Explain what the ±2 percent 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 believe 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 percent 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 percent? 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 is picked. In that class are 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 percent 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: lower limit a. b. upper limit c. error bound 89. The 92 percent confidence interval is _______. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 499 90. Fill in the blanks on the graph with the areas
, upper and lower limits of the confidence interval, and the sample proportion. Figure 8.11 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 percent? Why? 94. If you decreased the allowable error bound, why would the minimum sample size increase (keeping the same level of confidence)? 500 Chapter 8 \| Confidence Intervals 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 percent confidence interval for the mean height of male Swedes. 48 male Swedes are surveyed. The sample mean is 71 inches. The sample standard deviation is 2.8 in. x¯ =________ a. 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 percent 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 percent 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 =________ ¯. 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 percent 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, which 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 percent confident of the population mean length of time to within one hour. How would the number of people the firm surveys change? Why? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 501 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 =________ sx =________ iii. 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 percent 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 percent 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. x¯ =________ a. 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 percent 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 percent confident when constructing a confidence interval for a mean? a. b. c. d. If we took repeated samples, approximately 90 percent of the samples would produce the same confidence interval. If we took repeated samples, approximately 90 percent of the confidence intervals calculated from those samples would contain the sample mean. If we took repeated samples, approximately 90 percent 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 percent of the samples. 502 Chapter 8 \| Confidence Intervals 101. The Federal Election Commission collects information about campaign contributions and disbursements for candidates and political committees during 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. $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 percent confidence, calculate the error bound. c. Create a 95 percent 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 U.S. Census Bureau, conducts a yearly census similar to the one taken every 10 years, but with a smaller percentage of participants. The most recent survey estimates with 90 percent confidence that the mean household income in the United States 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 in. You want to estimate the mean height of students at your college or university to within 1 in. with 93 percent confidence. How many male students must you measure? 8.2 A Single Population Mean Using the Student's t-Distribution 104. In six packages of multicolored fruit snacks, there were five red snack pieces. The total number of snack pieces in the six bags was 68. We wish to calculate a 96 percent confidence interval for the population proportion of red 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 percent confidence interval for the population proportion of red 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? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 503 105. 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. d. Construct a 95 percent confidence interval for the population mean enrollment at community colleges in the in words. 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 are surveyed? Why? 106. Suppose that a committee is studying whether there is wasted 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 8 hr, with a sample standard deviation of 4 hr. 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 percent 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 a drug used during surgery. It is assumed that the distribution for the length of time the drug lasts is approximately normal. Researchers in a hospital used the drug on a random sample of nine patients. The effective period of the antibiotic drug 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 percent 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 percent confident in this problem? 504 Chapter 8 \| Confidence Intervals 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 percent 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 percent? 109. The Federal Election Commission (FEC) collects information about campaign contributions and disbursements for candidates and political committees during 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 operated during the 2011–2012 election cycle. The following table shows the total receipts during this cycle for a random selection of 20 Leadership PACs. $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 the sample data to construct a 96 percent 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. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 505 110. A major business magazine published data on the best small firms in 2012. These were firms that have 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. 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 the sample data to construct a 90 percent confidence interval for the mean age of CEOs 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 percent 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. 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 c. Construct a 95 percent confidence interval for the population mean cost of a used car. in words. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. d. Explain what a 95 percent 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 percent 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? 506 Chapter 8 \| Confidence Intervals 114. A survey of the mean number of cents off given by coupons was conducted by randomly surveying one coupon per page from the coupons section of a local newspaper. 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 percent 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 collected with 14 samples as the size,
which percentage 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 percent 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. Which of the following 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 percentage of drivers who always buckle up before riding in a car. a. When designing a study to determine this population proportion, what is the minimum number you would need b. to survey to be 95 percent 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 percent 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 conduct 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. 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 percent 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. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 507 119. According to a recent survey of 1,200 people, 61 percent believe that the president is doing an acceptable job. We are interested in the population proportion of
