Estimating the Binomial with the Normal Distribution We found earlier that various probability density functions are the limiting distributions of others; thus, we can estimate one with another under certain circumstances. We will find here that the normal distribution can be used to estimate a binomial process. The Poisson was used to estimate the binomial previously, and the binomial was used to estimate the hypergeometric distribution. In the case of the relationship between the hypergeometric distribution and the binomial, we had to recognize that a binomial process assumes that the probability of a success remains constant from trial to trial: a head on the last flip cannot have an effect on the probability of a head on the next flip. In the hypergeometric distribution this is the essence of the question because the experiment assumes that any "draw" is without replacement. If one draws without replacement, then all subsequent "draws" are conditional probabilities. We found that if the hypergeometric experiment draws only a small percentage of the total objects, then we can ignore the impact on the probability from draw to draw. Imagine that there are 312 cards in a deck comprised of 6 normal decks. If the experiment called for drawing only 10 cards, less than 5% of the total, than we will accept the binomial estimate of the probability, even though this is actually a hypergeometric distribution because the cards are presumably drawn without replacement. The Poisson likewise was considered an appropriate estimate of the binomial under certain circumstances. In Random Discrete Variables we found that if the number of trials of interest is large and the probability of success is small, such that μ = n p < 7 , the Poisson can be used to estimate the binomial with good results. Again, these rules of thumb do not in any way claim that the actual probability is what the estimate determines, only that the difference is in the third or fourth decimal and is thus de minimus . Here, again, we find that the normal distribution makes particularly accurate estimates of a binomial process under certain circumstances. is a frequency distribution of a binomial process for the experiment of flipping three coins where the random variable is the number of heads. The sample space is listed below the distribution. The experiment assumed that the probability of a success is 0.5; the probability of a failure, a tail, is thus also 0.5. In observing we are struck by the fact that the distribution is symmetrical. The root of this result is that the probabilities of success and failure are the same, 0.5. If the probability of success were smaller than 0.5, the distribution becomes skewed right. Indeed, as the probability of success diminishes, the degree of skewness increases. If the probability of success increases from 0.5, then the skewness increases in the lower tail, resulting in a left-skewed distribution. The reason the skewness of the binomial distribution is important is because if it is to be estimated with a normal distribution, then we need to recognize that the normal distribution is symmetrical. The closer the underlying binomial distribution is to being symmetrical, the better the estimate that is produced by the normal distribution. shows a symmetrical normal distribution transposed on a graph of a binomial distribution where p = 0.2 and n = 5. The discrepancy between the estimated probability using a normal distribution and the probability of the original binomial distribution is apparent. The criteria for using a normal distribution to estimate a binomial thus addresses this problem by requiring BOTH np AND n (1 − p ) are greater than five. Again, this is a rule of thumb, but is effective and results in acceptable estimates of the binomial probability. Imagine that it is known that only 10% of Australian Shepherd puppies are born with what is called "perfect symmetry" in their three colors, black, white, and copper. Perfect symmetry is defined as equal coverage on all parts of the dog when looked at in the face and measuring left and right down the centerline. A kennel would have a good reputation for breeding Australian Shepherds if they had a high percentage of dogs that met this criterion. During the past 5 years and out of the 100 dogs born to Dundee Kennels, 16 were born with this coloring characteristic. What is the probability that, in 100 births, more than 16 would have this characteristic? If we assume that one dog's coloring is independent of other dogs' coloring, a bit of a brave assumption, this becomes a classic binomial probability problem. The statement of the probability requested is 1 − [ p ( X = 0) + p ( X = 1) + p ( X = 2)+ … + p ( X = 16)]. This requires us to calculate 17 binomial formulas and add them together and then subtract from one to get the right hand part of the distribution. Alternatively, we can use the normal distribution to get an acceptable answer and in much less time. First, we need to check if the binomial distribution is symmetrical enough to use the normal distribution. We know that the binomial for this problem is skewed because the probability of success, 0.1, is not the same as the probability of failure, 0.9. Nevertheless, both np = 10 and n ( 1 − p ) = 90 are larger than 5, the cutoff for using the normal distribution to estimate the binomial. below shows the binomial distribution and marks the area we wish to know. The mean of the binomial, 10, is also marked, and the standard deviation is written on the side of the graph: σ = n p q = 3. The area under the distribution from zero to 16 is the probability requested, and has been shaded in. Below the binomial distribution is a normal distribution to be used to estimate this probability. That probability has also been shaded. Standardizing from the binomial to the normal distribution as done in the past shows where we are asking for the probability from 16 to positive infinity, or 100 in this case. We need to calculate the number of standard deviations 16 is away from the mean: 10. Z = x − μ σ = 16 − 10 3 = 2 We are asking for the probability beyond two standard deviations, a very unlikely event. We look up two standard deviations in the standard normal table and find the area from zero to two standard deviations is 0.4772. We are interested in the tail, however, so we subtract 0.4772 from 0.5 and thus find the area in the tail. Our conclusion is the probability of a kennel having 16 dogs with "perfect symmetry" is 0.0228. Dundee Kennels has an extraordinary record in this regard. Mathematically, we write this as: 1 − [ p ( X = 0 ) + p ( X = 1 ) + p ( X = 2 ) + … + p ( X = 16 ) ] = p ( X > 16 ) = p ( Z > 2 ) = 0.0228 Try It Let’s assume that only 10% of the wooden furniture made by a carpenter is made to the exact dimensions in the blueprint. During the last 3 years, the carpenter has made 100 pieces of wooden furniture, and 16 were found to be made to the exact dimensions in the blueprint. What is the probability that, in 100 wooden furniture items, more than 16 would be made to the exact dimensions of the blueprint? The required probability is 1 - [ p ( X = 0 ) + p ( X = 1 ) + p ( X = 2 ) + . . . + p ( X = 16 ) ] . The mean of the binomial is 10, since np = 10. The standard deviation is obtained as σ = n p q = 3 . The number of standard deviations 16 is away from 10 is obtained as follows: Z = x - μ σ = 16 - 10 3 = 2 From the standard normal table, the probability for Z = 2 is obtained as 0.4772. Since the required probability is only above 16, the required probability is obtained as 0.5 – 0.4772 = 0.0228. Chapter Review The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bell-shaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one. The parameters of the normal are the mean µ and the standard deviation σ . A special normal distribution, called the standard normal distribution is the distribution of z -scores. Its mean is zero, and its standard deviation is one. Formula Review Normal Distribution: X ~ N ( µ , σ ) where µ is the mean and σ is the standard deviation. Standard Normal Distribution: Z ~ N (0, 1). How would you represent the area to the left of one in a probability statement? P ( x < 1) What is the area to the right of one? Is P ( x < 1) equal to P ( x ≤ 1)? Why? Yes, because they are the same in a continuous distribution: P ( x = 1) = 0 How would you represent the area to the left of three in a probability statement? What is the area to the right of three? 1 – P ( x < 3) or P ( x > 3) If the area to the left of x in a normal distribution is 0.123, what is the area to the right of x ? If the area to the right of x in a normal distribution is 0.543, what is the area to the left of x ? 1 – 0.543 = 0.457 Use the following information to answer the next four exercises: X ~ N (54, 8) Find the probability that x > 56. Find the probability that x < 30. 