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of ˆp approximately normal? Justify your answer. c. What is the probability that the sample proportion ˆp exceeds 82%? d. What is the probability that the sample proportion lies between 83% and 88%? e. 99% of the time, the sample proportion would lie between what two limits? 7.71 What survey design is used in each of these situations? a. A random sample of n 50 city blocks is se- lected, and a census is done for each single-family dwelling on each block. b. The highway patrol stops every 10th vehicle on a given city artery between 9:00 A.M. and 3:00 P.M. to perform a routine traffic safety check. c. One hundred households in each of four city wards are surveyed concerning a pending city tax relief referendum. d. Every 10th tree in a managed slash pine plantation is checked for pine needle borer infestation. e. A random sample of n 1000 taxpayers from the city of San Bernardino is selected by the Internal Revenue Service and their tax returns are audited. 7.72 Elevator Loads The maximum load (with a generous safety factor) for the elevator in an office building is 2000 pounds. The relative frequency distribution of the weights of all men and women using the elevator is mound-shaped (slightly skewed to the heavy weights), with mean m equal to 150 pounds and standard deviation s equal to 35 pounds. What is the largest number of people you can allow on the elevator if you want their total weight to exceed the maximum weight with a small probability (say, near.01)? (HINT: If x1, x2,..., xn are independent observations made on a random variable x, and if x has mean m and variance s 2, then the mean and variance of Sxi are nm and ns 2, respectively. This result was given in Section 7.4.) 7.73 Wiring Packages The number of wiring packages that can be assembled by a company’s employees has a normal distribution, with a mean equal to 16.4 per hour and a standard deviation of 1.3 per hour. a. What are the mean and standard deviation of the number x of packages produced per worker in an 8-hour day? b. Do you expect the probability distribution for x to be mound-shaped and approximately normal? Explain. c. What is the probability that a worker will produce at least 135 packages per 8-hour day?
7.74 Wiring Packages, continued Refer to Exercise 7.73. Suppose the company employs 10 assemblers of wiring packages. a. Find the mean and standard deviation of the company’s daily (8-hour day) production of wiring packages. b. What is the probability that the company’s daily production is less than 1280 wiring packages per day? 7.75 Defective Lightbulbs The table lists the number of defective 60-watt lightbulbs EX0775 found in samples of 100 bulbs selected over 25 days from a manufacturing process. Assume that during these 25 days the manufacturing process was not producing an excessively large fraction of defectives. Day Defectives Day Defectives Day Defectives 1 4 11 2 21 2 2 2 12 4 22 2 3 5 13 3 23 3 4 8 14 4 24 5 5 3 15 0 25 3 6 4 7 4 8 5 9 6 16 2 17 3 18 19 1 4 10 1 20 0 a. Construct a p chart to monitor the manufacturing process, and plot the data. b. How large must the fraction of defective items be in a sample selected from the manufacturing process before the process is assumed to be out of control? c. During a given day, suppose a sample of 100 items is selected from the manufacturing process and 15 defective bulbs are found. If a decision is made to shut down the manufacturing process in an attempt to locate the source of the implied controllable variation, explain how this decision might lead to erroneous conclusions. 7.76 Lightbulbs, continued A hardware store chain purchases large shipments of lightbulbs from the manufacturer described in Exercise 7.75 and speciﬁes that each shipment must contain no more than 4% defectives. When the manufacturing process is in control, what is the probability that the hardware store’s speciﬁcations are met? 7.77 Lightbulbs, again Refer to Exercise 7.75. During a given week the number of defective bulbs in each of ﬁve samples of 100 were found to be 2, 4, 9, 7, and 11. Is there reason to believe that the production process has been producing an excessive proportion of defectives at any time during the week? 7.78 Canned Tomatoes During long production runs of canned tomatoes, the average EX0778 weights (in ounces) of samples of ﬁve cans of standard-grade tomatoes in pureed form were taken at 30 control points during an 11-
day period. These results are shown in the table.20 When the machine is performing normally, the average weight per can is 21 ounces with a standard deviation of 1.20 ounces. SUPPLEMENTARY EXERCISES ❍ 293 a. Compute the upper and lower control limits and the centerline for the x chart. b. Plot the sample data on the x chart and determine whether the performance of the machine is in control. Sample Average Number Weight Sample Average Number Weight 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 23.1 21.3 22.0 21.4 21.8 20.6 20.1 21.4 21.5 20.2 20.3 20.1 21.7 21.0 21.6 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 21.4 20.4 22.8 21.1 20.7 21.6 22.4 21.3 21.1 20.1 21.2 19.9 21.1 21.6 21.3 Source: Adapted from J. Hackl, Journal of Quality Technology, April 1991. Used with permission. 7.79 Pepsi or Coke? The battle for consumer preference continues between Pepsi and Coke. How can you make your preferences known? There is a web page where you can vote for one of these colas if you click on the link that says PAY CASH for your opinion. Explain why the respondents do not represent a random sample of the opinions of purchasers or drinkers of these drinks. Explain the types of distortions that could creep into an Internet opinion poll. 7.80 Strawberries An experimenter wants to ﬁnd an appropriate temperature at which to store fresh strawberries to minimize the loss of ascorbic acid. There are 20 storage containers, each with controllable temperature, in which strawberries can be stored. If two storage temperatures are to be used, how would the experimenter assign the 20 containers to one of the two storage temperatures? 7.81 Filling Soda Cans A bottler of soft drinks packages cans in six-packs. Suppose that the ﬁll per can has an approximate normal distribution with a mean of 12 ﬂuid ounces and a standard deviation of 0.2 ﬂuid ounces. a. What is the distribution of the total ﬁll for a case of 24 cans? b. What is the probability that the total ﬁll for a case is less than 286 �
��uid ounces? 294 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS c. If a six-pack of soda can be considered a random sample of size n 6 from the population, what is the probability that the average ﬁll per can for a six-pack of soda is less than 11.8 ﬂuid ounces? 7.82 Total Packing Weight Packages of food whose average weight is 16 ounces with a standard deviation of 0.6 ounces are shipped in boxes of 24 packages. If the package weights are approximately normally distributed, what is the probability that a box of 24 packages will weigh more than 392 ounces (24.5 pounds)? 7.83 Electronic Components A manufacturing process is designed to produce an electronic compo- Exercises 7.84 Dice Refer to the die-tossing experiment with n 1 in Section 7.4 in which x is the number on the upper face of a single balanced die. a. Use the formulas in Section 4.8 to verify that m 3.5 and s 1.71 for this population. b. Use the Central Limit Theorem applet to toss a single die at least 2000 times. (Your simulation can be done quickly by using the button.) What are the mean and standard deviation of these 2000 observations? What is the shape of the histogram? c. Compare the results of part b to the actual probability distribution shown in Figure 7.3 and the actual mean and standard deviation in part a. They should be similar! 7.85 Dice Two balanced dice are thrown, and the average number on the two upper faces is recorded. a. Use the values m 3.5 and s 1.71 from Exercise 7.84. What are the theoretical mean and standard deviation of the sampling distribution for x? b. Use the Central Limit Theorem applet to toss a single die at least 2000 times. (Your simulation can be done quickly by using the What are the mean and standard deviation of these 2000 observations? What is the shape of the histogram? button.) c. Compare the results of part b to the actual probability distribution shown in Figure 7.4 and the actual mean and standard deviation in part a. nent for use in small portable television sets. The components are all of standard size and need not conform to any measurable characteristic, but are sometimes inoperable when emerging from the manufacturing process. Fifteen samples were selected from the process at times when the process was known to be in statistical control. Fifty components
were observed within each sample, and the number of inoperable components was recorded. 6, 7, 3, 5, 6, 8, 4, 5, 7, 3, 1, 6, 5, 4, 5 Construct a p chart to monitor the manufacturing process. 7.86 Repeat the instructions in Exercise 7.85 when three dice are tossed. 7.87 Repeat the instructions in Exercise 7.85 when four dice are tossed. 7.88 Suppose a random sample of n 5 observations is selected from a population that is normally distributed, with mean equal to 1 and standard deviation equal to.36. a. Give the mean and the standard deviation of the sampling distribution of x. b. Find the probability that x exceeds 1.3, using the Normal Probabilities for Means applet. c. Find the probability that the sample mean x is less than.5. d. Find the probability that the sample mean deviates from the population mean m 1 by more than.4. 7.89 Batteries A certain type of automobile battery is known to last an average of 1110 days with a standard deviation of 80 days. If 400 of these batteries are selected, use the Normal Probabilities for Means applet to ﬁnd the following probabilities for the average length of life of the selected batteries: a. The average is between 1100 and 1110. b. The average is greater than 1120. c. The average is less than 900. CASE STUDY CASE STUDY ❍ 295 Sampling the Roulette at Monte Carlo The technique of simulating a process that contains random elements and repeating the process over and over to see how it behaves is called a Monte Carlo procedure. It is widely used in business and other ﬁelds to investigate the properties of an operation that is subject to random effects, such as weather, human behavior, and so on. For example, you could model the behavior of a manufacturing company’s inventory by creating, on paper, daily arrivals and departures of manufactured products from the company’s warehouse. Each day a random number of items produced by the company would be received into inventory. Similarly, each day a random number of orders of varying random sizes would be shipped. Based on the input and output of items, you could calculate the inventory—that is, the number of items on hand at the end of each day. The values of the random variables, the number of items produced, the number of orders, and the number of items per order needed
for each day’s simulation would be obtained from theoretical distributions of observations that closely model the corresponding distributions of the variables that have been observed over time in the manufacturing operation. By repeating the simulation of the supply, the shipping, and the calculation of daily inventory for a large number of days (a sampling of what might really happen), you can observe the behavior of the plant’s daily inventory. The Monte Carlo procedure is particularly valuable because it enables the manufacturer to see how the daily inventory would behave when certain changes are made in the supply pattern or in some other aspect of the operation that could be controlled. In an article entitled “The Road to Monte Carlo,” Daniel Seligman comments on the Monte Carlo method, noting that although the technique is widely used in business schools to study capital budgeting, inventory planning, and cash ﬂow management, no one seems to have used the procedure to study how well we might do if we were to gamble at Monte Carlo.21 To follow up on this thought, Seligman programmed his personal computer to simulate the game of roulette. Roulette involves a wheel with its rim divided into 38 pockets. Thirty-six of the pockets are numbered 1 to 36 and are alternately colored red and black. The two remaining pockets are colored green and are marked 0 and 00. To play the game, you bet a certain amount of money on one or more pockets. The wheel is spun and turns until it stops. A ball falls into a slot on the wheel to indicate the winning number. If you have money on that number, you win a speciﬁed amount. For example, if you were to play the number 20, the payoff is 35 to 1. If the wheel does not stop at that number, you lose your bet. Seligman decided to see how his nightly gains (or losses) would fare if he were to bet $5 on each turn of the wheel and repeat the process 200 times each night. He did this 365 times, thereby simulating the outcomes of 365 nights at the casino. Not surprisingly, the mean “gain” per $1000 evening for the 365 nights was a loss of $55, the average of the winnings retained by the gambling house. The surprise, according to Seligman, was the extreme variability of the nightly “winnings.” Seven times out of the 365 evenings, the ﬁctitious gambler lost the $1000 stake, and only once did he win
a maximum of $1160. On 141 nights, the loss exceeded $250. 1. To evaluate the results of Seligman’s Monte Carlo experiment, ﬁrst ﬁnd the prob- ability distribution of the gain x on a single $5 bet. 2. Find the expected value and variance of the gain x from part 1. 296 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS 3. Find the expected value and variance for the evening’s gain, the sum of the gains or losses for the 200 bets of $5 each. 4. Use the results of part 2 to evaluate the probability of 7 out of 365 evenings resulting in a loss of the total $1000 stake. 5. Use the results of part 3 to evaluate the probability that the largest evening’s win- nings were as great as $1160. Large-Sample Estimation 8 © Associated Press GENERAL OBJECTIVE In previous chapters, you learned about the probability distributions of random variables and the sampling distributions of several statistics that, for large sample sizes, can be approximated by a normal distribution according to the Central Limit Theorem. This chapter presents a method for estimating population parameters and illustrates the concept with practical examples. The Central Limit Theorem and the sampling distributions presented in Chapter 7 play a key role in evaluating the reliability of the estimates. CHAPTER INDEX ● Choosing the sample size (8.9) ● Estimating the difference between two binomial propor- tions (8.6) ● Estimating the difference between two population means (8.6) ● Interval estimation (8.5) ● Large-sample conﬁdence intervals for a population mean or proportion (8.5) ● One-sided conﬁdence bounds (8.8) ● Picking the best point estimator (8.4) ● Point estimation for a population mean or proportion (8.4) ● Types of estimators (8.3) How Reliable Is That Poll? Do the national polls conducted by the Gallup and Harris organizations, the news media, and others provide accurate estimates of the percentages of people in the United States who have a variety of eating habits? The case study at the end of this chapter examines the reliability of a poll conducted by CBS News using the theory of large-sample estimation. How Do I Estimate a Population Mean or Proportion? How Do I Choose the Sample Size? 297 298 ❍ CHAPTER 8 LAR
GE-SAMPLE ESTIMATION WHERE WE’VE BEEN 8.1 The ﬁrst seven chapters of this book have given you the building blocks you will need to understand statistical inference and how it can be applied in practical situations. The ﬁrst three chapters were concerned with using descriptive statistics, both graphical and numerical, to describe and interpret sets of measurements. In the next three chapters, you learned about probability and probability distributions—the basic tools used to describe populations of measurements. The binomial and the normal distributions were emphasized as important for practical applications. The seventh chapter provided the link between probability and statistical inference. Many statistics are either sums or averages calculated from sample measurements. The Central Limit Theorem states that, even if the sampled populations are not normal, the sampling distributions of these statistics will be approximately normal when the sample size n is large. These statistics are the tools you use for inferential statistics—making inferences about a population using information contained in a sample. WHERE WE’RE GOING— STATISTICAL INFERENCE 8.2 Parameter ⇔ Population Statistic ⇔ Sample Inference—speciﬁcally, decision making and prediction—is centuries old and plays a very important role in most peoples’ lives. Here are some applications: • The government needs to predict short- and long-term interest rates. • A broker wants to forecast the behavior of the stock market. • A metallurgist wants to decide whether a new type of steel is more resistant to high temperatures than the current type. • A consumer wants to estimate the selling price of her house before putting it on the market. There are many ways to make these decisions or predictions, some subjective and some more objective in nature. How good will your predictions or decisions be? Although you may feel that your own built-in decision-making ability is quite good, experience suggests that this may not be the case. It is the job of the mathematical statistician to provide methods of statistical inference making that are better and more reliable than just subjective guesses. Statistical inference is concerned with making decisions or predictions about parameters—the numerical descriptive measures that characterize a population. Three parameters you encountered in earlier chapters are the population mean m, the population standard deviation s, and the binomial proportion p. In statistical inference, a practical problem is restated in the framework of a population with a speciﬁc parameter of interest. For example, the metallurgist could measure the average coefficients
of expansion for both types of steel and then compare their values. Methods for making inferences about population parameters fall into one of two categories: • Estimation: Estimating or predicting the value of the parameter • Hypothesis testing: Making a decision about the value of a parameter based on some preconceived idea about what its value might be EXAMPLE 8.1 EXAMPLE 8.2 8.3 TYPES OF ESTIMATORS ❍ 299 The circuits in computers and other electronics equipment consist of one or more printed circuit boards (PCB), and computers are often repaired by simply replacing one or more defective PCBs. In an attempt to ﬁnd the proper setting of a plating process applied to one side of a PCB, a production supervisor might estimate the average thickness of copper plating on PCBs using samples from several days of operation. Since he has no knowledge of the average thickness m before observing the production process, his is an estimation problem. The supervisor in Example 8.1 is told by the plant owner that the thickness of the copper plating must not be less than.001 inch in order for the process to be in control. To decide whether or not the process is in control, the supervisor might formulate a test. He could hypothesize that the process is in control—that is, assume that the average thickness of the copper plating is.001 or greater—and use samples from several days of operation to decide whether or not his hypothesis is correct. The supervisor’s decision-making approach is called a test of hypothesis. Which method of inference should be used? That is, should the parameter be estimated, or should you test a hypothesis concerning its value? The answer is dictated by the practical question posed and is often determined by personal preference. Since both estimation and tests of hypotheses are used frequently in scientiﬁc literature, we include both methods in this and the next chapter. A statistical problem, which involves planning, analysis, and inference making, is incomplete without a measure of the goodness of the inference. That is, how accurate or reliable is the method you have used? If a stockbroker predicts that the price of a stock will be $80 next Monday, will you be willing to take action to buy or sell your stock without knowing how reliable her prediction is? Will the prediction be within $1, $2, or $10 of the actual price next Monday? Statistical procedures are important because they provide two types of information: • Methods for
making the inference • A numerical measure of the goodness or reliability of the inference TYPES OF ESTIMATORS 8.3 To estimate the value of a population parameter, you can use information from the sample in the form of an estimator. Estimators are calculated using information from the sample observations, and hence, by deﬁnition they are also statistics. Definition An estimator is a rule, usually expressed as a formula, that tells us how to calculate an estimate based on information in the sample. Estimators are used in two different ways: • Point estimation: Based on sample data, a single number is calculated to estimate the population parameter. The rule or formula that describes this calculation is called the point estimator, and the resulting number is called a point estimate. 300 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION EXAMPLE 8.3 • Interval estimation: Based on sample data, two numbers are calculated to form an interval within which the parameter is expected to lie. The rule or formula that describes this calculation is called the interval estimator, and the resulting pair of numbers is called an interval estimate or conﬁdence interval. A veterinarian wants to estimate the average weight gain per month of 4-month-old golden retriever pups that have been placed on a lamb and rice diet. The population consists of the weight gains per month of all 4-month-old golden retriever pups that are given this particular diet. The veterinarian wants to estimate the unknown parameter m, the average monthly weight gain for this hypothetical population. One possible estimator based on sample data is the sample mean, x Sxi/n. It could be used in the form of a single number or point estimate—for instance, 3.8 pounds—or you could use an interval estimate and estimate that the average weight gain will be between 2.7 and 4.9 pounds. Both point and interval estimation procedures use information provided by the sampling distribution of the speciﬁc estimator you have chosen to use. We will begin by discussing point estimation and its use in estimating population means and proportions. POINT ESTIMATION 8.4 In a practical situation, there may be several statistics that could be used as point estimators for a population parameter. To decide which of several choices is best, you need to know how the estimator behaves in repeated sampling, described by its sampling distribution. By way of analogy, think of ﬁring a revolver at
a target. The parameter of interest is the bull’s-eye, at which you are ﬁring bullets. Each bullet represents a single sample estimate, ﬁred by the revolver, which represents the estimator. Suppose your friend ﬁres a single shot and hits the bull’s-eye. Can you conclude that he is an excellent shot? Would you stand next to the target while he ﬁres a second shot? Probably not, because you have no measure of how well he performs in repeated trials. Does he always hit the bull’s-eye, or is he consistently too high or too low? Do his shots cluster closely around the target, or do they consistently miss the target by a wide margin? Figure 8.1 shows several target conﬁgurations. Which target would you pick as belonging to the best shot? parameter target’s bull’s-eye estimator bullet or arrow F IG URE 8. 1 Which marksman is best? ● Consistently below bulls-eye Consistently above bull’s-eye Off bull’s-eye by a wide margin Best marksmanship 8.4 POINT ESTIMATION ❍ 301 Sampling distributions provide information that can be used to select the best estimator. What characteristics would be valuable? First, the sampling distribution of the point estimator should be centered over the true value of the parameter to be estimated. That is, the estimator should not consistently underestimate or overestimate the parameter of interest. Such an estimator is said to be unbiased. Definition An estimator of a parameter is said to be unbiased if the mean of its distribution is equal to the true value of the parameter. Otherwise, the estimator is said to be biased. The sampling distributions for an unbiased estimator and a biased estimator are shown in Figure 8.2. The sampling distribution for the biased estimator is shifted to the right of the true value of the parameter. This biased estimator is more likely than an unbiased one to overestimate the value of the parameter. FIGURE 8.2 Distributions for biased and unbiased estimators ● Unbiased estimator Biased estimator True value of parameter The second desirable characteristic of an estimator is that the spread (as measured by the variance) of the sampling distribution should be as small as possible. This ensures that, with a high probability, an individual estimate will fall close to the true value of the parameter. The sampling distributions for