people who believe 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 percent confidence interval for the population proportion of people who believe 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 dating and marriage recently appeared in a major newspaper. Of the 1,709 randomly selected adults, 315 identified themselves as ethnicity A, 323 identified themselves as ethnicity B, 254 identified themselves as ethnicity C, and 779 identified themselves as ethnicity D. In this survey, 86 percent of ethnicity B said that they would welcome a person of ethnicity A into their families. Among ethnicity C, 77 percent would welcome a person of ethnicity D into their families, 71 percent would welcome a person of ethnicity A, and 66 percent would welcome a person of ethnicity B. a. We are interested in finding the 95 percent confidence interval for the percent of all ethnicity B adults who would welcome a person of ethnicity D 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 percent 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 percent confidence intervals: i. percentage of all ethnicity C who would welcome a person of ethnicity D into their families ii. percentage of all ethnicity C who would welcome a person of ethnicity A into their families iii. percentage of all ethnicity C who would welcome a person of ethnicity B 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 percent 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 percent confidence interval for the population proportion of people over 50 who ran and died in the same 8-year period. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. d. Explain what a 97 percent confidence interval means for this study. 508 Chapter 8 \| Confidence Intervals 123. A telephone poll of 1,000 adult Americans was reported in an issue of a national magazine. One of the questions asked, “What is the main problem facing the country?” Twenty percent responded "crime". We are interested in the population proportion of adult Americans who believe 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 percent confidence interval for the population proportion of adult Americans who believe 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. What is one way to accomplish that? e. The sampling error given by the group of researchers who conducted the poll is ±3 percent. In one to three complete sentences, explain what the ±3 percent represents. 124. Refer to Exercise 8.123. Another question in the poll asked, “[How much are] you worried about the quality of education in our schools?” Sixty-three percent responded “a lot”. We are interested in the population proportion of adult Americans who are worried a lot about the quality of education in our schools. 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 percent confidence interval for the population proportion of adult Americans who are worried a lot about the quality of education in our schools. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. d. The sampling error given by the group of researchers who conducted the poll is ±3 percent. In one to three complete sentences, explain what the ±3 percent represents. Use the following information to answer the next three exercises: According to a Field Poll, 79 percent of California adults (actual results are 400 out of 506 surveyed) believe
that education and our schools is one of the top issues facing California. We wish to construct a 90 percent confidence interval for the true proportion of California adults who believe that education and the schools is one of the top issues facing California. 125. A point estimate for the true population proportion is _______. a. 0.90 b. 1.27 c. 0.79 d. 400 126. A 90 percent confidence interval for the population proportion is _______. a. b. c. d. (0.761, 0.820) (0.125, 0.188) (0.755, 0.826) (0.130, 0.183) 127. The error bound is approximately _____. a. 1.581 b. 0.791 c. 0.059 d. 0.030 Use the following information to answer the next two exercises: Five hundred eleven (511) homes in a certain southern California community are randomly surveyed to determine whether they meet minimal earthquake preparedness recommendations. One hundred seventy-three (173) of the homes surveyed meet the minimum recommendations for earthquake preparedness, and 338 do not. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 509 128. Find the confidence interval at the 90 percent confidence level for the true population proportion of southern California community homes meeting at least the minimum recommendations for earthquake preparedness. a. b. c. d. (0.2975, 0.3796) (0.6270, 0.6959) (0.3041, 0.3730) (0.6204, 0.7025) 129. The point estimate for the population proportion of homes that do not meet the minimum recommendations for earthquake preparedness is ______. a. 0.6614 b. 0.3386 c. 173 d. 338 130. On May 23, 2013, a polling group reported that of the 1,005 people surveyed, 76 percent of U.S. workers believe that they will continue working past retirement age. The confidence level for this study was reported at 95 percent with a ±3 percent margin of error. a. Determine the estimated proportion from the sample. b. Determine the sample size. c. d. Calculate the error bound based on the information provided. e. Compare the error bound in Part d to the margin of error reported by the polling group. Explain
any differences Identify CL and α. between the values. f. Create a confidence interval for the results of this study. g. A reporter is covering the release of this study for a local news station. How should she explain the confidence interval to her audience? 131. A national survey of 1,000 adults was conducted on May 13, 2013, by a group of researchers. It concluded with 95 percent confidence that 49 percent to 55 percent of Americans believe that big-time college sports programs corrupt the process of higher education. a. Find the point estimate and the error bound for this confidence interval. b. Can we (with 95 percent confidence) conclude that more than half of all American adults believe this? c. Use the point estimate from Part a and n = 1,000 to calculate a 75 percent confidence interval for the proportion of American adults who believe that major college sports programs corrupt higher education. d. Can we (with 75 percent confidence) conclude that at least half of all American adults believe this? 132. A polling group recently conducted a survey asking adults across the United States about music preferences. When asked, 80 of the 571 participants download music weekly. a. Create a 99 percent confidence interval for the true proportion of American adults who download music weekly. b. This survey was conducted through automated telephone interviews on May 6 and 7, 2013. The error bound of the survey compensates for sampling error, or natural variability among samples. List some factors that could affect the survey’s outcome that are not covered by the margin of error. c. Without performing any calculations, describe how the confidence interval would change if the confidence level decreased from 99 percent to 90 percent. 133. You plan to conduct a survey on your college campus to learn about the political awareness of students. You want to estimate the true proportion of college students on your campus who voted in the 2012 presidential election with 95 percent confidence and a margin of error no greater than 5 percent. How many students must you interview? 134. In a recent poll, 9 of 48 respondents rated the likelihood of a certain event occurring in their community as likely or very likely. Use the plus-four method to create a 97 percent confidence interval for the proportion of American adults who believe that the event is likely or very likely. Explain what this confidence interval means in the context of the problem. A local poll in a New England town found that nine of 48 households think winter-proofing their cars is very important. Use the plus-four method to create a 97 percent confidence
interval for the proportion of town residents who think winterproofing their cars is very important. Explain what this confidence interval means in the context of this scenario. REFERENCES 510 Chapter 8 \| Confidence Intervals 8.1 A Single Population Mean Using the Normal Distribution Centers for Disease Control and Prevention. (n.d.). National health and nutrition examination survey. Retrieved from http://www.cdc.gov/nchs/nhanes.htm Foothill De Anza Community College District. (n.d.). Headcount enrollment trends by student demographics ten-year trends to most recently completed fall. Retrieved from http://research.fhda.edu/factbook/FH_Demo_Trends/ fall FoothillDemographicTrends.htm Kuczmarski, R. J., et al. (2002). 2000 CDC growth charts for the United States: Methods and development. Retrieved from http://www.cdc.gov/growthcharts/2000growthchart-us.pdf La, L., and; German, K. (n.d.). Cell phones with the highest radiation levels. CNET. Retrieved from http://reviews.cnet.com/ cell-phone-radiation-levels/ U.S. Census Bureau. (n.d.). American FactFinder. Retrieved from http://factfinder2.census.gov/faces/nav/jsf/pages/ searchresults.xhtml?refresh=t U.S. Census Bureau. (2011). Mean income in the past 12 months (in 2011 inflation-adjusted dollars): 2011 American Community from http://factfinder2.census.gov/faces/tableservices/jsf/pages/ productview.xhtml?pid=ACS_11_1YR_S1902&prodType=table estimates. Retrieved Survey 1-year U.S. Federal Election Commission. (n.d.). Disclosure data catalog: Candidate summary report 2012. Retrieved from http://www.fec.gov/data/CandidateSummary.do?format=html&election_yr=2012 U.S. Federal Election Commission. http://www.fec.gov/finance/disclosure/metadata/metadataforcandidatesummary.shtml (n.d.). Metadata description of candidate summary file. Retrieved from 8.2 A Single Population Mean Using the Student's t-Distribution Bloomberg Businessweek. (n.