0.0013 X ~ N (6, 2) Find the probability that x is between three and nine. X ~ N (–3, 4) Find the probability that x is between one and four. 0.1186 X ~ N (4, 5) Find the maximum of x in the bottom quartile. Use the following information to answer the next three exercise: The life of wearable fitness devices is normally distributed with a mean of 4.1 years and a standard deviation of 1.3 years. A wearable fitness device is guaranteed for three years. We are interested in the length of time a wearable fitness device lasts. Find the probability that a wearable fitness device will break down during the guarantee period. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability. P (0 < x < ____________) = ___________ (Use zero for the minimum value of x .) Answers may vary. 3, 0.1979 Find the probability that a wearable fitness device will last between 2.8 and six years. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability. P (__________ < x < __________) = __________ An experiment with a probability of success given as 0.40 is repeated 100 times. Use the normal distribution to approximate the binomial distribution, and find the probability the experiment will have at least 45 successes. 0.154 An experiment with a probability of success given as 0.30 is repeated 90 times. Use the normal distribution to approximate the binomial distribution, and find the probability the experiment will have at least 22 successes. 0.874 An experiment with a probability of success given as 0.40 is repeated 100 times. Use the normal distribution to approximate the binomial distribution, and find the probability the experiment will have from 35 to 45 successes. 0.693 An experiment with a probability of success given as 0.30 is repeated 90 times. Use the normal distribution to approximate the binomial distribution, and find the probability the experiment will have from 26 to 30 successes. 0.346 An experiment with a probability of success given as 0.40 is repeated 100 times. Use the normal distribution to approximate the binomial distribution, and find the probability the experiment will have at most 34 successes. 0.110 An experiment with a probability of success given as 0.30 is repeated 90 times. Use the normal distribution to approximate the binomial distribution, and find the probability the experiment will have at most 34 successes. 0.946 A multiple choice test has a probability any question will be guesses correctly of 0.25. There are 100 questions, and a student guesses at all of them. Use the normal distribution to approximate the binomial distribution, and determine the probability at least 30, but no more than 32, questions will be guessed correctly. 0.071 A multiple choice test has a probability any question will be guesses correctly of 0.25. There are 100 questions, and a student guesses at all of them. Use the normal distribution to approximate the binomial distribution, and determine the probability at least 24, but no more than 28, questions will be guessed correctly. 0.347 Homework Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. What is the probability of spending more than two days in recovery? 0.0580 0.8447 0.0553 0.9420 Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. Based upon the given information and numerically justified, would you be surprised if it took less than one minute to find a parking space? Yes No Unable to determine Find the probability that it takes at least eight minutes to find a parking space. 0.0001 0.9270 0.1862 0.0668 d Seventy percent of the time, it takes more than how many minutes to find a parking space? 1.24 2.41 3.95 6.05 According to a study done by De Anza students, the height for Asian adult males is normally distributed with an average of 66 inches and a standard deviation of 2.5 inches. Suppose one Asian adult male is randomly chosen. Let X = height of the individual. X ~ _____(_____,_____) Find the probability that the person is between 65 and 69 inches. Include a sketch of the graph, and write a probability statement. Would you expect to meet many Asian adult males over 72 inches? Explain why or why not, and justify your answer numerically. The middle 40% of heights fall between what two values? Sketch the graph, and write the probability statement. X ~ N (66, 2.5) 0.5404 No, the probability that an Asian male is over 72 inches tall is 0.0082 IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual. X ~ _____(_____,_____) Find the probability that the person has an IQ greater than 120. Include a sketch of the graph, and write a probability statement. MENSA is an organization whose members have the top 2% of all IQs. Find the minimum IQ needed to qualify for the MENSA organization. Sketch the graph, and write the probability statement. The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of 10. Suppose that one individual is randomly chosen. Let X = percent of fat calories. X ~ _____(_____,_____) Find the probability that the percent of fat calories a person consumes is more than 40. Graph the situation. Shade in the area to be determined. Find the maximum number for the lower quarter of percent of fat calories. Sketch the graph and write the probability statement. X ~ N (36, 10) The probability that a person consumes more than 40% of their calories as fat is 0.3446. Approximately 25% of people consume less than 29.26% of their calories as fat. 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. If X = distance in feet for a fly ball, then X ~ _____(_____,_____) If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 220 feet? Sketch the graph. Scale the horizontal axis X . Shade the region corresponding to the probability. Find the probability. In China, four-year-olds average three hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.5 hours and the amount of time spent alone is normally distributed. We randomly select one Chinese four-year-old living in a rural area. We are interested in the amount of time the child spends alone per day. In words, define the random variable X . X ~ _____(_____,_____) Find the probability that the child spends less than one hour per day unsupervised. Sketch the graph, and write the probability statement. What percent of the children spend over ten hours per day unsupervised? Seventy percent of the children spend at least how long per day unsupervised? X = number of hours that a Chinese four-year-old in a rural area is unsupervised during the day. X ~ N (3, 1.5) The probability that the child spends less than one hour a day unsupervised is 0.0918. The probability that a child spends over ten hours a day unsupervised is less than 0.0001. 2.21 hours In a certain presidential election, Alaska’s 40 election districts averaged 1,956.8 votes per district for Candidate A. The standard deviation was 572.3. (There are only 40 election districts in Alaska.) The distribution of the votes per district for Candidate A was bell-shaped. Let X = number of votes for Candidate A for an election district. State the approximate distribution of X . Is 1,956.8 a population mean or a sample mean? How do you know? Find the probability that a randomly selected district had fewer than 1,600 votes for Candidate A. Sketch the graph and write the probability statement. Find the probability that a randomly selected district had between 1,800 and 2,000 votes for Candidate A. Find the third quartile for votes for Candidate A. Suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 21 days and a standard deviation of seven days. In words, define the random variable X . X ~ _____(_____,_____) If one of the trials is randomly chosen, find the probability that it lasted at least 24 days. Sketch the graph and write the probability statement. Sixty percent of all trials of this type are completed within how many days? X = the distribution of the number of days a particular type of criminal trial will take X ~ N (21, 7) The probability that a randomly selected trial will last more than 24 days is 0.3336. 