two unbiased estimators, one with a small variance† and the other with a larger variance, are shown in Figure 8.3. FI GUR E 8. 3 Comparison of estimator variability ● Estimator with smaller variance Estimator with larger variance True value of parameter †Statisticians usually use the term variance of an estimator when in fact they mean the variance of the sampling distribution of the estimator. This contractive expression is used almost universally. 302 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION Naturally, you would prefer the estimator with the smaller variance because the estimates tend to lie closer to the true value of the parameter than in the distribution with the larger variance. In real-life sampling situations, you may know that the sampling distribution of an estimator centers about the parameter that you are attempting to estimate, but all you have is the estimate computed from the n measurements contained in the sample. How far from the true value of the parameter will your estimate lie? How close is the marksman’s bullet to the bull’s-eye? The distance between the estimate and the true value of the parameter is called the error of estimation. Definition The distance between an estimate and the estimated parameter is called the error of estimation. In this chapter, you may assume that the sample sizes are always large and, therefore, that the unbiased estimators you will study have sampling distributions that can be approximated by a normal distribution (because of the Central Limit Theorem). Remember that, for any point estimator with a normal distribution, the Empirical Rule states that approximately 95% of all the point estimates will lie within two (or more exactly, 1.96) standard deviations of the mean of that distribution. For unbiased estimators, this implies that the difference between the point estimator and the true value of the parameter will be less than 1.96 standard deviations or 1.96 standard errors (SE). This quantity, called the 95% margin of error (or simply the “margin of error”), provides a practical upper bound for the error of estimation (see Figure 8.4). It is possible that the error of estimation will exceed this margin of error, but that is very unlikely. F IG URE 8. 4 Sampling distribution of an unbiased estimator ● 95% 1.96SE 1.96SE True value Sample estimator Margin of error Margin of error A particular estimate 95% Margin of error 1.96 Standard error POINT
ESTIMATION OF A POPULATION PARAMETER • Point estimator: a statistic calculated using sample measurements 95% Margin of error: 1.96 Standard error of the estimator • The sampling distributions for two unbiased point estimators were discussed in Chapter 7. It can be shown that both of these point estimators have the minimum variability of all unbiased estimators and are thus the best estimators you can ﬁnd in each situation. 8.4 POINT ESTIMATION ❍ 303 The variability of the estimator is measured using its standard error. However, you might have noticed that the standard error usually depends on unknown parameters such as s or p. These parameters must be estimated using sample statistics such as s and ˆp. Although not exactly correct, experimenters generally refer to the estimated standard error as the standard error. How Do I Estimate a Population Mean or Proportion? • To estimate the population mean m for a quantitative population, the point estimator x is unbiased with standard error estimated as s SE † n The 95% margin of error when n 30 is estimated as s 1.96 n • To estimate the population proportion p for a binomial population, the point estimator ˆp x/n is unbiased, with standard error estimated as qˆ SE ˆp n The 95% margin of error is estimated as qˆ 1.96ˆp n Assumptions: npˆ 5 and n qˆ 5. EXAMPLE 8.4 An environmentalist is conducting a study of the polar bear, a species found in and around the Arctic Ocean. Their range is limited by the availability of sea ice, which they use as a platform to hunt seals, the mainstay of their diet. The destruction of its habitat on the Arctic ice, which has been attributed to global warming, threatens the bear’s survival as a species; it may become extinct within the century.1 A random sample of n 50 polar bears produced an average weight of 980 pounds with a standard deviation of 105 pounds. Use this information to estimate the average weight of all Arctic polar bears. Solution The random variable measured is weight, a quantitative random variable best described by its mean m. The point estimate of m, the average weight of all Arctic polar bears, is x 980 pounds. The margin of error is estimated as s 5 10 29.10 29 pounds 1.96 0 n 5 1.96 SE 1.96 †When you sample from
a normal distribution, the statistic (x m)/(s/ n) has a t distribution, which will be discussed in Chapter 10. When the sample is large, this statistic is approximately normally distributed whether the sampled population is normal or nonnormal. 304 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION You can be fairly conﬁdent that the sample estimate of 980 pounds is within 29 pounds of the population mean. In reporting research results, investigators often attach either the sample standard deviation s (sometimes called SD) or the standard error s/n (usually called SE or SEM) to the estimates of population means. You should always look for an explanation somewhere in the text of the report that tells you whether the investigator is reporting x SD or x SE. In addition, the sample means and standard deviations or standard errors are often presented as “error bars” using the graphical format shown in Figure 8.5. F IG URE 8. 5 Plot of treatment means and their standard errors ● 15 e 10 s n o p s e R 5 SE SE A B Treatments EXAMPLE 8.5 In addition to the average weight of the Arctic polar bear, the environmentalist from Example 8.4 is also interested in the opinions of adults on the subject of global warming. In particular, he wants to estimate the proportion of adults who think that global warming is a very serious problem. In a random sample of n 100 adults, 73% of the sample indicated that, from what they have heard or read, global warming is a very serious problem. Estimate the true population proportion of adults who believe that global warming is a very serious problem, and ﬁnd the margin of error for the estimate. Solution The parameter of interest is now p, the proportion of individuals in the population who believe that global warming is a very serious problem. The best estimator of p is the sample proportion pˆ, which for this sample is pˆ.73. In order to ﬁnd the margin of error, you can approximate the value of p with its estimate pˆ.73: 1.96 SE 1.96 pˆ qˆ 1.96.73 7).09 (.2 0 0 1 n With this margin of error, you can be fairly conﬁdent that the estimate of.73 is within.09 of the true value of p. Hence, you can conclude that the true value of p could be as small as.
64 or as large as.82. This margin of error is quite large when compared to the estimate itself and reﬂects the fact that large samples are required to achieve a small margin of error when estimating p. 8.4 POINT ESTIMATION ❍ 305 TABLE 8.1 ● Some Calculated Values of pq p.1.2.3.4.5 pq.09.16.21.24.25 pq.30.40.46.49.50 p.6.7.8.9 pq.24.21.16.09 pq.49.46.40.30 Table 8.1 shows how the numerator of the standard error of pˆ changes for various values of p. Notice that, for most values of p—especially when p is between.3 and.7—there is very little change in pq, the numerator of SE, reaching its maximum value when p.5. This means that the margin of error using the estimator pˆ will also be a maximum when p.5. Some pollsters routinely use the maximum margin of error—often called the sampling error—when estimating p, in which case they calculate 1.96 SE 1.96.5(.5) or sometimes 2 SE 2.5(.5) n n Gallup, Harris, and Roper polls generally use sample sizes of approximately 1000, so their margin of error is 1.96. ).031 5 (. 5 0 0 0 1 or approximately 3% In this case, the estimate is said to be within 3 percentage points of the true population proportion. 8.4 EXERCISES BASIC TECHNIQUES 8.1 Explain what is meant by “margin of error” in point estimation. 8.2 What are two characteristics of the best point estimator for a population parameter? 8.3 Calculate the margin of error in estimating a population mean m for these values: a. n 30, s 2.2 b. n 30, s 2.9 c. n 30, s 2 1.5 8.4 Refer to Exercise 8.3. What effect does a larger population variance have on the margin of error? 8.5 Calculate the margin of error in estimating a population mean m for these values: a. n 50, s2 4 b. n 500, s2 4 c. n 5000, s2 4 8.6 Refer to Exercise 8.5.
What effect does an increased sample size have on the margin of error? 8.7 Calculate the margin of error in estimating a binomial proportion for each of the following values of n. Use p.5 to calculate the standard error of the estimator. a. n 30 c. n 400 b. n 100 d. n 1000 8.8 Refer to Exercise 8.7. What effect does increasing the sample size have on the margin of error? 8.9 Calculate the margin of error in estimating a binomial proportion p using samples of size n 100 and the following values for p: b. p.3 a. p.1 d. p.7 e. p.9 f. Which of the values of p produces the largest mar- c. p.5 gin of error? 306 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION 8.10 Suppose you are writing a questionnaire for a sample survey involving n 100 individuals. The questionnaire will generate estimates for several different binomial proportions. If you want to report a single margin of error for the survey, which margin of error from Exercise 8.9 is the correct one to use? 8.11 A random sample of n 900 observations from a binomial population produced x 655 successes. Estimate the binomial proportion p and calculate the margin of error. 8.12 A random sample of n 50 observations from a quantitative population produced x 56.4 and s2 2.6. Give the best point estimate for the population mean m, and calculate the margin of error. APPLICATIONS 8.13 The San Andreas Fault Geologists are interested in shifts and movements of the earth’s surface indicated by fractures (cracks) in the earth’s crust. One of the most famous large fractures is the San Andreas fault in California. A geologist attempting to study the movement of the relative shifts in the earth’s crust at a particular location found many fractures in the local rock structure. In an attempt to determine the mean angle of the breaks, she sampled n 50 fractures and found the sample mean and standard deviation to be 39.8° and 17.2°, respectively. Estimate the mean angular direction of the fractures and ﬁnd the margin of error for your estimate. 8.14 Biomass Estimates of the earth’s biomass, the total amount of vegetation held by the earth’s forests, are important in determining the amount of unabsorbed carbon dioxide that is expected to remain in the earth
’s atmosphere.2 Suppose a sample of 75 one-square-meter plots, randomly chosen in North America’s boreal (northern) forests, produced a mean biomass of 4.2 kilograms per square meter (kg/m2), with a standard deviation of 1.5 kg/m2. Estimate the average biomass for the boreal forests of North America and ﬁnd the margin of error for your estimate. Source: Reprinted with permission from Science News, the weekly newsmagazine of Science, copyright 1989 by Science Services, Inc. 8.15 Consumer Conﬁdence An increase in the rate of consumer savings is frequently tied to a lack of conﬁdence in the economy and is said to be an indicator of a recessional tendency in the economy. A random sampling of n 200 savings accounts in a local community showed a mean increase in savings account values of 7.2% over the past 12 months, with a stan- dard deviation of 5.6%. Estimate the mean percent increase in savings account values over the past 12 months for depositors in the community. Find the margin of error for your estimate. 8.16 Multimedia Kids Do our children spend as much time enjoying the outdoors and playing with family and friends as previous generations did? Or are our children spending more and more time glued to the television, computer, and other multimedia equipment? A random sample of 250 children between the ages of 8 and 18 showed that 170 children had a TV in their bedroom and that 120 of them had a video game player in their bedroom. a. Estimate the proportion of all 8- to 18-year-olds who have a TV in their bedroom, and calculate the margin of error for your estimate. b. Estimate the proportion of all 8- to 18-year-olds who have a video game player in their bedroom, and calculate the margin of error for your estimate. 8.17 Legal Immigration At a time in U.S. history when there appears to be genuine concern about the number of illegal aliens living in the United States, there also appears to be concern over the number of legal immigrants allowed to move to the United States. In a recent poll that included questions about both legal and illegal immigrants to the United States, 51% of the n 900 registered voters interviewed indicated that the U.S. should decrease the number of legal immigrants entering the United States.3 a. What is a point estimate for the proportion of U.
S. registered voters who feel that the United States should decrease the number of legal immigrants entering the United States? Calculate the margin of error. b. The poll reports a margin of error of 3%. How was the reported margin of error calculated so that it can be applied to all of the questions in the survey? 8.18 Summer Vacations One of the major costs involved in planning a summer vacation is the cost of lodging. Even within a particular chain of hotels, costs can vary substantially depending on the type of room and the amenities offered.4 Suppose that we randomly select 50 billing statements from each of the computer databases of the Marriott, Radisson, and Wyndham hotel chains, and record the nightly room rates. Marriott Radisson Wyndham Sample average Sample standard deviation $170 17.5 $145 10 $150 16.5 a. Describe the sampled population(s). b. Find a point estimate for the average room rate for the Marriott hotel chain. Calculate the margin of error. c. Find a point estimate for the average room rate for the Radisson hotel chain. Calculate the margin of error. d. Find a point estimate for the average room rate for the Wyndham hotel chain. Calculate the margin of error. e. Display the results of parts b, c, and d graphically, using the form shown in Figure 8.5. Use this display to compare the average room rates for the three hotel chains. 8.19 “900” Numbers Radio and television stations often air controversial issues during broadcast time and ask viewers to indicate their agreement or disagreement with a given stand on the issue. A poll is conducted by asking those viewers who agree to call a certain 900 telephone number and those who disagree to call a second 900 telephone number. All respondents pay a fee for their calls. a. Does this polling technique result in a random sample? b. What can be said about the validity of the results of such a survey? Do you need to worry about a margin of error in this case? 8.5 INTERVAL ESTIMATION ❍ 307 8.20 Men On Mars? The Mars twin rovers, Spirit and Opportunity, which roamed the surface of Mars several years ago, found evidence that there was once water on Mars, raising the possibility that there was once life on the planet. Do you think that the United States should pursue a program to send humans to Mars? An opinion poll conducted by the Associated Press indicated that 49% of the 1034 adults surveyed
think that we should pursue such a program.5 a. Estimate the true proportion of Americans who think that the United States should pursue a program to send humans to Mars. Calculate the margin of error. b. The question posed in part a was only one of many questions concerning our space program that were asked in the opinion poll. If the Associated Press wanted to report one sampling error that would be valid for the entire poll, what value should they report? 8.21 Hungry Rats In an experiment to assess the strength of the hunger drive in rats, 30 previously trained animals were deprived of food for 24 hours. At the end of the 24-hour period, each animal was put into a cage where food was dispensed if the animal pressed a lever. The length of time the animal continued pressing the bar (although receiving no food) was recorded for each animal. If the data yielded a sample mean of 19.3 minutes with a standard deviation of 5.2 minutes, estimate the true mean time and calculate the margin of error. INTERVAL ESTIMATION 8.5 like lariat roping: parameter fence post interval estimate lariat An interval estimator is a rule for calculating two numbers—say, a and b—to create an interval that you are fairly certain contains the parameter of interest. The concept of “fairly certain” means “with high probability.” We measure this probability using the conﬁdence coefficient, designated by 1 a. Definition The probability that a conﬁdence interval will contain the estimated parameter is called the conﬁdence coefficient. For example, experimenters often construct 95% conﬁdence intervals. This means that the conﬁdence coefficient, or the probability that the interval will contain the estimated parameter, is.95. You can increase or decrease your amount of certainty by changing the conﬁdence coefficient. Some values typically used by experimenters are.90,.95,.98, and.99. Consider an analogy—this time, throwing a lariat at a fence post. The fence post represents the parameter that you wish to estimate, and the loop formed by the lariat represents the conﬁdence interval. Each time you throw your lariat, you hope to rope the fence post; however, sometimes your lariat misses. In the same way, each time 308 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION you draw a sample and
construct a conﬁdence interval for a parameter, you hope to include the parameter in your interval, but, just like the lariat, sometimes you miss. Your “success rate”—the proportion of intervals that “rope the post” in repeated sampling—is the conﬁdence coefficient. Constructing a Confidence Interval When the sampling distribution of a point estimator is approximately normal, an interval estimator or conﬁdence interval can be constructed using the following reasoning. For simplicity, assume that the conﬁdence coefficient is.95 and refer to Figure 8.6. F IG URE 8. 6 Parameter 1.96 SE ● 95% Parameter Parameter 1.96 SE Estimator • We know that, of all possible values of the estimator that we might select, 95% of them will be in the interval Parameter 1.96 SE shown in Figure 8.6. • Since the value of the parameter is unknown, consider constructing the interval estimator 1.96 SE which has the same width as the ﬁrst interval, but has a variable center. • How often will this interval work properly and enclose the parameter of interest? Refer to Figure 8.7. F IG URE 8. 7 Some 95% conﬁdence intervals ● 95% Interval 1 Interval 2 Interval 3 like a game of ring toss: parameter peg interval estimate ring 8.5 INTERVAL ESTIMATION ❍ 309 The ﬁrst two intervals work properly—the parameter (marked with a dotted line) is contained within both intervals. The third interval does not work, since it fails to enclose the parameter. This happened because the value of the estimator at the center of the interval was too far away from the parameter. Fortunately, values of the estimator only fall this far away 5% of the time—our procedure will work properly 95% of the time! You may want to change the conﬁdence coefficient from (1 a).95 to another conﬁdence level (1 a). To accomplish this, you need to change the value z 1.96, which locates an area.95 in the center of the standard normal curve, to a value of z that locates the area (1 a) in the center of the curve, as shown in Figure 8.8. Since the total area under the curve is
1, the remaining area in the two tails is a, and each tail contains area a/2. The value of z that has “tail area” a/2 to its right is called za/2, and the area between za/2 and za/2 is the conﬁdence coefficient (1 a). Values of za/2 that are typically used by experimenters will become familiar to you as you begin to construct conﬁdence intervals for different practical situations. Some of these values are given in Table 8.2. FI GUR E 8. 8 Location of za/2 ● f(z) (1 – α) α/2 α/2 –z α/2 0 z α/2 z A (1 a)100% LARGE-SAMPLE CONFIDENCE INTERVAL (Point estimator) za/2 (Standard error of the estimator) where za/2 is the z-value with an area a/2 in the right tail of a standard normal distribution. This formula generates two values; the lower conﬁdence limit (LCL) and the upper conﬁdence limit (UCL). TABLE 8.2 ● Values of z Commonly Used for Confidence Intervals Conﬁdence coefficient, (1 a).90.95.98.99 a.10.05.02.01 a/2.05.025.01.005 za/2 1.645 1.96 2.33 2.58 310 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION Large-Sample Confidence Interval for a Population Mean m Practical problems very often lead to the estimation of m, the mean of a population of quantitative measurements. Here are some examples: • The average achievement of college students at a particular university • The average strength of a new type of steel • The average number of deaths per age category • The average demand for a new cosmetics product When the sample size n is large, the sample mean x is the best point estimator for the population mean m. Since its sampling distribution is approximately normal, it can be used to construct a conﬁdence interval according to the general approach given earlier. A (1 a)100% LARGE-SAMPLE CONFIDENCE INTERVAL FOR A POPULATION MEAN m s x za/2 n where za/2 is the z
-value corresponding to an area a/2 in the upper tail of a standard normal z distribution, and n Sample size s Standard deviation of the sampled population If s is unknown, it can be approximated by the sample standard deviation s when the sample size is large (n 30) and the approximate conﬁdence interval is s x za/2 n Another way to ﬁnd the large-sample conﬁdence interval for a population mean m is to begin with the statistic m n / x z s which has a standard normal distribution. If you write za/2 as the value of z with area a/2 to its right, then you can write m x za/2 1 a Pza/2 n s / You can rewrite this inequality as s s za/2 x m za/2 n n s s m x za/2 x za/2 n n EXAMPLE 8.6 A 95% conﬁdence interval tells you that, if you were to construct many of these intervals (all of which would have slightly different endpoints), 95% of them would enclose the population mean. ● FI GUR E 8. 9 Twenty conﬁdence intervals for the mean for Example 8.6 8.5 INTERVAL ESTIMATION ❍ 311 so that Px za/2 s s 1 a m x za/2 n n Both x za/2(s/n) and x za/2(s/n), the lower and upper conﬁdence limits, are actually random quantities that depend on the sample mean x. Therefore, in repeated sampling, the random interval, x za/2(s/n), will contain the population mean m with probability (1 a). A scientist interested in monitoring chemical contaminants in food, and thereby the accumulation of contaminants in human diets, selected a random sample of n 50 male adults. It was found that the average daily intake of dairy products was x 756 grams per day with a standard deviation of s 35 grams per day. Use this sample information to construct a 95% conﬁdence interval for the mean daily intake of dairy products for men. Solution Since the sample size of n 50 is large, the distribution of the sample mean x is approximately normally distributed with mean m and standard error estimated by s/n. The approximate 95% conﬁdence interval is x 1.