d.). Retrieved from http://www.businessweek.com/ Environmental Working Group. (n.d.). Human toxome project: Mapping the pollution in people. Retrieved from http://www.ewg.org/sites/humantoxome/participants/participant-group.php?group=in+utero%2Fnewborn Federal Election Commission. (n.d.). Disclosure data catalog: 2012 leadership PACs and sponsors. Retrieved from http://www.fec.gov/data/index.jsp Federal Election Commission. (n.d.). Metadata description of leadership PAC list. Retrieved from http://www.fec.gov/ finance/disclosure/metadata/metadataLeadershipPacList.shtml Forbes. (2013). America’s best small companies. Retrieved from http://www.forbes.com/best-small-companies/list/ Forbes. (n.d.). Retrieved from http://www.forbes.com/ Microsoft Bookshelf. (n.d.). 8.3 A Population Proportion Jensen, http://www.publicpolicypolling.com/Day2MusicPoll.pdf (2013). Democrats, Republicans T. divided on opinion of music icons. Retrieved from Madden, M., et al. (2013). Teens, social media, and privacy. Retrieved from http://www.pewinternet.org/Reports/2013/ Teens-Social-Media-And-Privacy.aspx Princeton Survey Research Associates International. (2012). 2012 teens and privacy management survey. Retrieved from http://www.pewinternet.org/~/media//Files/Questionnaire/2013/ Methods%20and%20Questions_Teens%20and%20Social%20Media.pdf Rasmussen Reports. http://www.rasmussenreports.com/public_content/lifestyle/sports/may_2013/ 52_say_big_time_college_athletics_corrupt_education_process (2013). 52% say big-time college athletics corrupt education process. Retrieved from Saad, L. (2013). Three in four U.S. workers plan to work past retirement age. Retrieved from http://www.gallup.com/poll/ 162758/three-fourworkers-plan-work-past-retirement-age.aspx The Field Poll. (n.d.). Retrieved from http://field.com/fieldpollonline/subscribers/
This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 511 Zogby Analytics. (2013). New SUNYIT/Zogby analytics poll: Few Americans worry about emergency situations occurring in their community; Only one in three have an emergency plan; 70% support infrastructure “investment” for national security. Retrieved from http://www.zogbyanalytics.com/news/299-americans-neither-worried-norprepared-in-case-of-adisaster-sunyit-zogby-analytics-poll SOLUTIONS 1 a. 244 b. 15 c. 50 3 N ⎛ ⎝244, 15 50 ⎞ ⎠ 5 As the sample size increases, there will be less variability in the mean, so the interval size decreases. ¯ 7 X is the time in minutes it takes to complete the U.S. Census short form. X people to complete the U.S. Census short form. is the mean time it took a sample of 200 9 CI: (7.9441, 8.4559) Figure 8.12 EBM = 0.26 11 The level of confidence would decrease, because decreasing n makes the confidence interval wider, so at the same error bound, the confidence level decreases. 13 a. x¯ = 2.2 b. σ = 0.2 c. n = 20 ¯ 15 X is the mean weight of a sample of 20 heads of lettuce. 17 EBM = 0.07 CI: (2.1264, 2.2736) 512 Chapter 8 \| Confidence Intervals Figure 8.13 19 The interval is greater, because the level of confidence increased. If the only change made in the analysis is a change in confidence level, then all we are doing is changing how much area is being calculated for the normal distribution. Therefore, a larger confidence level results in larger areas and larger intervals. 21 The confidence level would increase. 23 30.4 25 σ 27 μ 29 normal 31 0.025 33 (24.52,36.28) 35 We are 95 percent confident that the true mean age for winter Foothill College students is between 24.52 and 36.28. 37 The error bound for the mean would decrease, because as the CL decreases, you need less area under the normal curve (which translates into
a smaller interval) to capture the true population mean. ¯ 39 X is the number of hours a patient waits in the emergency room before being called back to be examined. X mean wait time of 70 patients in the emergency room. is the 41 CI: (1.3808, 1.6192) Figure 8.14 EBM = 0.12 43 a. b. x¯ = 151 s x = 32 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 513 c. n = 108 d. n – 1 = 107 ¯ 45 X is the mean number of hours spent watching television per month from a sample of 108 Americans. 47 CI: (142.92, 159.08) Figure 8.15 EBM = 8.08 49 a. 3.26 b. 1.02 c. 39 51 μ 53 t38 55 0.025 57 (2.93, 3.59) 59 We are 95 percent confident that the true mean number of colors for national flags is between 2.93 colors and 3.59 colors. 60 The error bound would become EBM = 0.245. This error bound decreases, because as sample sizes increase, variability decreases, and we need less interval length to capture the true mean. 63 It would decrease, because the z-score would decrease, which would reduce the numerator and lower the number. 65 X is the number of successes where the woman makes the majority of the purchasing decisions for the household. P′ is the percentage of households sampled where the woman makes the majority of the purchasing decisions for the household. 67 CI: (0.5321, 0.6679) 514 Chapter 8 \| Confidence Intervals Figure 8.16 EBM: 0.0679 69 X is the number of successes where an executive prefers a truck. P′ is the percentage of executives sampled who prefer a truck. 71 CI: (0.19432, 0.33068) Figure 8.17 EBM: 0.0707 73 The sampling error means that the true mean can be 2 percent above or below the sample mean. 75 P′ is the proportion of voters sampled who said the economy is the most important issue in the upcoming election. 77 CI: (0.62735, 0.67265); EBM: 0.02265 79 the number of girls, ages 8 to 12, in the 5 p.m. Monday