22.77 A motorcycle racer averages 129.71 seconds per 2.5 mile lap (in a seven-lap race) with a standard deviation of 2.28 seconds. The distribution of their race times is normally distributed. We are interested in one of their randomly selected laps. In words, define the random variable X . X ~ _____(_____,_____) Find the percent of the racer's laps that are completed in less than 130 seconds. The fastest 3% of the racer's laps are under _____. The middle 80% of the racer's laps are from _______ seconds to _______ seconds. Thuy Dau, Ngoc Bui, Sam Su, and Lan Voung conducted a survey as to how long customers at Lucky claimed to wait in the checkout line until their turn. Let X = time in line. displays the ordered real data (in minutes): 0.50 4.25 5 6 7.25 1.75 4.25 5.25 6 7.25 2 4.25 5.25 6.25 7.25 2.25 4.25 5.5 6.25 7.75 2.25 4.5 5.5 6.5 8 2.5 4.75 5.5 6.5 8.25 2.75 4.75 5.75 6.5 9.5 3.25 4.75 5.75 6.75 9.5 3.75 5 6 6.75 9.75 3.75 5 6 6.75 10.75 Calculate the sample mean and the sample standard deviation. Construct a histogram. Draw a smooth curve through the midpoints of the tops of the bars. In words, describe the shape of your histogram and smooth curve. Let the sample mean approximate μ and the sample standard deviation approximate σ . The distribution of X can then be approximated by X ~ _____(_____,_____) Use the distribution in part e to calculate the probability that a person will wait fewer than 6.1 minutes. Determine the cumulative relative frequency for waiting less than 6.1 minutes. Why aren’t the answers to part f and part g exactly the same? Why are the answers to part f and part g as close as they are? If only ten customers has been surveyed rather than 50, do you think the answers to part f and part g would have been closer together or farther apart? Explain your conclusion. mean = 5.51, s = 2.15 Answers may vary. Answers may vary. Answers may vary. X ~ N (5.51, 2.15) 0.6029 The cumulative frequency for less than 6.1 minutes is 0.64. The answers to part f and part g are not exactly the same, because the normal distribution is only an approximation to the real one. The answers to part f and part g are close, because a normal distribution is an excellent approximation when the sample size is greater than 30. The approximation would have been less accurate, because the smaller sample size means that the data does not fit normal curve as well. Suppose that Ric and Anita attend different colleges. Ric’s GPA is the same as the average GPA at their school. Anita’s GPA is 0.70 standard deviations above her school average. In complete sentences, explain why each of the following statements may be false. Ric’s actual GPA is lower than Anita’s actual GPA. Ric is not passing because their z -score is zero. Anita is in the 70 th percentile of students at her college. An expert witness for a paternity lawsuit testifies that the length of a pregnancy is normally distributed with a mean of 280 days and a standard deviation of 13 days. The person who is being sued for lack of child support was out of the country from 240 to 306 days before the birth of the child, so the pregnancy would have been less than 240 days or more than 306 days long if he was the father. The birth was uncomplicated, and the child needed no medical intervention. What is the probability that he was NOT the father? What is the probability that he could be the father? Calculate the z -scores first, and then use those to calculate the probability. An automotive factory can build an average of 6,000 cars and trucks a week. Generally, 10% of the cars were defective coming off the assembly line. Suppose we draw a random sample of n = 100 cars. Let X represent the number of defective cars in the sample. What can we say about X in regard to the 68-95-99.7 empirical rule (one standard deviation, two standard deviations and three standard deviations from the mean are being referred to)? Assume a normal distribution for the defective cars in the sample. n = 100; p = 0.1; q = 0.9 μ = np = (100)(0.10) = 10 σ = n p q = (100)(0 .1)(0 .9) = 3 z = ± 1 : x 1 = µ + zσ = 10 + 1(3) = 13 and x 2 = µ – zσ = 10 – 1(3) = 7.68% of the defective cars will fall between seven and 13. z = ± 2 : x 1 = µ + zσ = 10 + 2(3) = 16 and x 2 = µ – zσ = 10 – 2(3) = 4. 95 % of the defective cars will fall between four and 16 z = ± 3 : x 1 = µ + zσ = 10 + 3(3) = 19 and x 2 = µ – zσ = 10 – 3(3) = 1. 99.7% of the defective cars will fall between one and 19. We flip a coin 100 times ( n = 100) and note that it only comes up heads 20% ( p = 0.20) of the time. The mean and standard deviation for the number of times the coin lands on heads is µ = 20 and σ = 4 (verify the mean and standard deviation). Solve the following: There is about a 68% chance that the number of heads will be somewhere between ___ and ___. There is about a ____chance that the number of heads will be somewhere between 12 and 28. There is about a ____ chance that the number of heads will be somewhere between eight and 32. A $1 scratch off lotto ticket will be a winner one out of five times. Out of a shipment of n = 190 lotto tickets, find the probability for the lotto tickets that there are somewhere between 34 and 54 prizes. somewhere between 54 and 64 prizes. more than 64 prizes. n = 190; p = 1 5 = 0.2; q = 0.8 μ = np = (190)(0.2) = 38 σ = n p q = (190)(0 .2)(0 .8) = 5.5136 For this problem: P (34 < x < 54) = 0.7641 For this problem: P (54 < x < 64) = 0.0018 For this problem: P ( x > 64) = 0.0000012 (approximately 0) On average, 28% of 18 to 34 year olds check social media before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of 5%. Find the probability that the percent of 18 to 34 year olds who check social media before getting out of bed in the morning is at least 30%. Find the 95th percentile, and express it in a sentence. A hospital has 49 births in a year. It is considered equally likely that a birth be a boy as it is the birth be a girl. What is the mean? What is the standard deviation? Can this binomial distribution be approximated with a normal distribution? If so, use the normal distribution to find the probability that at least 23 of the 49 births were boys. 24.5 3.5 Yes 0.67 Historically, a final exam in a course is passed with a probability of 0.9. The exam is given to a group of 70 students. What is the mean of the binomial distribution? What is the standard deviation? Can this binomial distribution be approximate with a normal distribution? If so, use the normal distribution to find the probability that at least 60 of the students pass the exam? 63 2.5 Yes 0.88 A tree in an orchard has 200 oranges. Of the oranges, 40 are not ripe. Use the normal distribution to approximate the binomial distribution, and determine the probability a box containing 35 oranges has at most two oranges that are not ripe. 0.02 In a large city one in ten fire hydrants are in need of repair. If a crew examines 100 fire hydrants in a week, what is the probability they will find nine of fewer fire hydrants that need repair? Use the normal distribution to approximate the binomial distribution. 0.37 On an assembly line it is determined 85% of the assembled products have no defects. If one day 50 items are assembled, what is the probability at least 4 and no more than 8 are defective. Use the normal distribution to approximate the binomial distribution. 0.50

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Introduction If you want to figure out the distribution of the change people carry in their pockets, using the Central Limit Theorem and assuming your sample is large enough, you will find that the distribution is the normal probability density function. (credit: modification of work “american coins” by Paula R. Lively/ Flickr, CC BY 2.0) Why are we so concerned with means? Two reasons are: they give us a middle ground for comparison, and they are easy to calculate. In this chapter, you will study means and the Central Limit Theorem . The Central Limit Theorem is one of the most powerful and useful ideas in all of statistics. The Central Limit Theorem is a theorem which means that it is NOT a theory or just somebody's idea of the way things work. As a theorem it ranks with the Pythagorean Theorem, or the theorem that tells us that the sum of the angles of a triangle must add to 180. These are facts of the ways of the world rigorously demonstrated with mathematical precision and logic. As we will see this powerful theorem will determine just what we can, and cannot say, in inferential statistics. The Central Limit Theorem is concerned with drawing finite samples of size n from a population with a known mean, μ , and a known standard deviation, σ . The conclusion is that if we collect samples of size n with a "large enough n ," calculate each sample's mean, and create a histogram (distribution) of those means, then the resulting distribution will tend to have an approximate normal distribution. The astounding result is that it does not matter what the distribution of the original population is, or whether you even need to know it. The important fact is that the distribution of sample means tend to follow the normal distribution. The size of the sample, n , that is required in order to be "large enough" depends on the original population from which the samples are drawn (the sample size should be at least 30 or the data should come from a normal distribution). If the original population is far from normal, then more observations are needed for the sample means. Sampling is done randomly and with replacement in the theoretical model. 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.