96 s n 3 5 756 1.96 0 5 756 9.70 Hence, the 95% conﬁdence interval for m is from 746.30 to 765.70 grams per day. Interpreting the Confidence Interval What does it mean to say you are “95% conﬁdent” that the true value of the population mean m is within a given interval? If you were to construct 20 such intervals, each using different sample information, your intervals might look like those shown in Figure 8.9. Of the 20 intervals, you might expect that 95% of them, or 19 out of 20, will perform as planned and contain m within their upper and lower bounds. 20 16 12 312 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION Remember that you cannot be absolutely sure that any one particular interval contains the mean m. You will never know whether your particular interval is one of the 19 that “worked,” or whether it is the one interval that “missed.” Your conﬁdence in the estimated interval follows from the fact that when repeated intervals are calculated, 95% of these intervals will contain m. You can try this experiment on your own using the Java applet called Interpreting Conﬁdence Intervals. The applet shown in Figure 8.10(a) shows the calculation of a 95% conﬁdence interval for m when n 50 and s 35. For this particular conﬁdence interval, we used the One Sample button. You can see the value of m shown as a vertical green line on your monitor (gray in Figure 8.10). Notice that this conﬁdence interval worked properly and enclosed the vertical line between its upper and lower limits. Figure 8.10(b) shows the calculation of 100 such intervals, using the 100 Samples button. The intervals that fail to work properly are shown in red on your monitor (black in Figure 8.10). How many intervals fail to work? Is it close to the 95% conﬁdence that we claim to have? You will use this applet again for the MyApplet Exercises section at the end of the chapter. F IG URE 8. 10 Interpeting Conﬁdence Intervals applet ● (a) (b) A good conﬁdence interval has two desirable
characteristics: • • It is as narrow as possible. The narrower the interval, the more exactly you have located the estimated parameter. It has a large conﬁdence coefficient, near 1. The larger the conﬁdence coefficient, the more likely it is that the interval will contain the estimated parameter. EXAMPLE 8.7 Construct a 99% conﬁdence interval for the mean daily intake of dairy products for adult men in Example 8.6. 8.5 INTERVAL ESTIMATION ❍ 313 Solution To change the conﬁdence level to.99, you must ﬁnd the appropriate value of the standard normal z that puts area (1 a).99 in the center of the curve. This value, with tail area a/2.005 to its right, is found from Table 8.2 to be z 2.58 (see Figure 8.11). The 99% conﬁdence interval is then x 2.58 s n 756 2.58(4.95) 756 12.77 or 743.23 to 768.77 grams per day. This conﬁdence interval is wider than the 95% conﬁdence interval in Example 8.6. FI GUR E 8. 11 Standard normal values for a 99% conﬁdence interval ● f(z) Right Tail Area.05.025.01.005 z-Value 1.645 1.96 2.33 2.58.005 –2.58.99 0 α/2 =.005 2.58 z The increased width is necessary to increase the conﬁdence, just as you might want a wider loop on your lariat to ensure roping the fence post! The only way to increase the conﬁdence without increasing the width of the interval is to increase the sample size, n. The standard error of x, s SE n measures the variability or spread of the values of x. The more variable the population data, measured by s, the more variable will be x, and the standard error will be larger. On the other hand, if you increase the sample size n, more information is available for estimating m. The estimates should fall closer to m and the standard error will be smaller. You can use the Exploring Conﬁdence Intervals applet, shown in Figure 8.12, to see the effect of changing the
sample size n, the standard deviation s, and the conﬁdence coefficient 1 a on the width of the conﬁdence interval. The conﬁdence intervals of Examples 8.6 and 8.7 are approximate because you substituted s as an approximation for s. That is, instead of the conﬁdence coefficient being.95, the value speciﬁed in the example, the true value of the coefficient may be.92,.94, or.97. But this discrepancy is of little concern from a practical point of view; as far as your “conﬁdence” is concerned, there is little difference among these conﬁdence coefficients. Most interval estimators used in statistics yield approximate conﬁdence intervals because the assumptions upon which they are based are not satisﬁed exactly. Having made this point, we will not continue to refer to conﬁdence intervals as “approximate.” It is of little practical concern as long as the actual conﬁdence coefficient is near the value speciﬁed. 314 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION F IG URE 8. 12 Exploring Conﬁdence Intervals applet ● Large-Sample Confidence Interval for a Population Proportion p Many research experiments or sample surveys have as their objective the estimation of the proportion of people or objects in a large group that possess a certain characteristic. Here are some examples: • The proportion of sales that can be expected in a large number of customer contacts • The proportion of seeds that germinate • The proportion of “likely” voters who plan to vote for a particular political candidate Each is a practical example of the binomial experiment, and the parameter to be estimated is the binomial proportion p. When the sample size is large, the sample proportion, x pˆ n Total number of successes Total number of trials is the best point estimator for the population proportion p. Since its sampling distribution is approximately normal, with mean p and standard error SE pq/n, pˆ can be used to construct a conﬁdence interval according to the general approach given in this section. A (1 a)100% LARGE-SAMPLE CONFIDENCE INTERVAL FOR A POPULATION PROPORTION p q pˆ za/
2p n where za/2 is the z-value corresponding to an area a/2 in the right tail of a standard normal z distribution. Since p and q are unknown, they are estimated using 8.5 INTERVAL ESTIMATION ❍ 315 the best point estimators: pˆ and qˆ. The sample size is considered large when the normal approximation to the binomial distribution is adequate—namely, when npˆ 5 and nqˆ 5. EXAMPLE 8.8 A random sample of 985 “likely” voters—those who are likely to vote in the upcoming election—were polled during a phone-athon conducted by the Republican Party. Of those surveyed, 592 indicated that they intended to vote for the Republican candidate in the upcoming election. Construct a 90% conﬁdence interval for p, the proportion of likely voters in the population who intend to vote for the Republican candidate. Based on this information, can you conclude that the candidate will win the election? Solution The point estimate for p is x 5 9 2.601 pˆ n 5 8 9 and the standard error is (5 qˆ (.601 pˆ ) 8 9 n.399).016 The z-value for a 90% conﬁdence interval is the value that has area a/2.05 in the upper tail of the z distribution, or z.05 1.645 from Table 8.2. The 90% conﬁdence interval for p is thus qˆ pˆ 1.645pˆ n.601.026 or.575 p.627. You estimate that the percentage of likely voters who intend to vote for the Republican candidate is between 57.5% and 62.7%. Will the candidate win the election? Assuming that she needs more than 50% of the vote to win, and since both the upper and lower conﬁdence limits exceed this minimum value, you can say with 90% conﬁdence that the candidate will win. There are some problems, however, with this type of sample survey. What if the voters who consider themselves “likely to vote” do not actually go to the polls? What if a voter changes his or her mind between now and election day? What if a surveyed voter does not respond truthfully when questioned by the campaign worker? The 90% conﬁdence interval you have constructed gives you 90%
conﬁdence only if you have selected a random sample from the population of interest. You can no longer be assured of “90% conﬁdence” if your sample is biased, or if the population of voter responses changes before the day of the election! You may have noticed that the point estimator with its 95% margin of error looks very similar to a 95% conﬁdence interval for the same parameter. This close relationship exists for most of the parameters estimated in this book, but it is not true in general. Sometimes the best point estimator for a parameter does not fall in the middle of the best conﬁdence interval; the best conﬁdence interval may not even be a function of the best point estimator. Although this is a theoretical distinction, you should remember that there is a difference between point and interval estimation, and that the choice between the two depends on the preference of the experimenter. 316 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION 8.5 EXERCISES BASIC TECHNIQUES 8.22 Find and interpret a 95% conﬁdence interval for a population mean m for these values: a. n 36, x 13.1, s2 3.42 b. n 64, x 2.73, s2.1047 8.23 Find a 90% conﬁdence interval for a population mean m for these values: a. n 125, x.84, s2.086 b. n 50, x 21.9, s2 3.44 c. Interpret the intervals found in parts a and b. 8.24 Find a (1 a)100% conﬁdence interval for a population mean m for these values: a. a.01, n 38, x 34, s2 12 b. a.10, n 65, x 1049, s2 51 c. a.05, n 89, x 66.3, s2 2.48 8.25 A random sample of n 300 observations from a binomial population produced x 263 successes. Find a 90% conﬁdence interval for p and interpret the interval. 8.26 Suppose the number of successes observed in n 500 trials of a binomial experiment is 27. Find a 95% conﬁdence interval for p. Why is the conﬁdence interval narrower than the
conﬁdence interval in Exercise 8.25? 8.27 A random sample of n measurements is selected from a population with unknown mean m and known standard deviation s 10. Calculate the width of a 95% conﬁdence interval for m for these values of n: c. n 400 a. n 100 8.28 Compare the conﬁdence intervals in Exercise 8.27. What effect does each of these actions have on the width of a conﬁdence interval? a. Double the sample size b. Quadruple the sample size b. n 200 8.29 Refer to Exercise 8.28. a. Calculate the width of a 90% conﬁdence interval for m when n 100. b. Calculate the width of a 99% conﬁdence interval for m when n 100. c. Compare the widths of 90%, 95%, and 99% conﬁdence intervals for m. What effect does increasing the conﬁdence coefficient have on the width of the conﬁdence interval? APPLICATIONS 8.30 A Chemistry Experiment Due to a variation in laboratory techniques, impurities in materials, and other unknown factors, the results of an experiment in a chemistry laboratory will not always yield the same numerical answer. In an electrolysis experiment, a class measured the amount of copper precipitated from a saturated solution of copper sulfate over a 30-minute period. The n 30 students calculated a sample mean and standard deviation equal to.145 and.0051 mole, respectively. Find a 90% conﬁdence interval for the mean amount of copper precipitated from the solution over a 30-minute period. 8.31 Acid Rain Acid rain, caused by the reaction of certain air pollutants with rainwater, appears to be a growing problem in the northeastern United States. (Acid rain affects the soil and causes corrosion on exposed metal surfaces.) Pure rain falling through clean air registers a pH value of 5.7 (pH is a measure of acidity: 0 is acid; 14 is alkaline). Suppose water samples from 40 rainfalls are analyzed for pH, and x and s are equal to 3.7 and.5, respectively. Find a 99% conﬁdence interval for the mean pH in rainfall and interpret the interval. What assumption must be made for the conﬁdence interval to be valid? 8.32
MP3 Players Do you own an iPod Nano or a Sony Walkman Bean? These and other brands of MP3 players are becoming more and more popular among younger Americans. An iPod survey reported that 54% of 12- to 17-year-olds, 30% of 18- to 34-year-olds, and 13% of 35- to 54-year-olds own MP3 players.6 Suppose that these three estimates are based on random samples of size 400, 350, and 362, respectively. a. Construct a 95% conﬁdence interval estimate for the proportion of 12- to 17-year-olds who own an MP3 player. b. Construct a 95% conﬁdence interval estimate for the proportion of 18- to 34-year-olds who own an MP3 player. 8.33 Hamburger Meat The meat department of a local supermarket chain packages ground beef using meat trays of two sizes: one designed to hold approximately 1 pound of meat, and one that holds approximately 3 pounds. A random sample of 35 packages in the smaller meat trays produced weight measurements with an average of 1.01 pounds and a standard deviation of.18 pound. a. Construct a 99% conﬁdence interval for the average weight of all packages sold in the smaller meat trays by this supermarket chain. b. What does the phrase “99% conﬁdent” mean? c. Suppose that the quality control department of this supermarket chain intends that the amount of ground beef in the smaller trays should be 1 pound on average. Should the conﬁdence interval in part a concern the quality control department? Explain. 8.34 Legal Abortions The results of a Newsweek poll concerning views on abortion given in Exercise 7.66 showed that of n 1002 adults, 39% favored the “right-to-life” stand, while 53% were “pro-choice.”7 The poll reported a margin of error of plus or minus 3%. a. Construct a 90% conﬁdence interval for the proportion of adult Americans who favor the “right-tolife” position. b. Construct a 90% conﬁdence interval for the proportion of adult Americans who favor the “pro-choice” position. 8.35 SUVs A sample survey is designed to estimate the proportion of sports utility vehicles being driven in the state of California. A random
sample of 500 registrations are selected from a Department of Motor Vehicles database, and 68 are classiﬁed as sports utility vehicles. a. Use a 95% conﬁdence interval to estimate the proportion of sports utility vehicles in California. b. How can you estimate the proportion of sports utility vehicles in California with a higher degree of accuracy? (HINT: There are two answers.) 8.36 e-Shopping In a report of why e-shoppers abandon their online sales transactions, Alison Stein Wellner8 found that “pages took too long to load” and “site was so confusing that I couldn’t ﬁnd the product” were the two complaints heard most often. Based on 8.5 INTERVAL ESTIMATION ❍ 317 customers’ responses, the average time to complete an online order form will take 4.5 minutes. Suppose that n 50 customers responded and that the standard deviation of the time to complete an online order is 2.7 minutes. a. Do you think that x, the time to complete the online order form, has a mound-shaped distribution? If not, what shape would you expect? b. If the distribution of the completion times is not normal, you can still use the standard normal distribution to construct a conﬁdence interval for m, the mean completion time for online shoppers. Why? c. Construct a 95% conﬁdence interval for m, the mean completion time for online orders. 8.37 What’s Normal? What is normal, when it comes to people’s body temperatures? A random sample of 130 human body temperatures, provided by Allen Shoemaker9 in the Journal of Statistical Education, had a mean of 98.25 degrees and a standard deviation of 0.73 degrees. a. Construct a 99% conﬁdence interval for the average body temperature of healthy people. b. Does the conﬁdence interval constructed in part a contain the value 98.6 degrees, the usual average temperature cited by physicians and others? If not, what conclusions can you draw? 8.38 Rocking the Vote How likely are you to vote in the next presidential election? A random sample of 300 adults was taken, and 192 of them said that they always vote in presidential elections. a. Construct a 95% conﬁdence interval for the proportion of adult Americans who say they always vote in presidential elections. b. An article
in American Demographics reports this percentage of 67%.10 Based on the interval constructed in part a, would you disagree with their reported percentage? Explain. c. Can we use the interval estimate from part a to estimate the actual proportion of adult Americans who vote in the 2008 presidential election? Why or why not? 318 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION 8.6 ESTIMATING THE DIFFERENCE BETWEEN TWO POPULATION MEANS A problem equally as important as the estimation of a single population mean m for a quantitative population is the comparison of two population means. You may want to make comparisons like these: • The average scores on the Medical College Admission Test (MCAT) for students whose major was biochemistry and those whose major was biology • The average yields in a chemical plant using raw materials furnished by two different suppliers • The average stem diameters of plants grown on two different types of nutrients For each of these examples, there are two populations: the ﬁrst with mean and vari1, and the second with mean and variance m2 and s 2 ance m1 and s 2 2. A random sample of n1 measurements is drawn from population 1, and a second random sample of size n2 is independently drawn from population 2. Finally, the estimates of the population parameters are calculated from the sample data using the estimators x1, s2 1, x2, and s 2 2 as shown in Table 8.3. TABLE 8.3 ● Samples From Two Quantitative Populations Mean Variance Population 1 m1 s 2 1 Population 2 m2 s 2 2 Sample 1 Sample 2 Mean Variance Sample Size x1 s2 1 n1 x2 s2 2 n2 Intuitively, the difference between two sample means would provide the maximum information about the actual difference between two population means, and this is in fact the case. The best point estimator of the difference (m1 m2) between the population means is (x1 x2). The sampling distribution of this estimator is not difficult to derive, but we state it here without proof. PROPERTIES OF THE SAMPLING DISTRIBUTION 1 x— OF (x— SAMPLE MEANS 2), THE DIFFERENCE BETWEEN TWO When independent random samples of n1 and n2 observations have been selected from populations with means m1 and m2 and variances s 2 the sampling distribution of the difference (x1 x2) has the following properties
: 2, respectively, 1 and s 2 1. The mean of (x1 x2) is m1 m2 8.6 ESTIMATING THE DIFFERENCE BETWEEN TWO POPULATION MEANS ❍ 319 and the standard error is SE 2 1 s s 2 2 n1 n2 which can be estimated as SE ns2 s2 1 n 1 2 2 when the sample sizes are large. 2. If the sampled populations are normally distributed, then the sampling distribution of (x1 x2) is exactly normally distributed, regardless of the sample size. 3. If the sampled populations are not normally distributed, then the sampling distribution of (x1 x2) is approximately normally distributed when n1 and n2 are both 30 or more, due to the Central Limit Theorem. Since (m1 m2) is the mean of the sampling distribution, it follows that (x1 x2) is an unbiased estimator of (m1 m2) with an approximately normal distribution when n1 and n2 are large. That is, the statistic z (x1 x2) (m1 m2) ns2 s2 1 n 1 2 2 has an approximately standard normal z distribution, and the general procedures of Section 8.5 can be used to construct point and interval estimates. Although the choice between point and interval estimation depends on your personal preference, most experimenters choose to construct conﬁdence intervals for two-sample problems. The appropriate formulas for both methods are given next. LARGE-SAMPLE POINT ESTIMATION OF (m1 m2) Point estimator: (x1 x2) 95% Margin of error: 1.96 SE 1.96 ns2 s2 1 n 1 2 2 A (1 a)100% LARGE-SAMPLE CONFIDENCE INTERVAL FOR (m1 m2) (x1 x2) za/2 2 ns2 s2 1 n 1 2 The wearing qualities of two types of automobile tires were compared by road-testing samples of n1 n2 100 tires for each type. The number of miles until wearout was deﬁned as a speciﬁc amount of tire wear. The test results are given in Table 8.4. Estimate (m1 m2), the difference in mean miles to wearout, using a 99% conﬁdence interval. Is there a difference in the average wearing quality for the two types