night beginning ice-skating class 81 a. x = 64 b. n = 80 c. p′ = 0.8 83 p 85 P′ ~ N ⎛ ⎝0.8, (0.8)(0.2) 80 ⎞ ⎠ CI = (0.72171, 0.87829). 87 0.04 89 (0.72; 0.88) 91 With 92 percent confidence, we estimate the proportion of girls, ages 8 to 12, in a beginning ice-skating class at the Ice Chalet to be between 72 percent and 88 percent. 93 The error bound would increase. Assuming all other variables are kept constant, as the confidence level increases, the area under the curve corresponding to the confidence level becomes larger, which creates a wider interval and thus a larger This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 515 error. 95 a. i. 71 ii. 3 iii. 48 b. X is the height of a Swedish male, and is the mean height from a sample of 48 Swedish males. c. Normal. We know the standard deviation for the population, and the sample size is greater than 30. d. i. CI: (70.151, 71.49) ii. Figure 8.18 iii. EBM = 0.849 e. The confidence interval will decrease in size, because the sample size increased. Recall, when all factors remain unchanged, an increase in sample size decreases variability. Thus, we do not need as large an interval to capture the true population mean. 97 a. i. ii. x¯ = 23.6 σ = 7 iii. n = 100 ¯ b. X is the time needed to complete an individual tax form. X is the mean time to complete tax forms from a sample of 100 customers. c. N ⎛ ⎝23.6, ⎞ ⎠ 7 100 because we know sigma. d. ii. i. (22.228, 24.972) 516 Chapter 8 \| Confidence Intervals Figure 8.19 iii. EBM = 1.372 e. It will need to change the sample size. The firm needs to determine what the confidence level should be and then apply the error bound formula to determine the necessary sample size. f. The confidence level would increase as a result of a larger interval
. Smaller sample sizes result in more variability. To capture the true population mean, we need to have a larger interval. g. According to the error bound formula, the firm needs to survey 206 people. Because we increase the confidence level, we need to increase either our error bound or the sample size. 99 a. i. 7.9 ii. 2.5 iii. 20 ¯ b. X is the number of letters a single camper will send home. X is the mean number of letters sent home from a sample of 20 campers. c. N 7.9 ⎛ ⎝ ⎞ ⎠ 2.5 20 d. i. CI: (6.98, 8.82) ii. Figure 8.20 iii. EBM: 0.92 e. The error bound and confidence interval will decrease. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 517 101 a. x¯ = $568,873 b. CL = 0.95, α = 1 – 0.95 = 0.05, z α 2 = 1.96 909200 EBM = z0.025 = 1.96 = $281,764 σ n 40 c. x¯ − EBM = 568,873 − 281,764 = 287,109 x¯ + EBM = 568,873 + 281,764 = 850,637 Alternate solution: 1. Press STAT and arrow over to TESTS. 2. Arrow down to 7:ZInterval. 3. Press ENTER. 4. Arrow to Stats and press ENTER. 5. Arrow down and enter the following values: σ : 909,200 x¯ : 568,873 n: 40 CL: 0.95 6. Arrow down to Calculate and press ENTER. 7. The confidence interval is ($287,114, $850,632). 8. Notice the small difference between the two solutions—these differences are simply due to rounding error in the hand calculations. d. We estimate with 95 percent confidence that the mean amount of contributions received from all individuals by House candidates is between $287,109 and $850,637. 103 Use the formula for EBM, solved for n: n = z2 σ 2 EBM 2 (This is the value of z for which the area under From the statement of
the problem, you know that σ = 2.5, and you need EBM = 1. z = z0.035 = ≈ 20.52. You need to measure at least 21 male students to achieve your goal. the density curve to the right of z is 0.035.) 1.812. n = z2 σ 2 EBM 2 = 1.8122 2.52 12 105 a. i. 8,629 ii. 6,944 iii. 35 iv. 34 t34 i. CI: (6244, 11,014) b. c. 518 Chapter 8 \| Confidence Intervals ii. Figure 8.21 iii. EB = 2385 d. It will become smaller. 107 a. i. ii. x¯ = 2.51 s x = 0.318 iii. n = 9 iv. n - 1 = 8 b. The effective length of time for a tranquilizer c. The mean effective length of time of tranquilizers from a sample of nine patients d. We need to use a Student’s t-distribution, because we do not know the population standard deviation. e. i. CI: (2.27, 2.76) ii. Check student's solution. iii. EBM: 0.25 f. If we were to sample many groups of nine patients, 95 percent of the samples would contain the true population mean length of time. 109 x¯ = $251, 854.23; s = $521, 130.41. Note that we are not given the population standard deviation, only the standard deviation of the sample. There are 30 measures in the sample, so n = 30, and df = 30 - 1 = 29. CL = 0.96, so α = x¯ ⎞ ⎠ = 2.150 ⎞ ⎠ ~ $204, 561.66. 1 - CL = 1 - 0.96 = 0.04. α 2 = 2.150. EBM = t α 2 521, 130.41 30 = 0.02t α 2 = t0.02 s n ⎛ ⎝ ⎛ ⎝ - EBM = $251,854.23 - $204,561.66 = $47,292.57. x¯ + EBM = $251,854.23 + $204,561.66 = $456,415.89. We estimate with 96
percent confidence that the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle lies between $47,292.57 and $456,415.89. Alternate Solution STATTESTS8:TIntervalENTERENTERFreqC-LevelCalculateEnter The difference between solutions arises from rounding differences. 111a. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 519 i. ii. x¯ = s x = iii. n = iv. n - 1 = ¯ b. X is the number of unoccupied seats on a single flight. X is the mean number of unoccupied seats from a sample of 225 flights. c. We will use a Student’s t-distribution, because we do not know the population standard deviation. d. i. CI: (11.12, 12.08) ii. Check student's solution. iii. EBM: 0.48 113 a. i. CI: (7.64, 9.36) ii. Figure 8.22 iii. EBM: 0.86 b. The sample should have been increased. c. Answers will vary. d. Answers will vary. e. Answers will vary. 115 b 117 a. 1,068 b. The sample size would need to be increased, because the critical value increases as the confidence level increases. 119 a. X = the number of people who believe that the president is doing an acceptable job; P′ = the proportion of people in a sample who believe that the president is doing an acceptable job. b. N ⎛ ⎝0.61, (0.61)(0.39) 1200 ⎞ ⎠ c. i. CI: (0.59, 0.63) ii. Check student’s solution. iii. EBM: 0.02 520 121 a. i. ii. (0.72, 0.82) (0.65, 0.76) iii. (0.60, 0.72) Chapter 8 \| Confidence Intervals b. Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap. c. We can say that there does not appear to be