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The Central Limit Theorem for Sample Means The sampling distribution is a theoretical distribution. It is created by taking many samples of size n from a population. Each sample mean is then treated like a single observation of this new distribution, the sampling distribution. The genius of thinking this way is that it recognizes that when we sample we are creating an observation and that observation must come from some particular distribution. The Central Limit Theorem answers the question: from what distribution did a sample mean come? If this is discovered, then we can treat a sample mean just like any other observation and calculate probabilities about what values it might take on. We have effectively moved from the world of statistics where we know only what we have from the sample, to the world of probability where we know the distribution from which the sample mean came and the parameters of that distribution. The reasons that one samples a population are obvious. The time and expense of checking every invoice to determine its validity or every shipment to see if it contains all the items may well exceed the cost of errors in billing or shipping. For some products, sampling would require destroying them, called destructive sampling. One such example is measuring the ability of a metal to withstand saltwater corrosion for parts on ocean going vessels. Sampling thus raises an important question; just which sample was drawn. Even if the sample were randomly drawn, there are theoretically an almost infinite number of samples. With just 100 items, there are more than 75 million unique samples of size five that can be drawn. If six are in the sample, the number of possible samples increases to just more than one billion. Of the 75 million possible samples, then, which one did you get? If there is variation in the items to be sampled, there will be variation in the samples. One could draw an "unlucky" sample and make very wrong conclusions concerning the population. This recognition that any sample we draw is really only one from a distribution of samples provides us with what is probably the single most important theorem is statistics: the Central Limit Theorem . Without the Central Limit Theorem it would be impossible to proceed to inferential statistics from simple probability theory. In its most basic form, the Central Limit Theorem states that regardless of the underlying probability density function of the population data, the theoretical distribution of the means of samples from the population will be normally distributed. In essence, this says that the mean of a sample should be treated like an observation drawn from a normal distribution. The Central Limit Theorem only holds if the sample size is "large enough" which has been shown to be only 30 observations or more. graphically displays this very important proposition. Notice that the horizontal axis in the top panel is labeled X. These are the individual observations of the population. This is the unknown distribution of the population values. The graph is purposefully drawn all squiggly to show that it does not matter just how odd ball it really is. Remember, we will never know what this distribution looks like, or its mean or standard deviation for that matter. The horizontal axis in the bottom panel is labeled X – 's. This is the theoretical distribution called the sampling distribution of the means. Each observation on this distribution is a sample mean. All these sample means were calculated from individual samples with the same sample size. The theoretical sampling distribution contains all of the sample mean values from all the possible samples that could have been taken from the population. Of course, no one would ever actually take all of these samples, but if they did this is how they would look. And the Central Limit Theorem says that they will be normally distributed. The Central Limit Theorem goes even further and tells us the mean and standard deviation of this theoretical distribution. Parameter Population distribution Sample Sampling distribution of X – 's Mean μ X ¯ μ x ¯ and E ( μ x ¯ ) = μ Standard deviation σ s σ x ¯ = σ n The practical significance of The Central Limit Theorem is that now we can compute probabilities for drawing a sample mean, X – , in just the same way as we did for drawing specific observations, X's, when we knew the population mean and standard deviation and that the population data were normally distributed.. The standardizing formula has to be amended to recognize that the mean and standard deviation of the sampling distribution, sometimes, called the standard error of the mean, are different from those of the population distribution, but otherwise nothing has changed. The new standardizing formula is Z = X – − μ X – σ X – = X – - μ σ n Notice that µ X – in the first formula has been changed to simply µ in the second version. The reason is that mathematically it can be shown that the expected value of µ X – is equal to µ. This was stated in above. Mathematically, the E(x) symbol read the “expected value of x”. This formula will be used in the next unit to provide estimates of the unknown population parameter μ. References Baran, Daya. “20 Percent of Americans Have Never Used Email.”WebGuild, 2010. Available online at http://www.webguild.org/20080519/20-percent-of-americans-have-never-used-email (accessed May 17, 2013). Data from The Flurry Blog, 2013. Available online at http://blog.flurry.com (accessed May 17, 2013). Data from the United States Department of Agriculture. Chapter Review In a population whose distribution may be known or unknown, if the size ( n ) of samples is sufficiently large, the distribution of the sample means will be approximately normal. The mean of the sample means will equal the population mean. The standard deviation of the distribution of the sample means, called the standard error of the mean, is equal to the population standard deviation divided by the square root of the sample size ( n ). Formula Review The Central Limit Theorem for Sample Means: X – ~ N ( μ x – , σ n ) Z = X – − μ X – σ X – = X – - μ σ / n The Mean X – : μ x ¯ Central Limit Theorem for Sample Means z-score z = x – − μ x ¯ ( σ n ) Standard Error of the Mean (Standard Deviation ( X – )): σ n Finite Population Correction Factor for the sampling distribution of means: Z = x ¯ − μ σ n · N − n N − 1 Finite Population Correction Factor for the sampling distribution of proportions: σ p' = p ( 1 − p ) n × N − n N − 1 Homework Previously, De Anza statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.88. Suppose that we randomly pick 25 daytime statistics students. In words, Χ = ____________ Χ ~ _____(_____,_____) In words, X – = ____________ X – ~ ______ (______, ______) Find the probability that an individual had between $0.80 and $1.00. Graph the situation, and shade in the area to be determined. Find the probability that the average of the 25 students was between $0.80 and $1.00. Graph the situation, and shade in the area to be determined. Explain why there is a difference in part e and part f. Χ = amount of change a student carries E x p ( 1 0 . 88 ) o r Χ ~ E x p ( 1 . 1364 ) X – = average amount of change carried by a sample of 25 students. X – ~ N (0.88, 0.176) 0.0819 0.4276 The distributions are different. Part e asks about the probability for an individual and part f asks about the probability of the mean value for a sample of 25. 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. If X – = average distance in feet for 49 fly balls, then X – ~ _______(_______,_______) What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for X – . Shade the region corresponding to the probability. Find the probability. Find the 80 th percentile of the distribution of the average of 49 fly balls. 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. In words, Χ = _____________ In words, X – = _____________ X – ~ _____(_____,_____) 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. Would you be surprised if one taxpayer finished their Form 1040 in more than 12 hours? In a complete sentence, explain why. length of time for an individual to complete IRS form 1040, in hours. mean length of time for a sample of 36 taxpayers to complete IRS form 1040, in hours. N ( 10 .53, 1 3 ) Yes. I would be surprised, because the probability is almost 0. No. I would not be totally surprised because the probability is 0.2312 Suppose that a category of world-class runners are known to run a marathon (26 miles) in an average of 145 minutes with a standard deviation of 14 minutes. Consider 49 of the races. Let X – the average of the 49 races. X – ~ _____(_____,_____) Find the probability that the runner will average between 142 and 146 minutes in these 49 marathons. Find the 80 th percentile for the average of these 49 marathons. Find the median of the average running times. The length of songs in a collector’s Apple Music album collection is uniformly distributed from two to 3.5 minutes. Suppose we randomly pick five albums from the collection. There are a total of 43 songs on the five albums. In words, Χ = _________ Χ ~ _____________ In words, X – = _____________ X – ~ _____(_____,_____) Find the first quartile for the average song length. The IQR(interquartile range) for the average song length is from _______–_______. the length of a song, in minutes, in the collection U (2, 3.5) the average length, in minutes, of the songs from a sample of five albums from the collection N (2.75, 0.066) 2.74 minutes 0.03 minutes 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 = _____________ In words, X – = _____________ X – ~ _____(_____,_____) The IQR for X – is from _______ acres to _______ acres. Determine which of the following are true and which are false. Then, in complete sentences, justify your answers. When the sample size is large, the mean of X – is approximately equal to the mean of Χ . When the sample size is large, X – is approximately normally distributed. When the sample size is large, the standard deviation of X – is approximately the same as the standard deviation of Χ . True. The mean of a sampling distribution of the means is approximately the mean of the data distribution. True. According to the Central Limit Theorem, the larger the sample, the closer the sampling distribution of the means becomes normal. False. The standard deviation of the sample distribution of the means will decrease as the sample size increases; however, the standard deviation of the sample distribution of the means will not equal the standard deviation of X. The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of about ten. Suppose that 16 individuals are randomly chosen. Let X – = average percent of fat calories. X – ~ ______(______, ______) For the group of 16, find the probability that the average percent of fat calories consumed is more than five. Graph the situation and shade in the area to be determined. Find the first quartile for the average percent of fat calories. The distribution of income in some developing countries is considered wedge shaped (many low income people, very few middle income people, and even fewer high income 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. In words, Χ = _____________ In words, X – = _____________ X – ~ _____(_____,_____) How is it possible for the standard deviation to be greater than the average? Why is it more likely that the average of the 1,000 residents will be from $2,000 to $2,100 than from $2,100 to $2,200? X = the yearly income of someone in a developing country the average salary from samples of 1,000 residents of a developing country X – ∼ N ( 2000, 8000 1000 ) Very wide differences in data values can have averages smaller than standard deviations. The distribution of the sample mean will have higher probabilities closer to the population mean. P (2000 < X – < 2100) = 0.1537 P (2100 < X – < 2200) = 0.1317 Which of the following is NOT TRUE about the distribution for averages? The mean, median, and mode are equal. The area under the curve is one. The curve never touches the x -axis. The curve is skewed to the right. 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: X – ~ N (4.59, 0.10) X – ~ N ( 4 .59, 0.10 16 ) X – ~ N ( 4 .59, 16 0.10 ) X – ~ N ( 4 .59, 16 0.10 ) b 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 with known mean μ and known standard deviation, σ , we are sampling with size n , and we are interested in two new RVs: the sample mean, X – . If the size ( n ) of the sample is sufficiently large, then X – ~ N ( μ , σ n ). If the size ( n ) of the sample is sufficiently large, then the distribution of the sample means will approximate a normal distributions regardless of the shape of the population. The mean of the sample means will equal the population mean. The standard deviation of the distribution of the sample means, σ n , is called the standard error of the mean. Standard Error of the Mean the standard deviation of the distribution of the sample means, or σ n .