of tires? Right Tail Area.05.025.01.005 z-Value 1.645 1.96 2.33 2.58 EXAMPLE 8.9 320 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION TABLE 8.4 ● Sample Data Summary for Two Types of Tires Tire 1 x1 26,400 miles 1 1,440,000 s2 Tire 2 x2 25,100 miles 2 1,960,000 s2 Solution The point estimate of (m1 m2) is (x1 x2) 26,400 25,100 1300 miles and the standard error of (x1 x2) is estimated as 1,44 SE 0, 0000 1,96 ns2 s2 000 184.4 miles The 99% conﬁdence interval is calculated as (x1 x2) 2.58 ns2 s2 1 n 1 2 2 1300 2.58(184.4) 1300 475.8 or 824.2 (m1 m2) 1775.8. The difference in the average miles to wearout for the two types of tires is estimated to lie between LCL 824.2 and UCL 1775.8 miles of wear. Based on this conﬁdence interval, can you conclude that there is a difference in the average miles to wearout for the two types of tires? If there were no difference in the two population means, then m1 and m2 would be equal and (m1 m2) 0. If you look at the conﬁdence interval you constructed, you will see that 0 is not one of the possible values for (m1 m2). Therefore, it is not likely that the means are the same; you can conclude that there is a difference in the average miles to wearout for the two types of tires. The conﬁdence interval has allowed you to make a decision about the equality of the two population means. The scientist in Example 8.6 wondered whether there was a difference in the average daily intakes of dairy products between men and women. He took a sample of n 50 adult women and recorded their daily intakes of dairy products in grams per day. He did the same for adult men. A summary of his sample results is listed in Table 8.5. Construct a 95% conﬁdence interval for the difference in the average daily intakes of dairy products for men and women. Can you conclude that there is a difference
in the average daily intakes for men and women? If 0 is not in the interval, you can conclude that there is a difference in the population means. EXAMPLE 8.10 TABLE 8.5 ● Sample Values for Daily Intakes of Dairy Products Sample Size Sample Mean Sample Standard Deviation Men Women 50 756 35 50 762 30 8.6 ESTIMATING THE DIFFERENCE BETWEEN TWO POPULATION MEANS ❍ 321 Solution The conﬁdence interval is constructed using a value of z with tail area a/2.025 to its right; that is, z.025 1.96. Using the sample standard deviations to approximate the unknown population standard deviations, the 95% conﬁdence interval is (x1 x2) 1.96 (756 762) 1.963 s2 1 n 1 ns2 12.78 or 18.78 (m1 m2) 6.78. Look at the possible values for (m1 m2) in the conﬁdence interval. It is possible that the difference (m1 m2) could be negative (indicating that the average for women exceeds the average for men), it could be positive (indicating that men have the higher average), or it could be 0 (indicating no difference between the averages). Based on this information, you should not be willing to conclude that there is a difference in the average daily intakes of dairy products for men and women. Examples 8.9 and 8.10 deserve further comment with regard to using sample esti- mates in place of unknown parameters. The sampling distribution of (x1 x2) (m1 m2) 2 1 2 s s 2 n1 n2 has a standard normal distribution for all sample sizes when both sampled populations are normal and an approximate standard normal distribution when the sampled 1 and s 2 populations are not normal but the sample sizes are large ( 30). When s 2 2 are not known and are estimated by the sample estimates s2 2, the resulting statistic will still have an approximate standard normal distribution when the sample sizes are large. The behavior of this statistic when the population variances are unknown and the sample sizes are small will be discussed in Chapter 10. 1 and s2 8.6 EXERCISES BASIC TECHNIQUES 8.39 Independent random samples were selected from populations 1 and 2. The sample sizes, means, and variances are as follows: Population 1 35 12.7 1.
38 2 49 7.4 4.14 Sample Size Sample Mean Sample Variance a. Find a 95% conﬁdence interval for estimating the difference in the population means (m1 m2). Sample Size Sample Mean Sample Variance Population 1 64 2.9 0.83 2 64 5.1 1.67 b. Based on the conﬁdence interval in part a, can you conclude that there is a difference in the means for the two populations? Explain. 8.40 Independent random samples were selected from populations 1 and 2. The sample sizes, means, and variances are as follows: 322 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION a. Find a 90% conﬁdence interval for the difference in the population means. What does the phrase “90% conﬁdent” mean? b. Find a 99% conﬁdence interval for the difference in the population means. Can you conclude that there is a difference in the two population means? Explain. APPLICATIONS 8.41 Selenium A small amount of the trace element selenium, 50–200 micrograms (mg) per day, is considered essential to good health. Suppose that random samples of n1 n2 30 adults were selected from two regions of the United States and that a day’s intake of selenium, from both liquids and solids, was recorded for each person. The mean and standard deviation of the selenium daily intakes for the 30 adults from region 1 were x1 167.1 and s1 24.3 mg, respectively. The corresponding statistics for the 30 adults from region 2 were x2 140.9 and s2 17.6. Find a 95% conﬁdence interval for the difference in the mean selenium intakes for the two regions. Interpret this interval. 8.42 9-1-1 A study was conducted to compare the mean numbers of police emergency calls per 8-hour shift in two districts of a large city. Samples of 100 8-hour shifts were randomly selected from the police records for each of the two regions, and the number of emergency calls was recorded for each shift. The sample statistics are listed here: Region 1 100 2.4 1.44 2 100 3.1 2.64 Sample Size Sample Mean Sample Variance Find a 90% conﬁdence interval for the difference in the mean numbers of police emergency calls per
shift between the two districts of the city. Interpret the interval. 8.43 Teaching Biology In developing a standard for assessing the teaching of precollege sciences in the United States, an experiment was conducted to evaluate a teacher-developed curriculum, “Biology: A Community Context” (BACC) that was standards-based, activity-oriented, and inquiry-centered. This approach was compared to the historical presentation through lecture, vocabulary, and memorized facts. Students were tested on biology concepts that featured biological knowledge and process skills in the traditional sense. The perhaps not-so-startling results from a test on biology concepts, published in The American Biology Teacher, are shown in the following table.11 Pretest: All BACC Classes Pretest: All Traditional Posttest: All BACC Classes Posttest: All Traditional Mean 13.38 14.06 18.5 16.5 Sample Size Standard Deviation 372 368 365 298 5.59 5.45 8.03 6.96 a. Find a 95% conﬁdence interval for the mean score for the posttest for all BACC classes. b. Find a 95% conﬁdence interval for the mean score for the posttest for all traditional classes. c. Find a 95% conﬁdence interval for the difference in mean scores for the posttest BACC classes and the posttest traditional classes. d. Does the conﬁdence interval in c provide evidence that there is a real difference in the posttest BACC and traditional class scores? Explain. Source: From “Performance Assessment of a Standards-Based High School Biology Curriculum,” by W. Leonard, B. Speziale, and J. Pernick in The American Biology Teacher, 2001, 63(5), 310–316. Reprinted by permission of National Association of Biology Teachers. 8.44 Are You Dieting? An experiment was conducted to compare two diets A and B designed for weight reduction. Two groups of 30 overweight dieters each were randomly selected. One group was placed on diet A and the other on diet B, and their weight losses were recorded over a 30-day period. The means and standard deviations of the weight-loss measurements for the two groups are shown in the table. Find a 95% conﬁdence interval for the difference in mean weight loss for the two diets. Interpret your conﬁdence interval. Diet A xA 21
.3 sA 2.6 Diet B xB 13.4 sB 1.9 8.45 Starting Salaries College graduates are getting more for their degrees as starting salaries rise. To compare the starting salaries of college graduates majoring in chemical engineering and computer science, random samples of 50 recent college graduates in each major were selected and the following information obtained. Major Chemical engineering Computer science Mean $53,659 51,042 SD 2225 2375 a. Find a point estimate for the difference in starting salaries of college students majoring in chemical engineering and computer science. What is the margin of error for your estimate? 8.6 ESTIMATING THE DIFFERENCE BETWEEN TWO POPULATION MEANS ❍ 323 b. Based upon the results in part a, do you think that there is a signiﬁcant difference in starting salaries for chemical engineers and computer scientists? Explain. c. Do the intervals in parts a and b contain the value (m1 m2) 0? Why is this of interest to the researcher? 8.46 Biology Skills Refer to Exercise 8.43. In addition to tests involving biology concepts, students were also tested on process skills. The results of pretest and posttest scores, published in The American Biology Teacher, are given below.11 Pretest: All BACC Classes Pretest: All Traditional Posttest: All BACC Classes Posstest: All Traditional Mean 10.52 11.97 14.06 12.96 Sample Size Standard Deviation 395 379 376 308 4.79 5.39 5.65 5.93 a. Find a 95% conﬁdence interval for the mean score on process skills for the posttest for all BACC classes. b. Find a 95% conﬁdence interval for the mean score on process skills for the posttest for all traditional classes. c. Find a 95% conﬁdence interval for the difference in mean scores on process skills for the posttest BACC classes and the posttest traditional classes. d. Does the conﬁdence interval in c provide evidence that there is a real difference in the mean process skills scores between posttest BACC and traditional class scores? Explain. Source: From “Peformance Assessment of a Standards-Based High School Biology Curriculum,” by W. Leonard, B. Speziale, and J. Pernick in The American Biology Teacher, 2001, 63(5),
310–316. Reprinted by permission of National Association of Biology Teachers. 8.47 Hotel Costs Refer to Exercise 8.18. The means and standard deviations for 50 billing statements from each of the computer databases of each of the three hotel chains are given in the table:4 Marriott Radisson Wyndham Sample Average Sample Standard Deviation $170 17.5 $145 10 $160 16.5 a. Find a 95% conﬁdence interval for the difference in the average room rates for the Marriott and the Wyndham hotel chains. b. Find a 99% conﬁdence interval for the difference in the average room rates for the Radisson and the Wyndham hotel chains. d. Do the data indicate a difference in the average room rates between the Marriott and the Wyndham chains? Between the Radisson and the Wyndham chains? 8.48 Noise and Stress To compare the effect of stress in the form of noise on the ability to perform a simple task, 70 subjects were divided into two groups. The ﬁrst group of 30 subjects acted as a control, while the second group of 40 were the experimental group. Although each subject performed the task in the same control room, each of the experimental group subjects had to perform the task while loud rock music was played. The time to ﬁnish the task was recorded for each subject and the following summary was obtained: Control Experimental n x s 30 15 minutes 4 minutes 40 23 minutes 10 minutes a. Find a 99% conﬁdence interval for the difference in mean completion times for these two groups. b. Based on the conﬁdence interval in part a, is there sufficient evidence to indicate a difference in the average time to completion for the two groups? Explain. 8.49 What’s Normal, continued Of the 130 people in Exercise 8.37, 65 were female and 65 were male.9 The means and standard deviation of their temperatures are shown below. Sample Mean Standard Deviation Men 98.11 0.70 Women 98.39 0.74 Find a 95% conﬁdence interval for the difference in the average body temperatures for males versus females. Based on this interval, can you conclude that there is a difference in the average temperatures for males versus females? Explain. 324 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION ESTIMATING THE DIFFERENCE BETWEEN TWO BINOMIAL PROPORT
IONS 8.7 A simple extension of the estimation of a binomial proportion p is the estimation of the difference between two binomial proportions. You may wish to make comparisons like these: • The proportion of defective items manufactured in two production lines • The proportion of female voters and the proportion of male voters who favor an equal rights amendment • The germination rates of untreated seeds and seeds treated with a fungicide These comparisons can be made using the difference ( p1 p2) between two binomial proportions, p1 and p2. Independent random samples consisting of n1 and n2 trials are drawn from populations 1 and 2, respectively, and the sample estimates pˆ1 and pˆ2 are calculated. The unbiased estimator of the difference ( p1 p2) is the sample difference ( pˆ1 pˆ2). PROPERTIES OF THE SAMPLING DISTRIBUTION OF THE DIFFERENCE (pˆ 1 pˆ 2) BETWEEN TWO SAMPLE PROPORTIONS Assume that independent random samples of n1 and n2 observations have been selected from binomial populations with parameters p1 and p2, respectively. The sampling distribution of the difference between sample proportions ( pˆ1 pˆ2) x2 x1 n n 1 2 has these properties: 1. The mean of ( pˆ1 pˆ2) is p1 p2 and the standard error is 2 1 p SE p 2q 1q n n 2 1 which is estimated as ˆ2 ˆ1 pˆ SE pˆ 2q 1q n n 1 2 2. The sampling distribution of ( pˆ1 pˆ2) can be approximated by a normal dis- tribution when n1 and n2 are large, due to the Central Limit Theorem. Although the range of a single proportion is from 0 to 1, the difference between two proportions ranges from 1 to 1. To use a normal distribution to approximate the distribution of ( pˆ1 pˆ2), both pˆ1 and pˆ2 should be approximately normal; that is, n1pˆ1 5, n1qˆ1 5, and n2 pˆ2 5, n2qˆ2 5. The appropriate formulas for point and interval estimation are given next. 8.7 ESTIMATING THE DIFFERENCE BETWEEN TWO BINOMIAL PROPORTIONS ❍
325 LARGE-SAMPLE POINT ESTIMATION OF (p1 p2) Point estimator: ( pˆ1 pˆ2) ˆ2 ˆ1 pˆ 95% Margin of error: 1.96 SE 1.96pˆ 2q 1q n n 1 2 A (1 a)100% LARGE-SAMPLE CONFIDENCE INTERVAL FOR ( p1 p2) ˆ2 ˆ1 pˆ ( pˆ1 pˆ2) za/2pˆ 2q 1q n n 1 2 Assumption: n1 and n2 must be sufficiently large so that the sampling distribution of ( pˆ1 pˆ2) can be approximated by a normal distribution—namely, if n1pˆ1, n1qˆ1, n2pˆ2, and n2qˆ2 are all greater than 5. EXAMPLE 8.11 A bond proposal for school construction will be submitted to the voters at the next municipal election. A major portion of the money derived from this bond issue will be used to build schools in a rapidly developing section of the city, and the remainder will be used to renovate and update school buildings in the rest of the city. To assess the viability of the bond proposal, a random sample of n1 50 residents in the developing section and n2 100 residents from the other parts of the city were asked whether they plan to vote for the proposal. The results are tabulated in Table 8.6. TABLE 8.6 ● Sample Values for Opinion on Bond Proposal Developing Section Rest of the City Sample Size Number Favoring Proposal Proportion Favoring Proposal 50 38.76 100 65.65 1. Estimate the difference in the true proportions favoring the bond proposal with 2. a 99% conﬁdence interval. If both samples were pooled into one sample of size n 150, with 103 in favor of the proposal, provide a point estimate of the proportion of city residents who will vote for the bond proposal. What is the margin of error? Solution 1. The best point estimate of the difference ( p1 p2) is given by ( pˆ1 pˆ2).76.65.11 and the standard error of ( pˆ1 pˆ2) is estimated as ˆ2 (.76 ˆ1 pˆ p
ˆ 2q 1q n n (0.24) (.65 For a 99% conﬁdence interval, z.005 2.58, and the approximate 99% conﬁdence interval is found as.35).0770 ) 0 1 )( 0 5 1 2 326 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION ˆ2 ˆ1 pˆ ( pˆ1 pˆ2) z.005pˆ 2q 1q n n 1 2.11 (2.58)(.0770).11.199 or (.089,.309). Since this interval contains the value ( p1 p2) 0, it is possible that p1 p2, which implies that there may be no difference in the proportions favoring the bond issue in the two sections of the city. If there is no difference in the two proportions, then the two samples are not really different and might well be combined to obtain an overall estimate of the proportion of the city residents who will vote for the bond issue. If both samples are pooled, then n 150 and 2. 0 3 pˆ 1.69 0 5 1 Therefore, the point estimate of the overall value of p is.69, with a margin of error given by 1.96(.69.31) 1.96(.0378).074 ( ) 0 5 1 Notice that.69.074 produces the interval.62 to.76, which includes only proportions greater than.5. Therefore, if voter attitudes do not change adversely prior to the election, the bond proposal should pass by a reasonable majority. 8.7 EXERCISES BASIC TECHNIQUES 8.50 Independent random samples of n1 500 and n2 500 observations were selected from binomial populations 1 and 2, and x1 120 and x2 147 successes were observed. a. What is the best point estimator for the difference ( p1 p2) in the two binomial proportions? b. Calculate the approximate standard error for the statistic used in part a. c. What is the margin of error for this point estimate? 8.51 Independent random samples of n1 800 and n2 640 observations were selected from binomial populations 1 and 2, and x1 337 and x2 374 successes were observed. a. Find a 90% conﬁdence interval for the difference ( p1 p2) in the two population proportions
. Interpret the interval. b. What assumptions must you make for the conﬁ- dence interval to be valid? Are these assumptions met? 8.52 Independent random samples of n1 1265 and n2 1688 observations were selected from binomial populations 1 and 2, and x1 849 and x2 910 successes were observed. a. Find a 99% conﬁdence interval for the difference ( p1 p2) in the two population proportions. What does “99% conﬁdence” mean? b. Based on the conﬁdence interval in part a, can you conclude that there is a difference in the two binomial proportions? Explain. APPLICATIONS 8.53 M&M’S Does Mars, Incorporated use the same proportion of red candies in its plain and peanut varieties? A random sample of 56 plain M&M’S contained 12 red candies, and another random sample of 32 peanut M&M’S contained 8 red candies. a. Construct a 95% conﬁdence interval for the difference in the proportions of red candies for the plain and peanut varieties. 8.7 ESTIMATING THE DIFFERENCE BETWEEN TWO BINOMIAL PROPORTIONS ❍ 327 b. Based on the conﬁdence interval in part a, can you conclude that there is a difference in the proportions of red candies for the plain and peanut varieties? Explain. 8.54 Different Realities The voters in the midterm 2006 elections showed that Democrat and Republican differences extend beyond matters of opinion, and actually include the way they see the world.12 The three most important issues mentioned by Democrats and Republicans are listed below. Republicans (n 995) Democrats (n 1094) 1st Terrorism 42% Iraq war 60% 2nd Economy 41% Economy 44% 3rd Iraq war 37% Health care 44% Use a large-sample estimation procedure to compare the proportions of Republicans and Democrats who mentioned the economy as an important issue in the elections. Explain your conclusions. 8.55 Baseball Fans The ﬁrst day of baseball comes in late March, ending in October with the World Series. Does fan support grow as the season goes on? Two CNN/USA Today/Gallup polls, one conducted in March and one in November, both involved random samples of 1001 adults aged 18 and older. In the March sample, 45% of the