a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a Latino person into their families. d. We can say that there is a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a black person into their families. 123 a. X = the number of adult Americans who believe that crime is the main problem; P′ = the proportion of adult Americans who believe that crime is the main problem. b. Because we are estimating a proportion, ⎞ ⎠ (0.2)(0.8) 1000 ⎛ ⎝0.2, N. that P′ = 0.2 and n = 1,000, the distribution we should use is c. i. CI: (0.18, 0.22) ii. Check student’s solution. iii. EBM: 0.02 d. One way to lower the sampling error is to increase the sample size. e. The stated ± 3 percent represents the maximum error bound. This means that those doing the study are reporting a maximum error of 3 percent. Thus, they estimate the percentage of adult Americans who the percentage of adult Americans who that crime is the main problem to be between 18 percent and 22 percent. 125 c 127 d 129 a 131 a. p′ = (0.55 + 0.49) 2 = 0.52; EBP = 0.55 – 0.52 = 0.03 b. No, the confidence interval includes values less than or equal to 0.50. It is possible that less than half of the population believe this. c. CL = 0.75, so α = 1 – 0.75 = 0.25 and α 2 = 0.125. z α 2 = 1.150. (The area to the right of this z is 0.125, so the area to the left is 1 – 0.125 = 0.875.) EBP = (1.150) 0.52(0.48) 1, 000 ≈ 0.018 (p′ - EBP, p′ + EBP) = (0.52 – 0.018, 0.52 + 0.018) = (0.502, 0.538) Alternate Solution STAT TESTS A: 1-PropZinterval
with x = (0.52)(1,000), n = 1,000, CL = 0.75. Answer is (0.502, 0.538). This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 8 \| Confidence Intervals 521 d. Yes, this interval does not fall below 0.50, so we can conclude that at least half of all American adults believe that major sports programs corrupt education – but we do so with only 75 percent confidence. 133 CL = 0.95; α = 1 – 0.95 = 0.05; α 2 = 0.025; z α 2 = 1.96. Use p′ = q′ = 0.5. n = 2 p′ q′ z α 2 EBP2 = 1.962(0.5)(0.5) 0.052 = 384.16. You need to interview at least 385 students to estimate the proportion to within 5 percent at 95 percent confidence. 522 Chapter 8 \| Confidence Intervals This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 523 9 \| HYPOTHESIS TESTING WITH ONE SAMPLE Figure 9.1 You can use a hypothesis test to decide if a dog breeder’s claim that every Dalmatian has 35 spots is statistically sound. (credit: Robert Neff) Introduction By the end of this chapter, the student should be able to do the following: Chapter Objectives • Differentiate between Type I and Type II errors • Describe hypothesis testing in general and in practice • Conduct and interpret hypothesis tests for a single population mean, population standard deviation known • Conduct and interpret hypothesis tests for a single population mean, population standard deviation unknown • Conduct and interpret hypothesis tests for a single population proportion One job of a statistician is to make statistical inferences about populations based on samples taken from the population. Confidence intervals are one way to estimate a population parameter. Another way to make a statistical inference is to 524 Chapter 9 \| Hypothesis Testing with One Sample make a decision about a parameter. For instance, a car dealer advertises that its new small truck gets 35 miles per gallon, on average. A tutoring service claims that its method of tutoring helps 90 percent of its students get an A or a B. A company says that
women managers in their company earn an average of $60,000 per year. A statistician will make a decision about these claims. This process is called hypothesis testing. A hypothesis test involves collecting data from a sample and evaluating the data. Then, the statistician makes a decision as to whether or not there is sufficient evidence, based upon analyses of the data, to reject the null hypothesis. In this chapter, you will conduct hypothesis tests on single means and single proportions. You will also learn about the errors associated with these tests. Hypothesis testing consists of two contradictory hypotheses or statements, a decision based on the data, and a conclusion. To perform a hypothesis test, a statistician will do the following: 1. Set up two contradictory hypotheses. 2. Collect sample data. In homework problems, the data or summary statistics will be given to you. 3. Determine the correct distribution to perform the hypothesis test. 4. Analyze sample data by performing the calculations that ultimately will allow you to reject or decline to reject the null hypothesis. 5. Make a decision and write a meaningful conclusion. NOTE To do the hypothesis test homework problems for this chapter and later chapters, make copies of the appropriate special solution sheets. See Appendix E. 9.1 \| Null and Alternative Hypotheses The actual test begins by considering two hypotheses. They are called the null hypothesis and the alternative hypothesis. These hypotheses contain opposing viewpoints. H0, the —null hypothesis: a statement of no difference between sample means or proportions or no difference between a sample mean or proportion and a population mean or proportion. In other words, the difference equals 0. Ha—, the alternative hypothesis: a claim about the population that is contradictory to H0 and what we conclude when we reject H0. Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data. After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are reject H0 if the sample information favors the alternative hypothesis or do not reject H0 or decline to reject H0 if the sample information is insufficient to reject the null hypothesis. Mathematical Symbols Used in H0 and Ha: H0 equal (=) Ha not equal (≠) or greater than (>) or less than (<) greater than or equal to (≥) less than (<) less than or equal to (≤) more