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Using the Central Limit Theorem The Law of Large Numbers , along with the Central Limit Theorem, provides another critical piece of information to allow us to engage in inferential statistics. In short, the Law of Large Numbers proves that the expected value of the sampling distribution of the sample mean is the population mean: E ( μ x ¯ ) = μ The proof is through the concept of large numbers. Suppose you were to take a sample and calculate a sample mean. Then you take another sample, combine it with the previous sample, and calculate the sample mean of the combined sample. Then you repeat this process over and over, creating bigger and bigger samples and calculating a sample mean each time along the way. The sample means from larger and larger samples will get closer and closer to the population mean, μ. shows the running average as more sample means are added and then averaged. The proof of the Law of Large Numbers mathematically was perfected during a period of 20 years and was presented by Jacob Bernoulli in 1713. Stated mathematically lim n → ∞ P ( | X ¯ n - μ | ) > ε = 0 or alternatively presented as P lim n → ∞ | X ¯ n = μ | < ε = 1 where X ¯ n is the running average as additional sample means are added to the previous sample means. In summary: There are three critical mathematical conclusions that flow from the Central Limit Theorem and the application of the Law of Large Numbers. By the Central Limit Theorem, for large enough sample sizes, the sampling distribution of sample means tends to be normally distributed regardless of the underlying distribution of the population data. As the sample size, n, gets larger and larger, the sampling distribution standard deviation gets smaller. Remember that the standard deviation for the sampling distribution of X ¯ n is σ n . The sample mean, x ¯ , is more likely to be closer to μ as n increases. By the Law of Large Numbers, the expected value of the sampling distribution of the sample mean is the population mean. Law of Large Numbers The law of large numbers says that if you take samples of larger and larger size from any population, then the mean of the sampling distribution, μ x – tends to get closer and closer to the true population mean, μ . From the Central Limit Theorem, we know that as n gets larger and larger, the sample means follow a normal distribution. The larger n gets, the smaller the standard deviation of the sampling distribution gets. (Remember that the standard deviation for the sampling distribution of X – is σ n .) This means that the sample mean x – must be closer to the population mean μ as n increases. We can say that μ is the value that the sample means approach as n gets larger. The Central Limit Theorem illustrates the law of large numbers. Examples of the Central Limit Theorem This concept is so important and plays such a critical role in what follows it deserves to be developed further. Indeed, there are two critical issues that flow from the Central Limit Theorem and the application of the Law of Large numbers to it. These are The probability density function of the sampling distribution of means is normally distributed regardless of the underlying distribution of the population observations and standard deviation of the sampling distribution decreases as the size of the samples that were used to calculate the means for the sampling distribution increases. Taking these in order. It would seem counterintuitive that the population may have any distribution and the distribution of means coming from it would be normally distributed. With the use of computers, experiments can be simulated that show the process by which the sampling distribution changes as the sample size is increased. These simulations show visually the results of the mathematical proof of the Central Limit Theorem. Here are three examples of very different population distributions and the evolution of the sampling distribution to a normal distribution as the sample size increases. The top panel in these cases represents the histogram for the original data. The three panels show the histograms for 1,000 randomly drawn samples for different sample sizes: n=10, n= 25 and n=50. As the sample size increases, and the number of samples taken remains constant, the distribution of the 1,000 sample means becomes closer to the smooth line that represents the normal distribution. is for a normal distribution of individual observations and we would expect the sampling distribution to converge on the normal quickly. The results show this and show that even at a very small sample size the distribution is close to the normal distribution. is a uniform distribution which, a bit amazingly, quickly approached the normal distribution even with only a sample of 10. is a skewed distribution. This last one could be an exponential, geometric, or binomial with a small probability of success creating the skew in the distribution. For skewed distributions our intuition would say that this will take larger sample sizes to move to a normal distribution and indeed that is what we observe from the simulation. Nevertheless, at a sample size of 50, not considered a very large sample, the distribution of sample means has very decidedly gained the shape of the normal distribution. The Central Limit Theorem provides more than the proof that the sampling distribution of means is normally distributed. It also provides us with the mean and standard deviation of this distribution. Further, as discussed above, the expected value of the mean, μ x – , is equal to the mean of the population of the original data which is what we are interested in estimating from the sample we took. We have already inserted this conclusion of the Central Limit Theorem into the formula we use for standardizing from the sampling distribution to the standard normal distribution. And finally, the Central Limit Theorem has also provided the standard deviation of the sampling distribution, σ x – = σ n , and this is critical to have to calculate probabilities of values of the new random variable, x ¯ . shows a sampling distribution. The mean has been marked on the horizontal axis of the x ¯ 's and the standard deviation has been written to the right above the distribution. Notice that the standard deviation of the sampling distribution is the original standard deviation of the population, divided by the sample size. We have already seen that as the sample size increases the sampling distribution becomes closer and closer to the normal distribution. As this happens, the standard deviation of the sampling distribution changes in another way; the standard deviation decreases as n increases. At very large n, the standard deviation of the sampling distribution becomes very small and at infinity it collapses on top of the population mean. This is what it means that the expected value of µ x – is the population mean, µ. At non-extreme values of n,this relationship between the standard deviation of the sampling distribution and the sample size plays a very important part in our ability to estimate the parameters we are interested in. shows three sampling distributions. The only change that was made is the sample size that was used to get the sample means for each distribution. As the sample size increases, n goes from 10 to 30 to 50, the standard deviations of the respective sampling distributions decrease because the sample size is in the denominator of the standard deviations of the sampling distributions. The implications for this are very important. shows the effect of the sample size on the confidence we will have in our estimates. These are two sampling distributions from the same population. One sampling distribution was created with samples of size 10 and the other with samples of size 50. All other things constant, the sampling distribution with sample size 50 has a smaller standard deviation that causes the graph to be higher and narrower. The important effect of this is that for the same probability of one standard deviation from the mean, this distribution covers much less of a range of possible values than the other distribution. One standard deviation is marked on the X ¯ axis for each distribution. This is shown by the two arrows that are plus or minus one standard deviation for each distribution. If the probability that the true mean is one standard deviation away from the mean, then for the sampling distribution with the smaller sample size, the possible range of values is much greater. A simple question is, would you rather have a sample mean from the narrow, tight distribution, or the flat, wide distribution as the estimate of the population mean? Your answer tells us why people intuitively will always choose data from a large sample rather than a small sample. The sample mean they are getting is coming from a more compact distribution. This concept will be the foundation for what will be called level of confidence in the next unit. Chapter Review 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 . Use the following information to answer the next ten exercises: A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of 100 weights is taken. What is the distribution for the weights of one 25-pound lifting weight? What is the mean and standard deviation? What is the distribution for the mean weight of 100 25-pound lifting weights? Find the probability that the mean actual weight for the 100 weights is less than 24.9. U (24, 26), 25, 0.5774 N (25, 0.0577) 0.0416 Draw the graph from Find the probability that the mean actual weight for the 100 weights is greater than 25.2. 0.0003 Draw the graph from Find the 90 th percentile for the mean weight for the 100 weights. 