adults claimed to be fans of professional baseball, while 51% of the adults in the November sample claimed to be fans.13 a. Construct a 99% conﬁdence interval for the difference in the proportion of adults who claim to be fans in March versus November. b. Does the data indicate that the proportion of adults who claim to be fans increases in November, around the time of the World Series? Explain. 8.56 Baseball and Steroids Refer to Exercise 8.55. In the March opinion poll, suppose that 451 adults identiﬁed themselves as baseball fans, while the other 550 were not fans. The following question was posed: Should major league baseball players be tested for steroids or other performance enhancing drugs, or not? Suppose that 410 of the baseball fans and 505 of those who are not fans answered yes to this question. a. Construct a 95% conﬁdence interval for the difference in the proportion of adults (fans versus nonfans) who favor mandatory drug testing for professional baseball players. b. Does the data indicate that there is a difference in the proportion of fans versus nonfans who favor mandatory drug testing for professional baseball players? Explain. 8.57 Catching a Cold Do well-rounded people get fewer colds? A study on the Chronicle of Higher Education was conducted by scientists at Carnegie Mellon University, the University of Pittsburgh, and the University of Virginia. They found that people who have only a few social outlets get more colds than those who are involved in a variety of social activities.14 Suppose that of the 276 healthy men and women tested, n1 96 had only a few social outlets and n2 105 were busy with six or more activities. When these people were exposed to a cold virus, the following results were observed: Few Social Outlets Many Social Outlets Sample Size Percent with Colds 96 62% 105 35% a. Construct a 99% conﬁdence interval for the difference in the two population proportions. b. Does there appear to be a difference in the population proportions for the two groups? c. You might think that coming into contact with more people would lead to more colds, but the data show the opposite effect. How can you explain this unexpected ﬁnding? 8.58 Union, Yes! A sampling of political candidates—200 randomly chosen from the West and 200 from the East—was classiﬁed according to whether the candidate received backing by a national labor union
and whether the candidate won. In the West, 120 winners had union backing, and in the East, 142 winners were backed by a national union. Find a 95% conﬁdence interval for the difference between the proportions of union-backed winners in the West versus the East. Interpret this interval. 8.59 Birth Order and College Success In a study of the relationship between birth order and college success, an investigator found that 126 in a sample of 180 college graduates were ﬁrstborn or only children. In a sample of 100 nongraduates of comparable age and socioeconomic background, the number of ﬁrstborn or only children was 54. Estimate the difference between the proportions of ﬁrstborn or only children in the two populations from which these samples were drawn. Use a 90% conﬁdence interval and interpret your results. 8.60 Generation Next Born between 1980 and 1990, Generation Next have lived in a post–Cold War world and a time of relative economic prosperity in America, but they have also experienced 328 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION September 11th and the fear of another attack, two Gulf Wars, the tragedy at Columbine High School, Hurricane Katrina, and the increasing polarization of public discourse. More than any who came before, Generation Next is engaged with technology, and the vast majority is dependent upon it.15 Suppose that a survey of 500 female and 500 male students in Generation Next, 345 of the females and 365 of the males reported that they decided to attend college in order to make more money. a. Construct a 98% conﬁdence interval for the difference in the proportions of female and male students who decided to attend college in order to make more money. b. What does it mean to say that you are “98% conﬁdent”? c. Based on the conﬁdence interval in part a, can you conclude that there is a difference in the proportions of female and male students who decided to attend college in order to make more money? 8.61 Excedrin or Tylenol? In a study to compare the effects of two pain relievers it was found that of n1 200 randomly selectd individuals instructed to use the ﬁrst pain reliever, 93% indicated that it relieved their pain. Of n2 450 randomly selected individuals instructed to use the second pain reliever, 96% indicated that it relieved
their pain. a. Find a 99% conﬁdence interval for the difference in the proportions experiencing relief from pain for these two pain relievers. b. Based on the conﬁdence interval in part a, is there sufficient evidence to indicate a difference in the proportions experiencing relief for the two pain relievers? Explain. 8.62 Auto Accidents Last year’s records of auto accidents occurring on a given section of highway were classiﬁed according to whether the resulting damage was $1000 or more and to whether a physical injury resulted from the accident. The data follows: Under $1000 $1000 or More Number of Accidents Number Involving Injuries 32 10 41 23 a. Estimate the true proportion of accidents involving injuries when the damage was $1000 or more for similar sections of highway and ﬁnd the margin of error. b. Estimate the true difference in the proportion of accidents involving injuries for accidents with damage under $1000 and those with damage of $1000 or more. Use a 95% conﬁdence interval. ONE-SIDED CONFIDENCE BOUNDS 8.8 The conﬁdence intervals discussed in Sections 8.5–8.7 are sometimes called two-sided conﬁdence intervals because they produce both an upper (UCL) and a lower (LCL) bound for the parameter of interest. Sometimes, however, an experimenter is interested in only one of these limits; that is, he needs only an upper bound (or possibly a lower bound) for the parameter of interest. In this case, you can construct a one-sided conﬁdence bound for the parameter of interest, such as m, p, m1 m2, or p1 p2. When the sampling distribution of a point estimator is approximately normal, an argument similar to the one in Section 8.5 can be used to show that one-sided conﬁdence bounds, constructed using the following equations when the sample size is large, will contain the true value of the parameter of interest (1 a)100% of the time in repeated sampling. A (1 a)100% LOWER CONFIDENCE BOUND (LCB) (Point estimator) za (Standard error of the estimator) A (1 a)100% UPPER CONFIDENCE BOUND (UCB) (Point estimator) za (Standard
error of the estimator) The z-value used for a (1 a)100% one-sided conﬁdence bound, za, locates an area a in a single tail of the normal distribution as shown in Figure 8.13. 8.9 CHOOSING THE SAMPLE SIZE ❍ 329 FI GUR E 8. 13 z-value for a one-sided conﬁdence bound ● f(z) EXAMPLE 8.12 α 0 zα z A corporation plans to issue some short-term notes and is hoping that the interest it will have to pay will not exceed 11.5%. To obtain some information about this problem, the corporation marketed 40 notes, one through each of 40 brokerage ﬁrms. The mean and standard deviation for the 40 interest rates were 10.3% and.31%, respectively. Since the corporation is interested in only an upper limit on the interest rates, ﬁnd a 95% upper conﬁdence bound for the mean interest rate that the corporation will have to pay for the notes. Solution Since the parameter of interest is m, the point estimator is x with stans dard error SE. The conﬁdence coefficient is.95, so that a.05 and n z.05 1.645. Therefore, the 95% upper conﬁdence bound is UCB x 1.645 s 1.3 10.3.0806 10.3806 10.3 1.645 0 n 4 Thus, you can estimate that the mean interest rate that the corporation will have to pay on its notes will be less than 10.3806%. The corporation should not be concerned about its interest rates exceeding 11.5%. How conﬁdent are you of this conclusion? Fairly conﬁdent, because intervals constructed in this manner contain m 95% of the time. CHOOSING THE SAMPLE SIZE 8.9 Designing an experiment is essentially a plan for buying a certain amount of information. Just as the price you pay for a video game varies depending on where and when you buy it, the price of statistical information varies depending on how and where the information is collected. As when you buy any product, you should buy as much statistical information as you can for the minimum possible cost. The total amount of relevant information in a sample is controlled by two factors: • The sampling plan or experimental design: the procedure for
collecting the information • The sample size n: the amount of information you collect 330 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION You can increase the amount of information you collect by increasing the sample size, or perhaps by changing the type of sampling plan or experimental design you are using. We will discuss the simplest sampling plan—random sampling from a relatively large population—and focus on ways to choose the sample size n needed to purchase a given amount of information. A researcher makes little progress in planning an experiment before encountering the problem of sample size. How many measurements should be included in the sample? How much information does the researcher want to buy? The total amount of information in the sample will affect the reliability or goodness of the inferences made by the researcher, and it is this reliability that the researcher must specify. In a statistical estimation problem, the accuracy of the estimate is measured by the margin of error or the width of the conﬁdence interval. Since both of these measures are a function of the sample size, specifying the accuracy determines the necessary sample size. For instance, suppose you want to estimate the average daily yield m of a chemical process and you need the margin of error to be less than 4 tons. This means that, approximately 95% of the time in repeated sampling, the distance between the sample mean x and the population mean m will be less than 1.96 SE. You want this quantity to be less than 4. That is, s 4 n or 1.96 1.96 SE 4 Solving for n, you obtain 962 n 1. s 2 or 4 n.24s 2 If you know s, the population standard deviation, you can substitute its value into the formula and solve for n. If s is unknown—which is usually the case—you can use the best approximation available: • An estimate s obtained from a previous sample • A range estimate based on knowledge of the largest and smallest possible measurements: s Range/4 For this example, suppose that a prior study of the chemical process produced a sample standard deviation of s 21 tons. Then n.24s 2.24(21)2 105.8 Using a sample of size n 106 or larger, you could be reasonably certain (with probability approximately equal to.95) that your estimate of the average yield will be within 4 tons of the actual average yield. The solution n 106 is only approximate because you had to use an approximate value for s to calculate the standard error of the mean. Although this may
bother you, it is the best method available for selecting the sample size, and it is certainly better than guessing! Sometimes researchers request a different conﬁdence level than the 95% conﬁdence speciﬁed by the margin of error. In this case, the half-width of the conﬁdence interval provides the accuracy measure for your estimate; that is, the bound B on the error of your estimate is s B n za/2 This method for choosing the sample size can be used for all four estimation procedures presented in this chapter. The general procedure is described next. 8.9 CHOOSING THE SAMPLE SIZE ❍ 331 How Do I Choose the Sample Size? Determine the parameter to be estimated and the standard error of its point estimator. Then proceed as follows: 1. Choose B, the bound on the error of your estimate, and a conﬁdence coefficient (1 a). 2. For a one-sample problem, solve this equation for the sample size n: za/2 (Standard error of the estimator) B where za/2 is the value of z having area a/2 to its right. 3. For a two-sample problem, set n1 n2 n and solve the equation in step 2. [NOTE: For most estimators (all presented in this textbook), the standard error is a function of the sample size n.] Exercise Reps Fill in the blanks in the table below and ﬁnd the necessary sample sizes. The ﬁrst problem has been done for you. p or s p.4 Bound, Solve This Inequality B q.1 1.96p n.1 Sample Size n ———93 Type of Data One or Two Samples Margin of Error Binomial 1 Quantitative 2 q 1.96p n s 1.96 n s 6 1 s1 s2 6 2 2 p1 p2.4.05 1 p 1.96p 2q 1q n n 2 1 1 2 n n1 n2.05 n1 n2 Progress Report • Still having trouble with sample sizes? Try again using the Exercise Reps at the end of this section. • Mastered the sample size problem? You can skip the Exercise Reps at the end of this section! Answers are located on the perforated card at the back of this book. EXAMPLE 8.13 Producers of polyvinyl plastic pipe
want to have a supply of pipes sufficient to meet marketing needs. They wish to survey wholesalers who buy polyvinyl pipe in order to estimate the proportion who plan to increase their purchases next year. What sample size is required if they want their estimate to be within.04 of the actual proportion with probability equal to.90? 332 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION Solution For this particular example, the bound B on the error of the estimate is.04. Since the conﬁdence coefficient is (1 a).90, a must equal.10 and a/2 is.05. The z-value corresponding to an area equal to.05 in the upper tail of the z distribution is z.05 1.645. You then require 1.645 SE 1.645p q.04 n In order to solve this equation for n, you must substitute an approximate value of p into the equation. If you want to be certain that the sample is large enough, you should use p.5 (substituting p.5 will yield the largest possible solution for n because the maximum value of pq occurs when p q.5). Then 1.645(.5) (.5).04 n or )(.5) 20.56 5 4 n (1.6 0 4. n (20.56)2 422.7 Therefore, the producers must include at least 423 wholesalers in its survey if it wants to estimate the proportion p correct to within.04. EXAMPLE 8.14 A personnel director wishes to compare the effectiveness of two methods of training industrial employees to perform a certain assembly operation. A number of employees are to be divided into two equal groups: the ﬁrst receiving training method 1 and the second training method 2. Each will perform the assembly operation, and the length of assembly time will be recorded. It is expected that the measurements for both groups will have a range of approximately 8 minutes. For the estimate of the difference in mean times to assemble to be correct to within 1 minute with a probability equal to.95, how many workers must be included in each training group? Solution Letting B 1 minute, you get 2 1 1.96 1 s s 2 2 n1 n2 Since you wish n1 to equal n2, you can let n1 n2 n and obtain the equation 1.96 1 2 1 s s 2 2 n n As noted above, the variability (range)
of each method of assembly is approximately 2 s 2. Since the range, equal to 8 minutes, is approx1 s 2 the same, and hence s 2 imately equal to 4s, you have 4s 8 or s 2 8.9 CHOOSING THE SAMPLE SIZE ❍ 333 Substituting this value for s1 and s2 in the earlier equation, you get 1 (2 1.96(2 )2 )2 n n 1.968 1 n n 1.968 Solving, you have n 31. Thus, each group should contain at least n 31 workers. Table 8.7 provides a summary of the formulas used to determine the sample sizes required for estimation with a given bound on the error of the estimate or conﬁdence interval width W (W 2B). Notice that to estimate p, the sample size formula uses s 2 pq, whereas to estimate ( p1 p2), the sample size formula uses s 2 1 p1q1 and s 2 2 p2q2. TABLE 8.7 ● Sample Size Formulas Parameter Estimator Sample Size Assumptions m x m1 m2 x1 x2 p pˆ p1 p2 pˆ1 pˆ2 s 2 n z 2 a /2 2 B a/2(s s 2 n z 2 2 2) n1 n2 n 1 2 B n z 2 pq a /2 2 B 2 )z n (.25 a/2 2 B p.5 p2q2) n1 n2 n n z 2 a/2(p1q 1 B 2 or or z 2 5) n 2(.2 a/2 2 B n1 n2 n p1 p2.5 and 8.9 EXERCISES EXERCISE REPS These exercises refer back to the MyPersonal Trainer section on page 331. 8.63 Fill in the blanks in the table below and ﬁnd the necessary sample sizes. Type of Data One or Two Samples Margin of Error Binomial 1 s 1.96 n p or s p.5 s 10 Bound, Solve This Inequality B Sample Size.05 2.05 n 2 n 334 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION 8.64 Fill in the blanks in the table below and ﬁnd the necessary sample sizes. Type of Data One or Two Samples Margin of Error Quantitative 2 Bound, Sol
ve This p or s Inequality B s1 s2 10 4 4 Sample Size n1 n2 2 p1 p2.5 1 p 1.96p 2q 1q n n 1 2.10.10 n1 n2 BASIC TECHNIQUES 8.65 Find a 90% one-sided upper conﬁdence bound for the population mean m for these values: a. n 40, s2 65, x 75 b. n 100, s 2.3, x 1.6 8.66 Find a 99% lower conﬁdence bound for the binomial proportion p when a random sample of n 400 trials produced x 196 successes. 8.67 Independent random samples of size 50 are drawn from two quantitative populations, producing the sample information in the table. Find a 95% upper conﬁdence bound for the difference in the two population means. Sample 1 Sample 2 Sample Size Sample Mean Sample Standard Deviation 50 12 5 50 10 7 8.68 Suppose you wish to estimate a population mean based on a random sample of n observations, and prior experience suggests that s 12.7. If you wish to estimate m correct to within 1.6, with probability equal to.95, how many observations should be included in your sample? 8.69 Suppose you wish to estimate a binomial parameter p correct to within.04, with probability equal to.95. If you suspect that p is equal to some value between.1 and.3 and you want to be certain that your sample is large enough, how large should n be? (HINT: When calculating the standard error, use the value of p in the interval.1 p.3 that will give the largest sample size.) 8.70 Independent random samples of n1 n2 n observations are to be selected from each of two populations 1 and 2. If you wish to estimate the difference between the two population means correct to within 1 s 2.17, with probability equal to.90, how large should n1 and n2 be? Assume that you know s 2 2 27.8. 8.71 Independent random samples of n1 n2 n observations are to be selected from each of two binomial populations 1 and 2. If you wish to estimate the difference in the two population proportions correct to within.05, with probability equal to.98, how large should n be? Assume that you have no prior information on the values of p1 and p2, but