than (>) Table 9.1 NOTE H0 always has a symbol with an equal in it. Ha never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers use = in the null hypothesis, even with This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 525 > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis. Example 9.1 H0: No more than 30 percent of the registered voters in Santa Clara County voted in the primary election. p ≤ 30 Ha: More than 30 percent of the registered voters in Santa Clara County voted in the primary election. p > 30 9.1 A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25 percent. State the null and alternative hypotheses. Example 9.2 We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are the following: H0: μ = 2.0 Ha: μ ≠ 2.0 9.2 We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. a. H0: μ __ 66 b. Ha: μ __ 66 Example 9.3 We want to test if college students take fewer than five years to graduate from college, on the average. The null and alternative hypotheses are the following: H0: μ ≥ 5 Ha: μ < 5 9.3 We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. a. H0: μ __ 45 b. Ha: μ __ 45 526 Chapter 9 \| Hypothesis Testing with One Sample Example 9.4 An article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third of the students pass. The same article stated that 6.6 percent of U.S. students take advanced placement exams and 4.
4 percent pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6 percent. State the null and alternative hypotheses. H0: p ≤ 0.066 Ha: p > 0.066 9.4 On a state driver’s test, about 40 percent pass the test on the first try. We want to test if more than 40 percent pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. a. H0: p __ 0.40 b. Ha: p __ 0.40 Bring to class a newspaper, some news magazines, and some internet articles. In groups, find articles from which your group can write null and alternative hypotheses. Discuss your hypotheses with the rest of the class. 9.2 \| Outcomes and the Type I and Type II Errors When you perform a hypothesis test, there are four possible outcomes depending on the actual truth, or falseness, of the null hypothesis H0 and the decision to reject or not. The outcomes are summarized in the following table: ACTION H0 IS ACTUALLY... True False Do not reject H0 Correct outcome Type II error Reject H0 Table 9.2 Type I error Correct outcome The four possible outcomes in the table are as follows: 1. The decision is not to reject H0 when H0 is true (correct decision). 2. The decision is to reject H0 when, in fact, H0 is true (incorrect decision known as a Type I error). 3. The decision is not to reject H0 when, in fact, H0 is false (incorrect decision known as a Type II error). 4. The decision is to reject H0 when H0 is false (correct decision whose probability is called the Power of the Test). Each of the errors occurs with a particular probability. The Greek letters α and β represent the probabilities. α = probability of a Type I error = P(Type I error) = probability of rejecting the null hypothesis when the null hypothesis is true. β = probability of a Type II error = P(Type II error) = probability of not rejecting the null hypothesis when the null hypothesis is false. α and β should be as small as possible because they are probabilities of errors. They are rarely zero. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \|
Hypothesis Testing with One Sample 527 The Power of the Test is 1 – β. Ideally, we want a high power that is as close to one as possible. Increasing the sample size can increase the Power of the Test. The following are examples of Type I and Type II errors. Example 9.5 Suppose the null hypothesis, H0, is: Frank's rock climbing equipment is safe. Type I error: Frank does not go rock climbing because he considers that the equipment is not safe, when in fact, the equipment is really safe. Frank is making the mistake of rejecting the null hypothesis, when the equipment is actually safe! Type II error: Frank goes climbing, thinking that his equipment is safe, but this is a mistake, and he painfully realizes that his equipment is not as safe as it should have been. Frank assumed that the null hypothesis was true, when it was not. α = probability that Frank thinks his rock climbing equipment may not be safe when, in fact, it really is safe. β = probability that Frank thinks his rock climbing equipment may be safe when, in fact, it is not safe. Notice that, in this case, the error with the greater consequence is the Type II error. (If Frank thinks his rock climbing equipment is safe, he will go ahead and use it.) 9.5 Suppose the null hypothesis, H0, is: the blood cultures contain no traces of pathogen X. State the Type I and Type II errors. Example 9.6 Suppose the null hypothesis, H0, is: a tomato plant is alive when a class visits the school garden. Type I error: The null hypothesis claims that the tomato plant is alive, and it is true, but the students make the mistake of thinking that the plant is already dead. Type II error: The tomato plant is already dead (the null hypothesis is false), but the students do not notice it, and believe that the tomato plant is alive. α = probability that the class thinks the tomato plant is dead when, in fact, it is alive = P(Type I error). β = probability that the class thinks the tomato plant is alive when, in fact, it is dead = P(Type II error). The error with the greater consequence is the Type I error. (If the class thinks the plant is dead, they will not water it.) 9.6 Suppose the null hypothesis, H0, is: a patient is not sick. Which type of error has the greater consequence, Type I or Type