25.07 Draw the graph from What is the distribution for the sum of the weights of 100 25-pound lifting weights? Find P ( Σx < 2,450). N (2,500, 5.7735) 0 Draw the graph from Find the 90 th percentile for the total weight of the 100 weights. 2,507.40 Draw the graph from Use the following information to answer the next five exercises: The length of time a particular smartphone's battery lasts follows an exponential distribution with a mean of ten months. A sample of 64 of these smartphones is taken. What is the standard deviation? What is the parameter m ? 10 1 10 What is the distribution for the length of time one battery lasts? What is the distribution for the mean length of time 64 batteries last? N ( 10, 10 8 ) What is the distribution for the total length of time 64 batteries last? Find the probability that the sample mean is between seven and 11. 0.7799 Find the 80 th percentile for the total length of time 64 batteries last. Find the IQR for the mean amount of time 64 batteries last. 1.69 Find the middle 80% for the total amount of time 64 batteries last. Use the following information to answer the next eight exercises: A uniform distribution has a minimum of six and a maximum of ten. A sample of 50 is taken. Find P ( Σx > 420). 0.0072 Find the 90 th percentile for the sums. Find the 15 th percentile for the sums. 391.54 Find the first quartile for the sums. Find the third quartile for the sums. 405.51 Find the 80 th percentile for the sums. A population has a mean of 25 and a standard deviation of 2. If it is sampled repeatedly with samples of size 49, what is the mean and standard deviation of the sample means? Mean = 25, standard deviation = 2/7 A population has a mean of 48 and a standard deviation of 5. If it is sampled repeatedly with samples of size 36, what is the mean and standard deviation of the sample means? Mean = 48, standard deviation = 5/6 A population has a mean of 90 and a standard deviation of 6. If it is sampled repeatedly with samples of size 64, what is the mean and standard deviation of the sample means? Mean = 90, standard deviation = 3/4 A population has a mean of 120 and a standard deviation of 2.4. If it is sampled repeatedly with samples of size 40, what is the mean and standard deviation of the sample means? Mean = 120, standard deviation = 0.38 A population has a mean of 17 and a standard deviation of 1.2. If it is sampled repeatedly with samples of size 50, what is the mean and standard deviation of the sample means? Mean = 17, standard deviation = 0.17 A population has a mean of 17 and a standard deviation of 0.2. If it is sampled repeatedly with samples of size 16, what is the expected value and standard deviation of the sample means? Expected value = 17, standard deviation = 0.05 A population has a mean of 38 and a standard deviation of 3. If it is sampled repeatedly with samples of size 48, what is the expected value and standard deviation of the sample means? Expected value = 38, standard deviation = 0.43 A population has a mean of 14 and a standard deviation of 5. If it is sampled repeatedly with samples of size 60, what is the expected value and standard deviation of the sample means? Expected value = 14, standard deviation = 0.65 Homework A large population of 5,000 students take a practice test to prepare for a standardized test. The population mean is 140 questions correct, and the standard deviation is 80. What size samples should a researcher take to get a distribution of means of the samples with a standard deviation of 10? 64 A large population has skewed data with a mean of 70 and a standard deviation of 6. Samples of size 100 are taken, and the distribution of the means of these samples is analyzed. Will the distribution of the means be closer to a normal distribution than the distribution of the population? Will the mean of the means of the samples remain close to 70? Will the distribution of the means have a smaller standard deviation? What is that standard deviation? Yes Yes Yes 0.6 A researcher is looking at data from a large population with a standard deviation that is much too large. In order to concentrate the information, the researcher decides to repeatedly sample the data and use the distribution of the means of the samples? The first effort used sample sized of 100. But the standard deviation was about double the value the researcher wanted. What is the smallest size samples the researcher can use to remedy the problem? 400 A researcher looks at a large set of data, and concludes the population has a standard deviation of 40. Using sample sizes of 64, the researcher is able to focus the mean of the means of the sample to a narrower distribution where the standard deviation is 5. Then, the researcher realizes there was an error in the original calculations, and the initial standard deviation is really 20. Since the standard deviation of the means of the samples was obtained using the original standard deviation, this value is also impacted by the discovery of the error. What is the correct value of the standard deviation of the means of the samples? 2.5 A population has a standard deviation of 50. It is sampled with samples of size 100. What is the variance of the means of the samples? 25 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 . Finite Population Correction Factor adjusts the variance of the sampling distribution if the population is known and more than 5% of the population is being sampled. Normal Distribution a continuous random variable with pdf f ( x ) = 1 σ 2 π e – ( x – μ ) 2 2 σ 2 , where μ is the mean of the distribution and σ is the standard deviation.; notation: X ~ N ( μ , σ ). If μ = 0 and σ = 1, the random variable, Z, is called the standard normal distribution . Standard Error of the Proportion the standard deviation of the sampling distribution of proportions

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The Central Limit Theorem for Proportions The Central Limit Theorem tells us that the point estimate for the sample mean, x ¯ , comes from a normal distribution of x ¯ 's. This theoretical distribution is called the sampling distribution of x ¯ 's. We now investigate the sampling distribution for another important parameter we wish to estimate; p from the binomial probability density function. If the random variable is discrete, such as for categorical data, then the parameter we wish to estimate is the population proportion. This is, of course, the probability of drawing a success in any one random draw. Unlike the case just discussed for a continuous random variable where we did not know the population distribution of X's, here we actually know the underlying probability density function for these data; it is the binomial distribution. The random variable is X = the number of successes and the parameter we wish to know is p, the probability of drawing a success which is of course the proportion of successes in the population. The question at issue is: from what distribution was the sample proportion, p' = x n drawn? The sample size is n and X is the number of successes found in that sample. This is a parallel question that was just answered by the Central Limit Theorem: from what distribution was the sample mean, x ¯ , drawn? We saw that once we knew that the distribution was the Normal distribution then we were able to create confidence intervals for the population parameter, µ. We will also use this same information to test hypotheses about the population mean later. We wish now to be able to develop confidence intervals for the population parameter "p" from the binomial probability density function. In order to find the distribution from which sample proportions come we need to develop the sampling distribution of sample proportions just as we did for sample means. So again imagine that we randomly sample say 50 people and ask them if they support the new school bond issue. From this we find a sample proportion, p', and graph it on the axis of p's. We do this again and again etc., etc. until we have the theoretical distribution of p's. Some sample proportions will show high favorability toward the bond issue and others will show low favorability because random sampling will reflect the variation of views within the population. What we have done can be seen in . The top panel is the population distributions of probabilities for each possible value of the random variable X. While we do not know what the specific distribution looks like because we do not know p, the population parameter, we do know that it must look something like this. In reality, we do not know either the mean or the standard deviation of this population distribution, the same difficulty we faced when analyzing the X's previously. places the mean on the distribution of population probabilities as µ = n p but of course we do not actually know the population mean because we do not know the population probability of success, p . Below the distribution of the population values is the sampling distribution of p 's. Again the Central Limit Theorem tells us that this distribution is normally distributed just like the case of the sampling distribution for x ¯ 's. This sampling distribution also has a mean, the mean of the p 's, and a standard deviation, σ p ' . Importantly, in the case of the analysis of the distribution of sample means, the Central Limit Theorem told us the expected value of the mean of the sample means in the sampling distribution, and the standard deviation of the sampling distribution. Again the Central Limit Theorem provides this information for the sampling distribution for proportions. The answers are: The expected value of the mean of sampling distribution of sample