you want to make certain that you have an adequate number of observations in the samples. APPLICATIONS 8.72 Operating Expenses A random sampling of a company’s monthly operating expenses for n 36 months produced a sample mean of $5474 and a standard deviation of $764. Find a 90% upper conﬁdence bound for the company’s mean monthly expenses. 8.73 Legal Immigration Exercise 8.17 discussed a research poll conducted for Fox News by Opinion Dynamics concerning opinions about the number of legal immigrants entering the United States.3 Suppose you were designing a poll of this type. a. Explain how you would select your sample. What problems might you encounter in this process? b. If you wanted to estimate the percentage of the population who agree with a particular statement in your survey questionnaire correct to within 1% with probability.95, approximately how many people would have to be polled? 8.74 Political Corruption A questionnaire is designed to investigate attitudes about political corruption in government. The experimenter would like to survey two different groups—Republicans and Democrats—and compare the responses to various “yes/no” questions for the two groups. The experimenter requires that the sampling error for the difference in the proportion of yes responses for the two groups is no more than 3 percentage points. If the two samples are the same size, how large should the samples be? 8.75 Less Red Meat! Americans are becoming more conscious of the importance of good nutrition, and some researchers believe that we may be altering our diets to include less red meat and more fruits and vegetables. To test this theory, a researcher decides to select hospital nutritional records for subjects surveyed 10 years ago and to compare the average amount of beef consumed per year to the amounts consumed by an equal number of subjects she will interview this year. She knows that the amount of beef consumed annually by Americans ranges from 0 to approximately 104 pounds. How many subjects should the researcher select for each group if she wishes to estimate the difference in the average annual per-capita beef consumption correct to within 5 pounds with 99% conﬁdence? 8.76 Red Meat, continued Refer to Exercise 8.75. The researcher selects two groups of 400 subjects each and collects the following sample information on the annual beef consumption now and 10 years ago: Ten Years Ago This Year Sample Mean Sample Standard Deviation 73 25 63 28 a. The researcher would like to show that per-capita beef consumption has decreased in the last 10 years, so
she needs to show that the difference in the averages is greater than 0. Find a 99% lower conﬁdence bound for the difference in the average per-capita beef consumptions for the two groups. b. What conclusions can the researcher draw using the conﬁdence bound from part a? 8.77 Hunting Season If a wildlife service wishes to estimate the mean number of days of hunting per hunter for all hunters licensed in the state during a given season, with a bound on the error of estimation equal to 2 hunting days, how many hunters must be 8.9 CHOOSING THE SAMPLE SIZE ❍ 335 included in the survey? Assume that data collected in earlier surveys have shown s to be approximately equal to 10. 8.78 Polluted Rain Suppose you wish to estimate the mean pH of rainfalls in an area that suffers heavy pollution due to the discharge of smoke from a power plant. You know that s is in the neighborhood of.5 pH, and you wish your estimate to lie within.1 of m, with a probability near.95. Approximately how many rainfalls must be included in your sample (one pH reading per rainfall)? Would it be valid to select all of your water specimens from a single rainfall? Explain. 8.79 pH in Rainfall Refer to Exercise 8.78. Suppose you wish to estimate the difference between the mean acidity for rainfalls at two different locations, one in a relatively unpolluted area along the ocean and the other in an area subject to heavy air pollution. If you wish your estimate to be correct to the nearest.1 pH, with probability near.90, approximately how many rainfalls (pH values) would have to be included in each sample? (Assume that the variance of the pH measurements is approximately.25 at both locations and that the samples will be of equal size.) 8.80 GPAs You want to estimate the difference in grade point averages between two groups of college students accurate to within.2 grade point, with probability approximately equal to.95. If the standard deviation of the grade point measurements is approximately equal to.6, how many students must be included in each group? (Assume that the groups will be of equal size.) 8.81 Selenium, again Refer to the comparison of the daily adult intake of selenium in two different regions of the United States in Exercise 8.41. Suppose you wish to estimate the difference in the mean daily intakes between the two regions correct to within
5 micrograms, with probability equal to.90. If you plan to select an equal number of adults from the two regions (i.e., n1 n2), how large should n1 and n2 be? 336 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION CHAPTER REVIEW Key Concepts and Formulas I. Types of Estimators 1. Point estimator: a single number is calculated to estimate the population parameter. 2. Interval estimator: two numbers are calculated to form an interval that, with a certain amount of conﬁdence, contains the parameter. II. Properties of Good Estimators 1. Unbiased: the average value of the estimator equals the parameter to be estimated. 2. Minimum variance: of all the unbiased estimators, the best estimator has a sampling distribution with the smallest standard error. 3. The margin of error measures the maximum distance between the estimator and the true value of the parameter. III. Large-Sample Point Estimators To estimate one of four population parameters when the sample sizes are large, use the following point estimators with the appropriate margins of error. Parameter Point Estimator m p x x pˆ n m1 m2 x1 x2 p1 p 2 (pˆ1 pˆ 2) x2 x1 n n 1 2 95% Margin of Error s 1.96 n 1.96pˆ qˆ n 2 s 1.96s ˆ2 ˆ1 pˆ 1.96pˆ 2 1 n n1 2 2 1q n 1 2q n 2 IV. Large-Sample Interval Estimators To estimate one of four population parameters when the sample sizes are large, use the following interval estimators. Parameter m p m1 m2 p1 p2 (1 a)100% Conﬁdence Interval s x za/2 n pˆ za/2pˆ qˆ n (x1 x2) za/2s (pˆ 1 pˆ 2) za/2pˆ 2 s ˆ2 ˆ1 pˆ 2 1 n1 n 2 2 2q n 2 1q n 1 1. All values in the interval are possible values for the unknown population parameter. 2. Any values outside the interval are unlikely to be the value of the unknown parameter. 3. To
compare two population means or propor- tions, look for the value 0 in the conﬁdence interval. If 0 is in the interval, it is possible that the two population means or proportions are equal, and you should not declare a difference. If 0 is not in the interval, it is unlikely that the two means or proportions are equal, and you can conﬁdently declare a difference. V. One-Sided Confidence Bounds Use either the upper () or lower () twosided bound, with the critical value of z changed from za/2 to za. VI. Choosing the Sample Size 1. Determine the size of the margin of error, B, that you are willing to tolerate. 2. Choose the sample size by solving for n or n n1 n2 in the inequality: za/2 B, where SE is a function of the sample size n. 3. For quantitative populations, estimate the population standard deviation using a previously calculated value of s or the range approximation s Range/4. 4. For binomial populations, use the conservative approach and approximate p using the value p.5. Supplementary Exercises 8.82 State the Central Limit Theorem. Of what value is the Central Limit Theorem in large-sample statistical estimation? 8.83 A random sample of n 64 observations has a mean x 29.1 and a standard deviation s 3.9. a. Give the point estimate of the population mean m and ﬁnd the margin of error for your estimate. b. Find a 90% conﬁdence interval for m. What does “90% conﬁdent” mean? c. Find a 90% lower conﬁdence bound for the population mean m. Why is this bound different from the lower conﬁdence limit in part b? d. How many observations do you need to estimate m to within.5, with probability equal to.95? 8.84 Independent random samples of n1 50 and n2 60 observations were selected from populations 1 and 2, respectively. The sample sizes and computed sample statistics are given in the table: Sample Size Sample Mean Sample Standard Deviation Population 1 5 100.4 0.8 2 60 96.2 1.3 Find a 90% conﬁdence interval for the difference in population means and interpret the interval. 8.85 Refer to Exercise 8.84. Suppose you wish
to estimate (m1 m2) correct to within.2, with probability equal to.95. If you plan to use equal sample sizes, how large should n1 and n2 be? 8.86 A random sample of n 500 observations from a binomial population produced x 240 successes. a. Find a point estimate for p, and ﬁnd the margin of error for your estimator. b. Find a 90% conﬁdence interval for p. Interpret this interval. 8.87 Refer to Exercise 8.86. How large a sample is required if you wish to estimate p correct to within.025, with probability equal to.90? 8.88 Independent random samples of n1 40 and n2 80 observations were selected from binomial populations 1 and 2, respectively. The number of successes in the two samples were x1 17 and x2 23. Find a 99% conﬁdence interval for the difference between the two binomial population proportions. Interpret this interval. 8.89 Refer to Exercise 8.88. Suppose you wish to estimate (p1 p2) correct to within.06, with probability equal to.99, and you plan to use equal sample sizes—that is, n1 n2. How large should n1 and n2 be? 8.90 Ethnic Cuisine Ethnic groups in America buy differing amounts of various food products because of their ethnic cuisine. Asians buy fewer canned vegetables than do other groups, and Hispanics purchase more cooking oil. A researcher interested in market segmentation for these two groups would like to estimate the proportion of households that select certain brands for various products. If the researcher wishes these estimates to be within.03 with probability.95, SUPPLEMENTARY EXERCISES ❍ 337 how many households should she include in the samples? Assume that the sample sizes are equal. 8.91 Women on Wall Street Women on Wall Street can earn large salaries, but may need to make sacriﬁces in their personal lives. In fact, many women in the securities industry have to make signiﬁcant personal sacriﬁces. A survey of 482 women and 356 men found that only half of the women have children, compared to three-quarters of the men surveyed.16 a. What are the values of pˆ1 and pˆ2 for the women and men in this survey? b. Find a 95% conﬁdence interval for the
difference in the proportion of women and men on Wall Street who have children. c. What conclusions can you draw regarding the groups compared in part b? 8.92 Smoking and Blood Pressure An experiment was conducted to estimate the effect of smoking on the blood pressure of a group of 35 cigarette smokers. The difference for each participant was obtained by taking the difference in the blood pressure readings at the beginning of the experiment and again ﬁve years later. The sample mean increase, measured in millimeters of mercury, was x 9.7. The sample standard deviation was s 5.8. Estimate the mean increase in blood pressure that one would expect for cigarette smokers over the time span indicated by the experiment. Find the margin of error. Describe the population associated with the mean that you have estimated. 8.93 Blood Pressure, continued Using a conﬁdence coefficient equal to.90, place a conﬁdence interval on the mean increase in blood pressure for Exercise 8.92. 8.94 Iodine Concentration Based on repeated measurements of the iodine concentration in a solution, a chemist reports the concentration as 4.614, with an “error margin of.006.” a. How would you interpret the chemist’s “error margin”? b. If the reported concentration is based on a random sample of n 30 measurements, with a sample standard deviation s.017, would you agree that the chemist’s “error margin” is.006? 8.95 Heights If it is assumed that the heights of men are normally distributed with a standard deviation of 2.5 inches, how large a sample should be taken to be fairly sure (probability.95) that the sample mean does 338 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION not differ from the true mean (population mean) by more than.50 in absolute value? 8.96 Chicken Feed An experimenter fed different rations, A and B, to two groups of 100 chicks each. Assume that all factors other than rations are the same for both groups. Of the chicks fed ration A, 13 died, and of the chicks fed ration B, 6 died. a. Construct a 98% conﬁdence interval for the true difference in mortality rates for the two rations. b. Can you conclude that there is a difference in the mortality rates for the two rations? 8.97 Antibiotics You want to estimate the
mean hourly yield for a process that manufactures an antibiotic. You observe the process for 100 hourly periods chosen at random, with the results x 34 ounces per hour and s 3. Estimate the mean hourly yield for the process using a 95% conﬁdence interval. 8.98 Cheese and Soda The average American has become accustomed to eating away from home, especially at fast-food restaurants. Partly as a result of this fast-food habit, the per-capita consumption of cheese (the main ingredient in pizza) and nondiet soft drinks has risen dramatically from a decade ago. A study in American Demographics reports that the average American consumes 25.7 pounds of cheese and drinks 40 gallons (or approximately 645 8-ounce servings) of nondiet soft drinks per year.17 To test the accuracy of these reported averages, a random sample of 40 consumers is selected, and these summary statistics are recorded: Cheese (lb/yr) Soft Drinks (gal/yr) Sample Mean Sample Standard Deviation 28.1 3.8 39.2 4.5 Use your knowledge of statistical estimation to estimate the average per-capita annual consumption for these two products. Does this sample cause you to support or to question the accuracy of the reported averages? Explain. 8.99 Healthy Eating Don’t Americans know that eating pizza and french fries leads to being overweight? In the same American Demographics article referenced in Exercise 8.98, a survey of women who are the main meal preparers in their households reported these results: • 90% know that obesity causes health problems. • 80% know that high fat intake may lead to health problems. • 86% know that cholesterol is a health problem. • 88% know that sodium may have negative effects on health. a. Suppose that this survey was based on a random sample of 750 women. How accurate do you expect the percentages given above to be in estimating the actual population percentages? (HINT: If these are the only four percentages for which you need a margin of error, a conservative estimate for p is p.80.) b. If you want to decrease your sampling error to 1%, how large a sample should you take? 8.100 Sunﬂowers In an article in the Annals of Botany, a researcher reported the basal stem diameters of two groups of dicot sunﬂowers: those that were left to sway freely in the wind and those that were artiﬁcially supported.18 A similar
experiment was conducted for monocot maize plants. Although the authors measured other variables in a more complicated experimental design, assume that each group consisted of 64 plants (a total of 128 sunﬂower and 128 maize plants). The values shown in the table are the sample means plus or minus the standard error. Free-Standing Supported Sunﬂower Maize 35.3.72 32.1.72 16.2.41 14.6.40 Use your knowledge of statistical estimation to compare the free-standing and supported basal diameters for the two plants. Write a paragraph describing your conclusions, making sure to include a measure of the accuracy of your inference. 8.101 A Female President? For a number of years, nearly all Americans say that they would vote for a woman for president IF she were qualiﬁed, and IF she were from their own political party. But is America ready for a female president? A CBS/New York Times poll asked this question of a random sample of 1229 adults, with the following results: 19 % Responding “Yes” Now 1999 55% 60 51 48 61 55 48% 46 49 47 44 54 Total Men Women Republicans Democrats Independents a. Construct a 95% conﬁdence interval for the proportion of all Americans who now believe that America is ready for a female president. b. If there were n1 610 men and n2 619 women in the sample, construct a 95% conﬁdence interval for the difference in the proportion of men and women who now believe that America is ready for a female president. Can you conclude that the proportion of men who now believe that America is ready for a female president is larger than the proportion of women? Explain. c. Look at the percentages of “yes” responses for Republicans, Democrats and Independents now compared to the percentages in 1999. Can you think of a reason why the percentage of Democrats might have changed so dramatically? 8.102 College Costs A dean of men wishes to estimate the average cost of the freshman year at a particular college correct to within $500, with a probability of.95. If a random sample of freshmen is to be selected and each asked to keep ﬁnancial data, how many must be included in the sample? Assume that the dean knows only that the range of expenditures will vary from approximately $4800 to $13,000. 8.103 Quality Control A quality-control engineer wants to estimate
the fraction of defectives in a large lot of ﬁlm cartridges. From previous experience, he feels that the actual fraction of defectives should be somewhere around.05. How large a sample should he take if he wants to estimate the true fraction to within.01, using a 95% conﬁdence interval? 8.104 Circuit Boards Samples of 400 printed circuit boards were selected from each of two production lines A and B. Line A produced 40 defectives, and line B produced 80 defectives. Estimate the difference in the actual fractions of defectives for the two lines with a conﬁdence coefficient of.90. 8.105 Circuit Boards II Refer to Exercise 8.104. Suppose 10 samples of n 400 printed circuit boards were tested and a conﬁdence interval was constructed for p for each of the ten samples. What is the probability that exactly one of the intervals will not contain the true value of p? That at least one interval will not contain the true value of p? 8.106 Ice Hockey The ability to accelerate rapidly is an important attribute for an ice hockey player. G. Wayne Marino investigated some of the variables related to the acceleration and speed of a hockey player from a stopped position.20 Sixty-nine hockey players, varsity and intramural, from the University of Illinois were included in the experiment. Each player SUPPLEMENTARY EXERCISES ❍ 339 was required to move as rapidly as possible from a stopped position to cover a distance of 6 meters. The means and standard deviations of some of the variables recorded for each of the 69 skaters are shown in the table: Mean SD Weight (kilograms) Stide Length (meters) Stride Rate (strides/second) Average Acceleration (meters/second2) Instantaneous Velocity (meters/second) Time to Skate (seconds) 75.270 1.110 3.310 2.962 5.753 1.953 9.470.205.390.529.892.131 a. Give the formula that you would use to construct a 95% conﬁdence interval for one of the population means (e.g., mean time to skate the 6-meter distance). b. Construct a 95% conﬁdence interval for the mean time to skate. Interpret this interval. 8.107 Ice Hockey, continued Exercise 8.106 presented statistics from a study of fast starts by ice hockey sk
aters. The mean and standard deviation of the 69 individual average acceleration measurements over the 6-meter distance were 2.962 and.529 meters per second, respectively. a. Find a 95% conﬁdence interval for this population mean. Interpret the interval. b. Suppose you were dissatisﬁed with the width of this conﬁdence interval and wanted to cut the interval in half by increasing the sample size. How many skaters (total) would have to be included in the study? 8.108 Ice Hockey, continued The mean and standard deviation of the speeds of the sample of 69 skaters at the end of the 6-meter distance in Exercise 8.106 were 5.753 and.892 meters per second, respectively. a. Find a 95% conﬁdence interval for the mean velocity at the 6-meter mark. Interpret the interval. b. Suppose you wanted to repeat the experiment and you wanted to estimate this mean velocity correct to within.1 second, with probability.99. How many skaters would have to be included in your sample? 8.109 School Workers In addition to teachers and administrative staff, schools also have many other employees, including bus drivers, custodians, and cafeteria workers. The average hourly wage is $14.18 for bus drivers, $12.61 for custodians, and $10.33 for cafeteria workers.21 Suppose that a school district 340 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION employs n 36 bus drivers who earn an average of $11.45 per hour with a standard deviation of s $2.84. Find a 95% conﬁdence interval for the average hourly wage of bus drivers in school districts similar to this one. Does your conﬁdence interval enclose the stated average of $14.18? What can you conclude about the hourly wages for bus drivers in this school district? 8.110 Recidivism An experimental rehabilitation technique was used on released convicts. It was shown that 79 of 121 men subjected to the technique pursued useful and crime-free lives for a three-year period following prison release. Find a 95% conﬁdence interval for p, the probability that a convict subjected to the rehabilitation technique will follow a crime-free existence for at least three years after prison release. 8.111 Speciﬁc Gravity If 36 measurements of the speciﬁc gravity of aluminum