II? Example 9.7 It’s a Boy Genetic Labs, a genetics company, claims to be able to increase the likelihood that a pregnancy will result in a boy being born. Statisticians want to test the claim. Suppose that the null hypothesis, H0, is: It’s a Boy 528 Chapter 9 \| Hypothesis Testing with One Sample Genetic Labs has no effect on gender outcome. Type I error: This error results when a true null hypothesis is rejected. In the context of this scenario, we would state that we believe that It’s a Boy Genetic Labs influences the gender outcome, when in fact it has no effect. The probability of this error occurring is denoted by the Greek letter alpha, α. Type II error: This error results when we fail to reject a false null hypothesis. In context, we would state that It’s a Boy Genetic Labs does not influence the gender outcome of a pregnancy when, in fact, it does. The probability of this error occurring is denoted by the Greek letter beta, β. The error with the greater consequence would be the Type I error since couples would use the It’s a Boy Genetic Labs product in hopes of increasing the chances of having a boy. 9.7 Red tide is a bloom of poison-producing algae—a few different species of a class of plankton called dinoflagellates. When the weather and water conditions cause these blooms, shellfish such as clams living in the area develop dangerous levels of a paralysis-inducing toxin. In Massachusetts, the Division of Marine Fisheries montors levels of the toxin in shellfish by regular sampling of shellfish along the coastline. If the mean level of toxin in clams exceeds 800 μg (micrograms) of toxin per kilogram of clam meat in any area, clam harvesting is banned there until the bloom is over and levels of toxin in clams subside. Describe both a Type I and a Type II error in this context, and state which error has the greater consequence. Example 9.8 A certain experimental drug claims a cure rate of at least 75 percent for males with a disease. Describe both the Type I and Type II errors in context. Which error is the more serious? Type I: A patient believes the cure rate for the drug is less than 75 percent when it actually is at least 75 percent. Type II: A patient believes the experimental drug has at least a 75 percent cure rate when it has a cure rate that
is less than 75 percent. In this scenario, the Type II error contains the more severe consequence. If a patient believes the drug works at least 75 percent of the time, this most likely will influence the patient’s (and doctor’s) choice about whether to use the drug as a treatment option. 9.8 Determine both Type I and Type II errors for the following scenario: Assume a null hypothesis, H0, that states the percentage of adults with jobs is at least 88 percent. Identify the Type I and Type II errors from these four possible choices. a. Not to reject the null hypothesis that the percentage of adults who have jobs is at least 88 percent when that percentage is actually less than 88 percent b. Not to reject the null hypothesis that the percentage of adults who have jobs is at least 88 percent when the percentage is actually at least 88 percent c. Reject the null hypothesis that the percentage of adults who have jobs is at least 88 percent when the percentage is actually at least 88 percent d. Reject the null hypothesis that the percentage of adults who have jobs is at least 88 percent when that percentage is actually less than 88 percent This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 529 9.3 \| Distribution Needed for Hypothesis Testing Earlier in the course, we discussed sampling distributions. Particular distributions are associated with hypothesis testing. Perform tests of a population mean using a normal distribution or a Student's t-distribution. (Remember, use a Student's t-distribution when the population standard deviation is unknown and the distribution of the sample mean is approximately normal.) We perform tests of a population proportion using a normal distribution (usually n is large). Assumptions When you perform a hypothesis test of a single population mean μ using a Student's t-distribution (often called a t-test), there are fundamental assumptions that need to be met in order for the test to work properly. Your data should be a simple random sample that comes from a population that is approximately normally distributed. You use the sample standard deviation to approximate the population standard deviation. Note that if the sample size is sufficiently large, a t-test will work even if the population is not approximately normally distributed. When you perform a hypothesis test of a single population mean μ using a normal distribution (often called a z-test), you take a simple random sample from the population.