proportions, µ p' , is the population proportion, p. The standard deviation of the sampling distribution of sample proportions, σ p' , is the population standard deviation divided by the square root of the sample size, n. Both these conclusions are the same as we found for the sampling distribution for sample means. However in this case, because the mean and standard deviation of the binomial distribution both rely upon p , the formula for the standard deviation of the sampling distribution requires algebraic manipulation to be useful. We will take that up in the next chapter. The proof of these important conclusions from the Central Limit Theorem is provided below. E ( p ' ) = E ( x n ) = ( 1 n ) E ( x ) = ( 1 n ) n p = p (The expected value of X, E(x), is simply the mean of the binomial distribution which we know to be np.) σ p' 2 = Var ( p ' ) = Var ( x n ) = 1 n 2 ( Var ( x ) ) = 1 n 2 ( n p ( 1 − p ) ) = p ( 1 − p ) n The standard deviation of the sampling distribution for proportions is thus: σ p' = p ( 1 − P ) n Parameter Population distribution Sample Sampling distribution of p's Mean µ = np p ' = x n p' and E(p') = p Standard Deviation σ = n p q σ p' = p ( 1 − p ) n summarizes these results and shows the relationship between the population, sample and sampling distribution. Notice the parallel between this Table and Table 7.1 for the case where the random variable is continuous and we were developing the sampling distribution for means. Reviewing the formula for the standard deviation of the sampling distribution for proportions we see that as n increases the standard deviation decreases. This is the same observation we made for the standard deviation for the sampling distribution for means. Again, as the sample size increases, the point estimate for either µ or p is found to come from a distribution with a narrower and narrower distribution. We concluded that with a given level of probability, the range from which the point estimate comes is smaller as the sample size, n, increases. Figure 7.8 shows this result for the case of sample means. Simply substitute p ' for x ¯ and we can see the impact of the sample size on the estimate of the sample proportion. Chapter Review The Central Limit Theorem can also be used to illustrate that the sampling distribution of sample proportions is normally distributed with the expected value of p and a standard deviation of σ p' = p ( 1 − p ) n A question is asked of a class of 200 first-year students, and 23% of the students know the correct answer. If a sample of 50 students is taken repeatedly, what is the expected value of the mean of the sampling distribution of sample proportions? 0.23 A question is asked of a class of 200 first-year students, and 23% of the students know the correct answer. If a sample of 50 students is taken repeatedly, what is the standard deviation of the mean of the sampling distribution of sample proportions? 0.060 A game is played repeatedly. A player wins one-fifth of the time. If samples of 40 times the game is played are taken repeatedly, what is the expected value of the mean of the sampling distribution of sample proportions? 1/5 A game is played repeatedly. A player wins one-fifth of the time. If samples of 40 times the game is played are taken repeatedly, what is the standard deviation of the mean of the sampling distribution of sample proportions? 0.063 A virus attacks one in three of the people exposed to it. An entire large city is exposed. If samples of 70 people are taken, what is the expected value of the mean of the sampling distribution of sample proportions? 1/3 A virus attacks one in three of the people exposed to it. An entire large city is exposed. If samples of 70 people are taken, what is the standard deviation of the mean of the sampling distribution of sample proportions? 0.056 A company inspects products coming through its production process, and rejects detected products. One-tenth of the items are rejected. If samples of 50 items are taken, what is the expected value of the mean of the sampling distribution of sample proportions? 1/10 A company inspects products coming through its production process, and rejects detected products. One-tenth of the items are rejected. If samples of 50 items are taken, what is the standard deviation of the mean of the sampling distribution of sample proportions? 0.042 Homework A farmer picks pumpkins from a large field. The farmer makes samples of 260 pumpkins and inspects them. If one in fifty pumpkins are not fit to market and will be saved for seeds, what is the standard deviation of the mean of the sampling distribution of sample proportions? 0.0087 A store surveys customers to see if they are satisfied with the service they received. Samples of 25 surveys are taken. One in five people are unsatisfied. What is the variance of the mean of the sampling distribution of sample proportions for the number of unsatisfied customers? What is the variance for satisfied customers? 0.0064, 0.0064 A company gives an anonymous survey to its employees to see what percent of its employees are happy. The company is too large to check each response, so samples of 50 are taken, and the tendency is that three-fourths of the employees are happy. For the mean of the sampling distribution of sample proportions, answer the following questions, if the sample size is doubled. How does this affect the mean? How does this affect the standard deviation? How does this affect the variance? It has no effect. It is divided by 2 . It is divided by 2. A pollster asks a single question with only yes and no as answer possibilities. The poll is conducted nationwide, so samples of 100 responses are taken. There are four yes answers for each no answer overall. For the mean of the sampling distribution of sample proportions, find the following for yes answers. The expected value. The standard deviation. The variance. 4/5 0.04 0.0016 The mean of the sampling distribution of sample proportions has a value of p of 0.3, and sample size of 40. Is there a difference in the expected value if p and q reverse roles? Is there a difference in the calculation of the standard deviation with the same reversal? Yes No

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Finite Population Correction Factor We saw that the sample size has an important effect on the variance and thus the standard deviation of the sampling distribution. Also of interest is the proportion of the total population that has been sampled. We have assumed that the population is extremely large and that we have sampled a small part of the population. As the population becomes smaller and we sample a larger number of observations the sample observations are not independent of each other. To correct for the impact of this, the Finite Correction Factor can be used to adjust the variance of the sampling distribution. It is appropriate when more than 5% of the population is being sampled and the population has a known population size. There are cases when the population is known, and therefore the correction factor must be applied. The issue arises for both the sampling distribution of the means and the sampling distribution of proportions. The Finite Population Correction Factor for the variance of the means shown in the standardizing formula is: Z = x ¯ − µ σ n · N − n N − 1 and for the variance of proportions is: σ p' = p ( 1 − p ) n × N − n N − 1 The following examples show how to apply the factor. Sampling variances get adjusted using the above formula. It is learned that the population of White German Shepherds in the USA is 4,000 dogs, and the mean weight for German Shepherds is 75.45 pounds. It is also learned that the population standard deviation is 10.37 pounds. If the sample size is 225 dogs, then find the probability that a sample will have a mean that differs from the true probability mean by less than 2 pounds. N = 4000 , n = 225 , σ = 10.37 , µ = 75.45 , ( x ¯ − µ ) = ± 2 Z = x ¯ − µ σ n · N − n N − 1 = ± 2 10.37 225 · 4000 − 225 4000 − 1 ≈ 2.978 f ( Z ) ≈ 2.978 · 2 = 0.997 Note that "differs by less" references the area on both sides of the mean within 2 pounds right or left. Try It For the 5,000 students in University A, the average height of the students is 170 cm. The population standard deviation is 5 cm. If the sample size is 100 students, find the probability that the sample mean is within 1 cm of the true mean. N = 5000 , n = 100 , σ = 5 , μ = 170 , ( x ¯ - μ ) = ± 1 Z = x ¯ - μ σ n * N - n N - 1 = ± 1 5 100 * 5000 - 100 5000 - 1 = ± 2 . 02 P ( - 1 ≤ x ¯ - μ ≤ 1 ) = P ( - 2 . 02 ≤ Z ≤ 2 . 02 ) = 2 · P ( 0 ≤ Z ≤ 2 . 02 ) = 2 · 0 . 4783 = 0 . 9566 So, the probability that the sample mean is within 1 cm of the true mean is 0.9566. When a customer places an order with Rudy's On-Line Office Supplies, a computerized accounting information system (AIS) automatically checks to see if the customer has exceeded their credit limit. Past records indicate that the probability of customers exceeding their credit limit is .06. Suppose that on a given day, 3,000 orders are placed in total. If we randomly select 360 orders, what is the probability that between 10 and 20 customers will exceed their credit limit? N = 3000 , n = 360 , p = 0.06 σ p' = p ( 1 − p ) n × N − n N − 1 = 0.06 ( 1 − 0.06 ) 360 × 3000 − 360 3000 − 1 = 0.0117 p 1 = 10 360 = 0.0278 , p 2 = 20 360 = 0.0556 Z = p ' − p p ( 1 − p ) n · N − n N − 1 = 0.0278 − 0.06 0.011744 = −2.74 Z = p ' − p p ( 1 − p ) n · N − n N − 1 = 0.0556 − 0.06 0.011744 = −0.38 p ( 0.0278 − 0.06 0.011744 < z < 0.0556 − 0.06 0.011744 ) = p ( −2.74 < z < −0.38 ) = 0.4969 − 0.1480 = 0.3489 Try It There is a 0.01 probability that a student gets food poisoning eating food in a cafeteria. Suppose there are 2,000 orders in a day in the cafeteria. If we select 200 orders randomly, what is the probability that between 5 and 10 students will get food poisoning? N = 2000 , n = 200 , p = 0 . 