had a mean of 2.705 and a standard deviation of.028, construct a 98% conﬁdence interval for the actual speciﬁc gravity of aluminum. 8.112 Audiology Research In a study to establish the absolute threshold of hearing, 70 male college freshmen were asked to participate. Each subject was seated in a soundproof room and a 150 H tone was presented at a large number of stimulus levels in a Exercises 8.116 Refer to the Interpreting Conﬁdence Intervals applet. a. Suppose that you have a random sample of size n 50 from a population with unknown mean m and known standard deviation s 35. Calculate the half width of a 95% conﬁdence interval for m. What would the width of this interval be? b. Use the button to create a single conﬁdence interval for m. What is the width of this interval? Compare your results to the calculation you did in part a. 8.117 Refer to the Interpreting Conﬁdence Intervals applet. a. Use the button to create 10 conﬁdence intervals for m. b. What do you notice about the widths of these intervals? randomized order. The subject was instructed to press a button if he detected the tone; the experimenter recorded the lowest stimulus level at which the tone was detected. The mean for the group was 21.6 db with s 2.1. Estimate the mean absolute threshold for all college freshmen and calculate the margin of error. 8.113 Right- or Left-Handed A researcher classiﬁed his subjects as innately right-handed or lefthanded by comparing thumbnail widths. He took a sample of 400 men and found that 80 men could be classiﬁed as left-handed according to his criterion. Estimate the proportion of all males in the population who would test to be left-handed using a 95% conﬁdence interval. 8.114 The Citrus Red Mite An entomologist wishes to estimate the average development time of the citrus red mite correct to within.5 day. From previous experiments it is known that s is in the neighborhood of 4 days. How large a sample should the entomologist take to be 95% conﬁdent of her estimate? 8.115 The Citrus Red Mite, continued A grower believes that one in
ﬁve of his citrus trees are infected with the citrus red mite mentioned in Exercise 8.114. How large a sample should be taken if the grower wishes to estimate the proportion of his trees that are infected with citrus red mite to within.08? c. How many of the intervals work properly and en- close the true value of m? d. Try this simulation again by clicking the button a few more times and counting the number of intervals that work correctly. Is it close to our 95% conﬁdence level? 8.118 Refer to the Interpreting Conﬁdence Intervals applet. a. Use the button to create one hundred conﬁdence intervals for m. b. What do you notice about the widths of these intervals? c. How many of the intervals work properly and en- close the true value of m? d. Try this simulation again by clicking the button a few more times and counting the number of intervals that work correctly. Is it close to our 95% conﬁdence level? 8.119 Suppose that a random sample of size n is selected from a population with mean m 750 and standard deviation s. The Exploring Conﬁdence Intervals applet shows the sampling distribution of x and a representative conﬁdence interval, calculated as s x za/2 n a. The applet loads with n 50, s 35, and x 756. Calculate the half-width of a 95% conﬁdence interval for m. b. Calculate the upper and lower conﬁdence limits and compare these limits to the endpoints of the interval shown in the applet. c. Does the conﬁdence interval work properly? That is, does it enclose the true value of m 750? 8.120 Refer to the Exploring Conﬁdence Intervals applet. a. Use the applet to ﬁnd the values of za/2 for a 99% conﬁdence interval. For a 95% conﬁdence interval? For a 90% conﬁdence interval? b. What effect does reducing the conﬁdence level have on the width of the conﬁdence interval? CASE STUDY ❍ 341 c. A narrower interval indicates a more precise estimate of m, consisting of a smaller range
of values. To obtain a more precise estimate by using a smaller z-value, what has been sacriﬁced? 8.121 Refer to the Exploring Conﬁdence Intervals applet. a. Move the slider marked “n” from bottom to top. b. What is the effect of increasing the sample size on the standard error of x? On the width of the conﬁdence interval? c. Can you think of a practical explanation for the phenomena you observe in part b? 8.122 Refer to the Exploring Conﬁdence Intervals applet. a. Move the slider marked “sigma” from bottom to top. b. What is the effect of increasing the variability on the standard error of x? On the width of the conﬁdence interval? c. Can you think of a practical explanation for the phenomena you observe in part b? CASE STUDY How Reliable Is That Poll? CBS News: How and Where America Eats When Americans eat out at restaurants, most choose American food; however, tastes for Mexican, Chinese, and Italian food vary from region to region of the U.S. In a CBS telephone survey22 conducted October 30 through November 1, 2005, it was found that 39% of families ate together 7 nights a week, slightly less than the 46% of families who reported eating together 7 nights a week in a 1990 survey by CBS. Most Americans, both men and women, do some of the cooking when meals are cooked at home, as reported in the following table where we compare the number of evening meals personally cooked per week by men and women. Number of Meals Cooked 3 or less 4 or more Men Women 76 33 24 67 How often Americans eat out at restaurants is largely a function of income. “While most households earning over $50,000 got restaurant food for dinner at least once in the last week, 75% of those earning under $15,000 did not do so at all.” Income All Under $15,000 $15–$30,000 $30–$50,000 Over $50,000 None 1–3 Nights 4 or More Nights 47 75 58 59 31 49 19 39 38 64 4 6 3 3 5 342 ❍ CHAPTER 8 LARGE-SAMPLE ESTIMATION In spite of all the negative publicity about obesity and high calories associated with burgers and fries, many Americans continue to eat fast food to save time
within busy schedules. Fast Food Nights With Kids Without Kids Fast Food Nights Men Women 0 47 59 0 46 63 1 30 20 1 28 20 2–3 19 16 2–3 20 15 4 4 5 4 6 2 Fifty-three percent of families with kids ate fast food at least once last week, compared with 41% of families without kids. Furthermore, 54% of men ate fast food at least once last week, compared with only 37% of women. The description of the survey methods that gave rise to this data was stated as follows: “This poll was conducted among a nationwide random sample of 936 adults, interviewed by telephone October 30 through November 1, 2005. The error due to sampling for results based on the entire sample could be plus or minus three percentage points.” 1. Verify the margin of error of 3 percentage points as stated for the sample of n 936 adults. Suppose that the sample contained an equal number of men and women or 468 men and 468 women. What is the margin of error for men and for women? 2. Do the numbers in the tables indicate the number of people/families in the cate- gories? If not, what do those numbers represent? 3. a. Construct a 95% conﬁdence interval for the proportion of Americans who ate together seven nights a week. b. Construct a 95% conﬁdence interval for the difference in the proportion of women and men who personally cook at least 4 meals per week. c. Construct a 95% conﬁdence interval for the proportion of Americans who eat out at restaurants at least once a week. 4. If these questions were asked today, would you expect the responses to be similar to those reported here or would you expect them to differ signiﬁcantly? 9 Large-Sample Tests of Hypotheses © Scott Olson/Getty Images An Aspirin a Day...? Will an aspirin a day reduce the risk of heart attack? A very large study of U.S. physicians showed that a single aspirin taken every other day reduced the risk of heart attack in men by one-half. However, three days later, a British study reported a completely opposite conclusion. How could this be? The case study at the end of this chapter looks at how the studies were conducted, and you will analyze the data using large-sample techniques. GENERAL OBJECTIVE In this chapter, the concept of a statistical test of hypothesis is formally introduced
. The sampling distributions of statistics presented in Chapters 7 and 8 are used to construct largesample tests concerning the values of population parameters of interest to the experimenter. CHAPTER INDEX ● Large-sample test about (m1 m2) (9.4) ● Large-sample test about a population mean m (9.3) ● A statistical test of hypothesis (9.2) ● Testing a hypothesis about (p1 p2) (9.6) ● Testing a hypothesis about a population proportion p (9.5) Rejection Regions, p-Values, and Conclusions How Do I Calculate b? 343 344 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES TESTING HYPOTHESES ABOUT POPULATION PARAMETERS 9.1 In practical situations, statistical inference can involve either estimating a population parameter or making decisions about the value of the parameter. For example, if a pharmaceutical company is fermenting a vat of antibiotic, samples from the vat can be used to estimate the mean potency m for all of the antibiotic in the vat. In contrast, suppose that the company is not concerned about the exact mean potency of the antibiotic, but is concerned only that it meet the minimum government potency standards. Then the company can use samples from the vat to decide between these two possibilities: • The mean potency m does not exceed the minimum allowable potency. • The mean potency m exceeds the minimum allowable potency. The pharmaceutical company’s problem illustrates a statistical test of hypothesis. The reasoning used in a statistical test of hypothesis is similar to the process in a court trial. In trying a person for theft, the court must decide between innocence and guilt. As the trial begins, the accused person is assumed to be innocent. The prosecution collects and presents all available evidence in an attempt to contradict the innocent hypothesis and hence obtain a conviction. If there is enough evidence against innocence, the court will reject the innocence hypothesis and declare the defendant guilty. If the prosecution does not present enough evidence to prove the defendant guilty, the court will ﬁnd him not guilty. Notice that this does not prove that the defendant is innocent, but merely that there was not enough evidence to conclude that the defendant was guilty. We use this same type of reasoning to explain the basic concepts of hypothesis testing. These concepts are used to test the four population parameters discussed in Chapter 8: a single population mean or proportion (m or p) and the difference between two
population means or proportions (m1 m2 or p1 p2). When the sample sizes are large, the point estimators for each of these four parameters have normal sampling distributions, so that all four large-sample statistical tests follow the same general pattern. A STATISTICAL TEST OF HYPOTHESIS 9.2 A statistical test of hypothesis consists of ﬁve parts: 1. The null hypothesis, denoted by H0 2. The alternative hypothesis, denoted by Ha 3. The test statistic and its p-value 4. The rejection region 5. The conclusion When you specify these ﬁve elements, you deﬁne a particular test; changing one or more of the parts creates a new test. Let’s look at each part of the statistical test of hypothesis in more detail. 1–2 Definition The two competing hypotheses are the alternative hypothesis Ha, generally the hypothesis that the researcher wishes to support, and the null hypothesis H0, a contradiction of the alternative hypothesis. 9.2 A STATISTICAL TEST OF HYPOTHESIS ❍ 345 As you will soon see, it is easier to show support for the alternative hypothesis by proving that the null hypothesis is false. Hence, the statistical researcher always begins by assuming that the null hypothesis H0 is true. The researcher then uses the sample data to decide whether the evidence favors Ha rather than H0 and draws one of these two conclusions: • Reject H0 and conclude that Ha is true. • Accept (do not reject) H0 as true. EXAMPLE 9.1 You wish to show that the average hourly wage of carpenters in the state of California is different from $14, which is the national average. This is the alternative hypothesis, written as 2 1 Ha : m 14 The null hypothesis is H0 : m 14 You would like to reject the null hypothesis, thus concluding that the California mean is not equal to $14. EXAMPLE 9.2 A milling process currently produces an average of 3% defectives. You are interested in showing that a simple adjustment on a machine will decrease p, the proportion of defectives produced in the milling process. Thus, the alternative hypothesis is 2 1 Ha : p.03 and the null hypothesis is H0 : p.03 If you can reject H0, you can conclude that the adjusted process produces fewer than 3% defectives. There is a difference in the forms of the alternative hypotheses given in Examples 9.1 and 9
.2. In Example 9.1, no directional difference is suggested for the value of m; that is, m might be either larger or smaller than $14 if Ha is true. This type of test is called a two-tailed test of hypothesis. In Example 9.2, however, you are speciﬁcally interested in detecting a directional difference in the value of p; that is, if Ha is true, the value of p is less than.03. This type of test is called a one-tailed test of hypothesis. The decision to reject or accept the null hypothesis is based on information contained in a sample drawn from the population of interest. This information takes these forms: Two-tailed ⇔ Look for a sign in Ha. One-tailed ⇔ Look for a or sign in Ha. 3 • Test statistic: a single number calculated from the sample data • p-value: a probability calculated using the test statistic Either or both of these measures act as decision makers for the researcher in deciding whether to reject or accept H0. 346 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES EXAMPLE 9.3 For the test of hypothesis in Example 9.1, the average hourly wage x for a random sample of 100 California carpenters might provide a good test statistic for testing H0 : m 14 versus Ha : m 14 If the null hypothesis H0 is true, then the sample mean should not be too far from the population mean m 14. Suppose that this sample produces a sample mean x 15 with standard deviation s 2. Is this sample evidence likely or unlikely to occur, if in fact H0 is true? You can use two measures to ﬁnd out. Since the sample size is large, the sampling distribution of x is approximately normal with mean m 14 and standard error s/n, estimated as 3 4 s 2 SE.2 00 n 1 • The test statistic x 15 lies m 15 x 14 5 z n s /.2 standard deviations from the population mean m. • The p-value is the probability of observing a test statistic as extreme as or more extreme than the observed value, if in fact H0 is true. For this example, we deﬁne “extreme” as far below or far above what we would have expected. That is, p-value P(z 5) P(z 5) 0 The large value of the test statistic and the small p-value mean
that you have observed a very unlikely event, if indeed H0 is true and m 14. How do you decide whether to reject or accept H0? The entire set of values that the test statistic may assume is divided into two sets, or regions. One set, consisting of values that support the alternative hypothesis and lead to rejecting H0, is called the rejection region. The other, consisting of values that support the null hypothesis, is called the acceptance region. For example, in Example 9.1, you would be inclined to believe that California’s average hourly wage was different from $14 if the sample mean is either much less than $14 or much greater than $14. The two-tailed rejection region consists of very small and very large values of x, as shown in Figure 9.1. In Example 9.2, since you want to prove that the percentage of defectives has decreased, you would be inclined to reject H0 for values of pˆ that are much smaller than.03. Only small values of pˆ belong in the left-tailed rejection region shown in Figure 9.2. When the rejection region is in the left tail of the distribution, the test is called a left-tailed test. A test with its rejection region in the right tail is called a right-tailed test. F IG URE 9. 1 Rejection and acceptance regions for Example 9.1 ● Rejection region Acceptance region Rejection region x Critical value $14 Critical value 9.3 A LARGE-SAMPLE TEST ABOUT A POPULATION MEAN ❍ 347 FI GUR E 9.2 Rejection and acceptance regions for Example 9.2 ● Rejection region Acceptance region Critical value.03 p 5 If the test statistic falls into the rejection region, then the null hypothesis is rejected. If the test statistic falls into the acceptance region, then either the null hypothesis is accepted or the test is judged to be inconclusive. We will clarify the different types of conclusions that are appropriate as we consider several practical examples of hypothesis tests. Finally, how do you decide on the critical values that separate the acceptance and rejection regions? That is, how do you decide how much statistical evidence you need before you can reject H0? This depends on the amount of conﬁdence that you, the researcher, want to attach to the test conclusions and the signiﬁcance level a, the risk you are willing to take of making an incorrect decision. Definition A Type I error
for a statistical test is the error of rejecting the null hypothesis when it is true. The level of signiﬁcance (signiﬁcance level) for a statistical test of hypothesis is a P(Type I error) P(falsely rejecting H0) P(rejecting H0 when it is true) This value a represents the maximum tolerable risk of incorrectly rejecting H0. Once this signiﬁcance level is ﬁxed, the rejection region can be set to allow the researcher to reject H0 with a ﬁxed degree of conﬁdence in the decision. In the next section, we will show you how to use a test of hypothesis to test the value of a population mean m. As we continue, we will clarify some of the computational details and add some additional concepts to complete your understanding of hypothesis testing. A LARGE-SAMPLE TEST ABOUT A POPULATION MEAN 9.3 Consider a random sample of n measurements drawn from a population that has mean m and standard deviation s. You want to test a hypothesis of the form† 1 2 The null hypothesis will always have an “equals” sign attached. H0 : m m0 where m0 is some hypothesized value for m, versus a one-tailed alternative hypothesis: Ha : m m0 The subscript zero indicates the value of the parameter speciﬁed by H0. Notice that H0 provides an exact value for the parameter to be tested, whereas Ha gives a range of possible values for m. †Note that if the test rejects the null hypothesis m m0 in favor of the alternative hypothesis m m0, then it will certainly reject a null hypothesis that includes m m0, since this is even more contradictory to the alternative hypothesis. For this reason, in this text we state the null hypothesis for a one-tailed test as m m0 rather than m m0. 348 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES The Essentials of the Test The sample mean x is the best estimate of the actual value of m, which is presently in question. What values of x would lead you to believe that H0 is false and m is, in fact, greater than the hypothesized value? The values of x that are extremely large would imply that m is larger than hypothesized. Hence, you should reject H0 if x is too large. The next problem is