The population you are testing is normally distributed or your sample size is sufficiently large. You know the value of the population standard deviation which, in reality, is rarely known. When you perform a hypothesis test of a single population proportion p, you take a simple random sample from the population. You must meet the conditions for a binomial distribution, which are the following: there are a certain number n of independent trials, the outcomes of any trial are success or failure, and each trial has the same probability of a success 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). Then the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = p and σ = pq n. Remember that q = 1 – p. 9.4 \| Rare Events, the Sample, and the Decision and Conclusion Establishing the type of distribution, sample size, and known or unknown standard deviation can help you figure out how to go about a hypothesis test. However, there are several other factors you should consider when working out a hypothesis test. Rare Events The thinking process in hypothesis testing can be summarized as follows: You want to test whether or not a particular property of the population is true. You make an assumption about the true population mean for numerical data or the true population proportion for categorical data. This assumption is the null hypothesis. Then you gather sample data that is representative of the population. From this sample data you compute the sample mean (or the sample proportion). If the value that you observe is very unlikely to occur (a rare event) if the null hypothesis is true, then you wonder why this is happening. A plausible explanation is that the null hypothesis is false. For example, Didi and Ali are at a birthday party of a very wealthy friend. They hurry to be first in line to grab a prize from a tall basket that they cannot see inside because they will be blindfolded. There are 200 plastic bubbles in the basket, and Didi and Ali have been told that there is only one with a $100 bill. Didi is the first person to reach into the basket and pull out a bubble. Her bubble contains a $100 bill. The probability of this happening is 1 200 = 0.005. Because this is so unlikely, Ali is hoping that what the two of them were told is
wrong and there are more $100 bills in the basket. A rare event has occurred (Didi getting the $100 bill) so Ali doubts the assumption about only one $100 bill being in the basket. Using the Sample to Test the Null Hypothesis After you collect data and obtain the test statistic (the sample mean, sample proportion, or other test statistic), you can determine the probability of obtaining that test statistic when the null hypothesis is true. This probability is called the p-value. When the p-value is very small, it means that the observed test statistic is very unlikely to happen if the null hypothesis is true. This gives significant evidence to suggest that the null hypothesis is false, and to reject it in favor of the alternative hypothesis. In practice, to reject the null hypothesis we want the p-value to be smaller than 0.05 (5 percent) or sometimes even smaller than 0.01 (1 percent). 530 Chapter 9 \| Hypothesis Testing with One Sample Example 9.9 Suppose a baker claims that his bread height is more than 15 cm, on average. Several of his customers do not believe him. To persuade his customers that he is right, the baker decides to do a hypothesis test. He bakes 10 loaves of bread. The mean height of the sample loaves is 17 cm. The baker knows from baking hundreds of loaves of bread that the standard deviation for the height is 0.5 cm and the distribution of heights is normal. The null hypothesis could be H0: μ ≤ 15. The alternate hypothesis is Ha: μ > 15. The words is more than translates as a ">" so "μ > 15" goes into the alternate hypothesis. The null hypothesis must contradict the alternate hypothesis. Since σ is known (σ = 0.5 cm), the distribution for the population is known to be normal with mean μ = 15 and standard deviation σ n = 0.16. = 0.5 10 Suppose the null hypothesis is true (which is that the mean height of the loaves is no more than 15 cm). Then is the mean height (17 cm) calculated from the sample unexpectedly large? The hypothesis test works by asking the question how unlikely the sample mean would be if the null hypothesis were true. The graph shows how far out the sample mean is on the normal curve. The p-value is the probability that, if we were to take other samples, any other sample mean would fall at least as far out as 17 cm. The p-
value, then, is the probability that a sample mean is the same or greater than 17 cm when the population mean is, in fact, 15 cm. We can calculate this probability using the normal distribution for means. In Figure 9.2, the p-value is the area under the normal curve to the right of 17. Using a normal distribution table or a calculator, we can compute that this probability is practically zero. Figure 9.2 p-value = P( x¯ > 17), which is approximately zero. Because the p-value is almost 0, we conclude that obtaining a sample height of 17 cm or higher from 10 loaves of bread is very unlikely if the true mean height is 15 cm. We reject the null hypothesis and conclude that there is sufficient evidence to claim that the true population mean height of the baker’s loaves of bread is higher than 15 cm. 9.9 A normal distribution has a standard deviation of 1. We want to verify a claim that the mean is greater than 12. A sample of 36 is taken with a sample mean of 12.5. H0: μ ≤ 12 Ha: μ > 12 The p-value is 0.0013. Draw a graph that shows the p-value. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 \| Hypothesis Testing with One Sample 531 Decision and Conclusion A systematic way to make a decision of whether to reject or not reject the null hypothesis is to compare the p-value and a preset or preconceived α, also called the level of significance of the test. A preset α is the probability of a Type I error (rejecting the null hypothesis when the null hypothesis is true). It may or may not be given to you at the beginning of the problem. When you make a decision to reject or not reject H0, do as follows: • • If p-value < α, reject H0. The results of the sample data are significant. There is sufficient evidence to conclude that H0 is an incorrect belief and that the alternative hypothesis, Ha, may be correct. If p-value ≥ α, do not reject H0. The results of the sample data are not significant.There is not sufficient evidence to conclude that the alternative hypothesis, Ha, may be correct. • When you do not reject H0, it does not mean that you should believe that H0 is true. It simply means that the sample
data have failed to provide sufficient evidence to cast serious doubt about the truthfulness of H0. Conclusion: After you make your decision, write a thoughtful conclusion about the hypotheses in terms of the given problem. Example 9.10 When using the p-value to evaluate a hypothesis test, you might find it useful to use the following mnemonic device: If the p-value is low, the null must go. If the p-value is high, the null must fly. This memory aid relates a p-value less than the established alpha (the p is low) as rejecting the null hypothesis and, likewise, relates a p-value higher than the established alpha (the p is high) as not rejecting the null hypothesis. Fill in the blanks. Reject the null hypothesis when ______________________________________. The results of the sample data _____________________________________. Do not reject the null hypothesis when __________________________________________. The results of the sample data ____________________________________________. Solution 9.10 Reject the null hypothesis when the p-value is less than the established alpha value. The results of the sample data support the alternative hypothesis. Do not reject the null hypothesis when the p-value is greater or equal to the established alpha value. The results of the sample data do not support the alternative hypothesis. 9.10 It’s a Boy Genetics Labs, a genetics company, claims their procedures improve the chances of a boy being born. The results for a test of a single population proportion are as follows: H0: p = 0.50, Ha: p > 0.50 α = 0.01 p-value = 0.025 Interpret the results and state a conclusion in simple, nontechnical terms. 532 Chapter 9 \| Hypothesis Testing with One Sample 9.5 \| Additional Information and Full Hypothesis Test Examples • In a hypothesis test problem, you may see words such as "the level of significance is 1 percent". The "1 percent" is the preconceived or preset α. • The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data. • If no level of significance is given, a common standard to use is α = 0.05. • When you calculate the p-value and draw the picture, the p-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed. • The alternative hypothesis,

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