01 σ p ' = p ( 1 - p ) n × N - n N - 1 = 0 . 01 ( 1 - 0 . 01 ) 200 × 2000 - 200 2000 - 1 = 0 . 00668 p 1 = 5 200 = 0 . 025 , p 2 = 10 200 = 0 . 05 Z = p ' - p σ p ' = 0 . 025 - 0 . 01 0 . 00668 = 2 . 2455 Z = p ' - p σ p ' = 0 . 05 - 0 . 01 0 . 00668 = 5 . 9880 P ( 2 . 2455 ≤ z ≤ 5 . 9880 ) = 0 . 0124 So, the probability that between 5 and 10 students will get food poisoning is 0.0124. A fishing boat has 1,000 fish on board, with an average weight of 120 pounds and a standard deviation of 6.0 pounds. If sample sizes of 50 fish are checked, what is the probability the fish in a sample will have mean weight within 2.8 pounds the true mean of the population? 0.999 An experimental garden has 500 sunflowers plants. The plants are being treated so they grow to unusual heights. The average height is 9.3 feet with a standard deviation of 0.5 foot. If sample sizes of 60 plants are taken, what is the probability the plants in a given sample will have an average height within 0.1 foot of the true mean of the population? 0.901 A company has 800 employees. The average number of workdays between absence for illness is 123 with a standard deviation of 14 days. Samples of 50 employees are examined. What is the probability a sample has a mean of workdays with no absence for illness of at least 124 days? 0.301 Cars pass an automatic speed check device that monitors 2,000 cars on a given day. This population of cars has an average speed of 67 miles per hour with a standard deviation of 2 miles per hour. If samples of 30 cars are taken, what is the probability a given sample will have an average speed within 0.50 mile per hour of the population mean? 0.832 A town keeps weather records. From these records it has been determined that it rains on an average of 37% of the days each year. If 30 days are selected at random from one year, what is the probability that at least 5 and at most 11 days had rain? 0.483 A maker of yardsticks has an ink problem that causes the markings to smear on 4% of the yardsticks. The daily production run is 2,000 yardsticks. What is the probability if a sample of 100 yardsticks is checked, there will be ink smeared on at most 4 yardsticks? 0.500 A school has 300 students. Usually, there are an average of 21 students who are absent. If a sample of 30 students is taken on a certain day, what is the probability that at most 2 students in the sample will be absent? 0.502 A college gives a placement test to 5,000 incoming students each year. On the average 1,213 place in one or more developmental courses. If a sample of 50 is taken from the 5,000, what is the probability at most 12 of those sampled will have to take at least one developmental course? 0.519 Homework A company has 1,000 employees. The average number of workdays between absence for illness is 80 with a standard deviation of 11 days. Samples of 80 employees are examined. What is the probability a sample has a mean of workdays with no absence for illness of at least 78 days and at most 84 days? 0.955 Trucks pass an automatic scale that monitors 2,000 trucks. This population of trucks has an average weight of 20 tons with a standard deviation of 2 tons. If a sample of 50 trucks is taken, what is the probability the sample will have an average weight within one-half ton of the population mean? 0.927 A town keeps weather records. From these records it has been determined that it rains on an average of 12% of the days each year. If 30 days are selected at random from one year, what is the probability that at most 3 days had rain? 0.648 A maker of greeting cards has an ink problem that causes the ink to smear on 7% of the cards. The daily production run is 500 cards. What is the probability that if a sample of 35 cards is checked, there will be ink smeared on at most 5 cards? 0.101 A school has 500 students. Usually, there are an average of 20 students who are absent. If a sample of 30 students is taken on a certain day, what is the probability that at least 2 students in the sample will be absent? 0.273

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Introduction Have you ever wondered what the average number of M&Ms in a bag at the grocery store is? You can use confidence intervals to answer this question. (credit: modification of work “sweet, orange, food, green, red, color, brown, blue, colorful, yellow, chocolate, snack, dessert, toy, plain, candy, sweetness, treat, confectionery, coated, m ms, hard shell, snack food, jelly bean”/ Pxhere, Public Domain) Suppose you were trying to determine the rent of a two-bedroom apartment in your town. You might look in the classified section of the newspaper, write down several rents listed, and average them together. You would have obtained a point estimate of the true mean. If you are trying to determine the percentage of times you make a basket when shooting a basketball, you might count the number of shots you make and divide that by the number of shots you attempted. In this case, you would have obtained a point estimate for the true proportion the parameter p in the binomial probability density function. 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. What statistics provides us beyond a simple average, or point estimate, is an estimate to which we can attach a probability of accuracy, what we will call a confidence level. We make inferences with a known level of probability. In this chapter, you will learn to construct and interpret confidence intervals. You will also learn a new distribution, the Student's-t, and how it is used with these intervals. Throughout the chapter, it is important to keep in mind that the confidence interval is a random variable. It is the population parameter that is fixed. If you worked in the marketing department of an entertainment company, you might be interested in the mean number of songs a consumer downloads a month from Apple Music. 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 . x – and s are each 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 the unknown population parameter. Suppose, for the Apple Music example, we do not know the population mean μ , but we do know the sample mean and that the population standard deviation is σ = 1 and our sample size is 100. Then, by the central limit theorem, the standard deviation of the sampling distribution of the sample means is σ n = 1 100 = 0.1 The Empirical Rule , which applies to the normal distribution, says that in approximately 95% of the samples, the sample mean, x – , will be within two standard deviations of the population mean μ . For our Apple Music example, two standard deviations is (2)(0.1) = 0.2. The sample mean x – is likely to be within 0.2 units of μ . Because x – is within 0.2 units of μ , which is unknown, then μ is likely to be within 0.2 units of x – with 95% probability. The population mean μ is contained in an interval whose lower number is calculated by taking the sample mean and subtracting two standard deviations (2)(0.1) and whose upper number is calculated by taking the sample mean and adding two standard deviations. In other words, μ is between x ¯ − 0 .2 and x ¯ + 0 .2 in 95% of all the samples. For the Apple Music example, suppose that a sample produced a sample mean x ¯ = 2 . Then with 95% probability the unknown population mean μ is between x ¯ − 0.2 = 2 − 0.2 = 1.8 and x ¯ + 0.2 = 2 + 0.2 = 2.2 We say that we are 95% confident that the unknown population mean number of songs downloaded from Apple Music per month is between 1.8 and 2.2. The 95% confidence interval is (1.8, 2.2). Please note that we talked in terms of 95% confidence using the empirical rule. The empirical rule for two standard deviations is only approximately 95% of the probability under the normal distribution. To be precise, two standard deviations under a normal distribution is actually 95.44% of the probability. To calculate the exact 95% confidence level we would use 1.96 standard deviations. The 95% confidence interval implies two possibilities. Either the interval (1.8, 2.2) contains the true mean μ, or our sample produced an x – that is not within 0.2 units of the true mean μ . The first possibility happens for 95% of well-chosen samples. It is important to remember that the second possibility happens for 5% of samples, even though correct procedures are followed. Remember that a confidence interval is created for an unknown population parameter like the population mean, μ . For the confidence interval for a mean the formula would be: μ = X ¯ ± Z α σ n Or written another way as: X ¯ − Z α σ n ≤ μ ≤ X ¯ + Z α σ n Where X ¯ is the sample mean. Z α is determined by the level of confidence desired by the analyst, and σ n is the standard deviation of the sampling distribution for means given to us by the Central Limit Theorem. Confidence Interval (CI) an interval estimate for an unknown population parameter. This depends on: the desired confidence level, information that is known about the distribution (for example, known standard deviation), the sample and its size. Inferential Statistics also called statistical inference or inductive statistics; this facet of statistics deals with estimating a population parameter based on a sample statistic. For example, if four out of the 100 calculators sampled are defective we might infer that four percent of the production is defective. Parameter a numerical characteristic of a population Point Estimate a single number computed from a sample and used to estimate a population parameter

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