to deﬁne what is meant by “too large.” Values of x that lie too many standard deviations to the right of the mean are not very likely to occur. Those values have very little area to their right. Hence, you can deﬁne “too large” as being too many standard deviations away from m0. But what is “too many”? This question can be answered using the signiﬁcance level a, the probability of rejecting H0 when H0 is true. Remember that the standard error of x is estimated as s SE n Since the sampling distribution of the sample mean x is approximately normal when n is large, the number of standard deviations that x lies from m0 can be measured using the standardized test statistic, m x 0 z n / s which has an approximate standard normal distribution when H0 is true and m m0. The signiﬁcance level a is equal to the area under the normal curve lying above the rejection region. Thus, if you want a.01, you will reject H0 when x is more than 2.33 standard deviations to the right of m0. Equivalently, you will reject H0 if the standardized test statistic z is greater than 2.33 (see Figure 9.3). 3 4 F IG URE 9. 3 The rejection region for a right-tailed test with a.01 ● f(z) α =.01 0 2.33 z Acceptance region Rejection region EXAMPLE 9.4 The average weekly earnings for female social workers is $670. Do men in the same positions have average weekly earnings that are higher than those for women? A random sample of n 40 male social workers showed x $725 and s $102. Test the appropriate hypothesis using a.01. 9.3 A LARGE-SAMPLE TEST ABOUT A POPULATION MEAN ❍ 349 For one-tailed tests, look for directional words like “greater,” “less than,” “higher,” “lower,” etc. Solution You would like to show that the average weekly earnings for men are higher than $670, the women’s average. Hence, if m is the average weekly earnings for male social workers, you can set out the formal test of hypothesis in steps: 1–2 Null and alternative hypotheses: H0: m 670 versus Ha: m 670 3 4 5 If
the test is two-tailed, you will not see any directional words. The experimenter is only looking for a “difference” from the hypothesized value. Test statistic: Using the sample information, with s as an estimate of the population standard deviation, calculate x 0 6 7 6 70 z 3.41 Rejection region: For this one-tailed test, values of x much larger than 670 would lead you to reject H0; or, equivalently, values of the standardized test statistic z in the right tail of the standard normal distribution. To control the risk of making an incorrect decision as a.01, you must set the critical value separating the rejection and acceptance regions so that the area in the right tail is exactly a.01. This value is found in Table 3 of Appendix I to be z 2.33, as shown in Figure 9.3. The null hypothesis will be rejected if the observed value of the test statistic, z, is greater than 2.33. Conclusion: Compare the observed value of the test statistic, z 3.41, with the critical value necessary for rejection, z 2.33. Since the observed value of the test statistic falls in the rejection region, you can reject H0 and conclude that the average weekly earnings for male social workers are higher than the average for female social workers. The probability that you have made an incorrect decision is a.01. If you wish to detect departures either greater or less than m0, then the alternative hypothesis is two-tailed, written as Ha : m m0 which implies either m m0 or m m0. Values of x that are either “too large” or “too small” in terms of their distance from m0 are placed in the rejection region. If you choose a.01, the area in the rejection region is equally divided between the two tails of the normal distribution, as shown in Figure 9.4. Using the standardized test statistic z, you can reject H0 if z 2.58 or z 2.58. For different values of a, the critical values of z that separate the rejection and acceptance regions will change accordingly. FI GUR E 9.4 The rejection region for a two-tailed test with a.01 ● f(z) α/2 =.005 α/2 =.005 –2.58 0 Rejection region 2.58 z Rejection region 350 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES EXAM
PLE 9.5 The daily yield for a local chemical plant has averaged 880 tons for the last several years. The quality control manager would like to know whether this average has changed in recent months. She randomly selects 50 days from the computer database and computes the average and standard deviation of the n 50 yields as x 871 tons and s 21 tons, respectively. Test the appropriate hypothesis using a.05. Solution 1–2 Null and alternative hypotheses: H0: m 880 versus Ha: m 880 3 4 5 Test statistic: The point estimate for m is x. Therefore, the test statistic is m x 80 8 0 3.03 z 0 n 5 / 1/ 2 87 1 s Rejection region: For this two-tailed test, you use values of z in both the right and left tails of the standard normal distribution. Using a.05, the critical values separating the rejection and acceptance regions cut off areas of a/2.025 in the right and left tails. These values are z 1.96 and the null hypothesis will be rejected if z 1.96 or z 1.96. Conclusion: Since z 3.03 and the calculated value of z falls in the rejection region, the manager can reject the null hypothesis that m 880 tons and conclude that it has changed. The probability of rejecting H0 when H0 is true and a.05, a fairly small probability. Hence, she is reasonably conﬁdent that the decision is correct. LARGE-SAMPLE STATISTICAL TEST FOR m 1. Null hypothesis: H0 : m m0 2. Alternative hypothesis: Two-Tailed Test Ha : m m0 One-Tailed Test Ha : m m0 (or, Ha : m m0) m m x 0 0 estimated as z 3. Test statistic. Rejection region: Reject H0 when One-Tailed Test z za (or z za when the alternative hypothesis is Ha : m m0) Two-Tailed Test z za/2 or z za/2 α α/2 0 zα –zα/2 0 α/2 zα/2 9.3 A LARGE-SAMPLE TEST ABOUT A POPULATION MEAN ❍ 351 Assumptions: The n observations in the sample are randomly selected from the population and n is large—say, n 30. Calculating the p-Value In the previous examples, the decision to reject or accept H0 was made by comparing the
calculated value of the test statistic with a critical value of z based on the significance level a of the test. However, different signiﬁcance levels may lead to different conclusions. For example, if in a right-tailed test, the test statistic is z 2.03, you can reject H0 at the 5% level of signiﬁcance because the test statistic exceeds z 1.645. However, you cannot reject H0 at the 1% level of signiﬁcance, because the test statistic is less than z 2.33 (see Figure 9.5). To avoid any ambiguity in their conclusions, some experimenters prefer to use a variable level of signiﬁcance called the p-value for the test. Definition The p-value or observed signiﬁcance level of a statistical test is the smallest value of a for which H0 can be rejected. It is the actual risk of committing a Type I error, if H0 is rejected based on the observed value of the test statistic. The p-value measures the strength of the evidence against H0. In the right-tailed test with observed test statistic z 2.03, the smallest critical value you can use and still reject H0 is z 2.03. For this critical value, the risk of an incorrect decision is P(z 2.03) 1.9788.0212 This probability is the p-value for the test. Notice that it is actually the area to the right of the calculated value of the test statistic. FI GUR E 9.5 Variable rejection regions ● f(z).0500.0212.0100 0 1.645 2.03 2.33 z p-value tail area (one or two tails) “beyond” the observed value of the test statistic A small p-value indicates that the observed value of the test statistic lies far away from the hypothesized value of m. This presents strong evidence that H0 is false and should be rejected. Large p-values indicate that the observed test statistic is not far from the hypothesized mean and does not support rejection of H0. How small does the p-value need to be before H0 can be rejected? 352 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES EXAMPLE 9.6 Definition If the p-value is less than or equal to a preassigned signiﬁcance
level a, then the null hypothesis can be rejected, and you can report that the results are statistically signiﬁcant at level a. In the previous instance, if you choose a.05 as your signiﬁcance level, H0 can be rejected because the p-value is less than.05. However, if you choose a.01 as your signiﬁcance level, the p-value (.0212) is not small enough to allow rejection of H0. The results are signiﬁcant at the 5% level, but not at the 1% level. You might see these results reported in professional journals as signiﬁcant (p.05).† Refer to Example 9.5. The quality control manager wants to know whether the daily yield at a local chemical plant—which has averaged 880 tons for the last several years—has changed in recent months. A random sample of 50 days gives an average yield of 871 tons with a standard deviation of 21 tons. Calculate the p-value for this two-tailed test of hypothesis. Use the p-value to draw conclusions regarding the statistical test. Solution The rejection region for this two-tailed test of hypothesis is found in both tails of the normal probability distribution. Since the observed value of the test statistic is z 3.03, the smallest rejection region that you can use and still reject H0 is z 3.03. For this rejection region, the value of a is the p-value: p-value P(z 3.03) P(z 3.03) (1.9988).0012.0024 Notice that the two-tailed p-value is actually twice the tail area corresponding to the calculated value of the test statistic. If this p-value.0024 is less than or equal to the preassigned level of signiﬁcance a, H0 can be rejected. For this test, you can reject H0 at either the 1% or the 5% level of signiﬁcance. If you are reading a research report, how small should the p-value be before you decide to reject H0? Many researchers use a “sliding scale” to classify their results. • • • • If the p-value is less than.01, H0 is rejected. The results are highly signiﬁcant. If the p-value
is between.01 and.05, H0 is rejected. The results are statistically signiﬁcant. If the p-value is between.05 and.10, H0 is usually not rejected. The results are only tending toward statistical signiﬁcance. If the p-value is greater than.10, H0 is not rejected. The results are not statistically signiﬁcant. EXAMPLE 9.7 Standards set by government agencies indicate that Americans should not exceed an average daily sodium intake of 3300 milligrams (mg). To ﬁnd out whether Americans are exceeding this limit, a sample of 100 Americans is selected, and the mean and standard deviation of daily sodium intake are found to be 3400 mg and 1100 mg, respectively. Use a.05 to conduct a test of hypothesis. †In reporting statistical signiﬁcance, many researchers write (p.05) or (P.05) to mean that the p-value of the test was smaller than.05, making the results signiﬁcant at the 5% level. The symbol p or P in the expression has no connection with our notation for probability or with the binomial parameter p. 9.3 A LARGE-SAMPLE TEST ABOUT A POPULATION MEAN ❍ 353 Solution The hypotheses to be tested are versus Ha : m 3300 H0 : m 3300 and the test statistic is 91 small p-value ⇔ large z-value small p-value ⇒ reject H0 How small? p-value a The two approaches developed in this section yield the same conclusions. • The critical value approach: Since the signiﬁcance level is a.05 and the test is one-tailed, the rejection region is determined by a critical value with tail area equal to a.05; that is, H0 can be rejected if z 1.645. Since z.91 is not greater than the critical value, H0 is not rejected (see Figure 9.6). • The p-value approach: Calculate the p-value, the probability that z is greater than or equal to z.91: p-value P(z.91) 1.8186.1814 The null hypothesis can be rejected only if the p-value is less than or equal to the speciﬁed 5% signiﬁcance level. Therefore, H0 is
not rejected and the results are not statistically signiﬁcant (see Figure 9.6). There is not enough evidence to indicate that the average daily sodium intake exceeds 3300 mg. F IG URE 9. 6 Rejection region and p-value for Example 9.7 ● f(z) p-value =.1814 α =.05 0.91 1.645 z Reject H0 (z > 1.645) You can use the Large-Sample Test of a Population Mean applet to visualize the p-values for either one- or two-tailed tests of the population mean m (Figure 9.7). Remember, however, that these large-sample z-tests are restricted to samples of size n 30. The applet does not prohibit you from entering a value of n 30; you’ll have to be careful to check the sample size before you start! The procedure follows the same pattern as with previous applets. You enter the values of x, n, and s—remember to press “Enter” after each entry to record the changes. The applet will calculate z (using full accuracy) and give you the option of choosing one- or two-tailed p-values (Area to Left, Area to Right, or Two Tails), as well as a Middle area that you will not need. 354 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES F IG URE 9. 7 Large-Sample Test of a Population Mean applet ● For the data of Example 9.7, the p-value is the one-tailed area to the right of z.909. Do the results shown in the applet conﬁrm our conclusions in Example 9.7? Remember that the applet uses full accuracy for the calculation of z and its corresponding probability. This means that the probability we calculate using Table 3 in Appendix I may be slightly different from the probability shown in the applet. Notice that these two approaches are actually the same, as shown in Figure 9.6. As soon as the calculated value of the test statistic z becomes larger than the critical value, za, the p-value becomes smaller than the signiﬁcance level a. You can use the most convenient of the two methods; the conclusions you reach will always be the same! The p-value approach does have two advantages, however: • Statistical output from packages such as MINITAB usually reports
the p-value of the test. • Based on the p-value, your test results can be evaluated using any signiﬁcance level you wish to use. Many researchers report the smallest possible signiﬁcance level for which their results are statistically signiﬁcant. Sometimes it is easy to confuse the signiﬁcance level a with the p-value (or observed signiﬁcance level). They are both probabilities calculated as areas in the tails of the sampling distribution of the test statistic. However, the signiﬁcance level a is preset by the experimenter before collecting the data. The p-value is linked directly to the data and actually describes how likely or unlikely the sample results are, assuming that H0 is true. The smaller the p-value, the more unlikely it is that H0 is true! 9.3 A LARGE-SAMPLE TEST ABOUT A POPULATION MEAN ❍ 355 Rejection Regions, p-Values, and Conclusions The signiﬁcance level, a, lets you set the risk that you are willing to take of making an incorrect decision in a test of hypothesis. • To set a rejection region, choose a critical value of z so that the area in the tail(s) of the z-distribution is (are) either a for a one-tailed test or a/2 for a two-tailed test. Use the right tail for an upper-tailed test and the left tail for a lower-tailed test. Reject H0 when the test statistic exceeds the critical value and falls in the rejection region. • To ﬁnd a p-value, ﬁnd the area in the tail “beyond” the test statistic. If the test is one-tailed, this is the p-value. If the test is two-tailed, this is only half the p-value and must be doubled. Reject H0 when the p-value is less than a. Exercise Reps A. Critical value approach: Fill in the blanks in the table below. The ﬁrst prob- lem has been done for you. Signiﬁcance One or Test Level Statistic a.05 z 1.4 z 2.46 a.01 z 0.74 a.05 z 6.12 a.01 Two-tailed Two-tailed Two-Tailed Test?
One-tailed (upper) One-tailed (upper) Rejection Critical Value Region Conclusion 1.645 z 1.645 Do not reject H0 B. p-value approach: Fill in the blanks in the table below. The ﬁrst problem has been done for you. Test Statistic z 1.4 z 2.46 z 0.74 z 6.12 Signiﬁcance Level a.05 a.01 a.05 a.01 One or Two-Tailed Test? p-Value One-tailed (upper).0808 One-tailed (upper) Two-tailed Two-tailed p-Value b a? No Conclusion Do not reject H0 Progress Report • Still having trouble with p-values and rejection regions? Try again using the Excercise Reps at the end of this section. • Mastered p-values and rejection regions? You can skip the Exercise Reps at the end of this section! Answers are located on the perforated card at the back of this book. 356 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES Two Types of Errors You might wonder why, when H0 was not rejected in the previous example, we did not say that H0 was deﬁnitely true and m 3300. This is because, if we choose to accept H0, we must have a measure of the probability of error associated with this decision. Since there are two choices in a statistical test, there are also two types of errors that can be made. In the courtroom trial, a defendant could be judged not guilty when he’s really guilty, or vice versa—the same is true in a statistical test. In fact, the null hypothesis may be either true or false, regardless of the decision the experimenter makes. These two possibilities, along with the two decisions that can be made by the researcher, are shown in Table 9.1. TABLE 9.1 ● Decision Table Null Hypothesis Decision True False Reject H0 Accept H0 Type I error Correct decision Correct decision Type II error a P (reject H0 when H0 true) b P (accept H0 when H0 false) In addition to the Type I error with probability a deﬁned earlier in this section, it is possible to commit a second error, called a Type II error, which has probability b. Definition A Type I error for a statistical test is the error of rejecting the null
hypothesis when it is true. The probability of making a Type I error is denoted by the symbol a. A Type II error for a statistical test is the error of accepting the null hypothesis when it is false and some alternative hypothesis is true. The probability of making a Type II error is denoted by the symbol b. Notice that the probability of a Type I error is exactly the same as the level of signiﬁcance a and is therefore controlled by the researcher. When H0 is rejected, you have an accurate measure of the reliability of your inference; the probability of an incorrect decision is a. However, the probability b of a Type II error is not always controlled by the experimenter. In fact, when H0 is false and Ha is true, you may not be able to specify an exact value for m, but only a range of values. This makes it difficult, if not impossible, to calculate b. Without a measure of reliability, it is not wise to conclude that H0 is true. Rather than risk an incorrect decision, you should withhold judgment, concluding that you do not have enough evidence to reject H0. Instead of accepting H0, you should not reject or fail to reject H0. Keep in mind that “accepting” a particular hypothesis means deciding in its favor. Regardless of the outcome of a test, you are never certain that the hypothesis you “accept” is true. There is always a risk of being wrong (measured by a or b). Consequently, you never “accept” H0 if b is unknown or its value is unacceptable to you. When this situation occurs, you should withhold judgment and collect more data. The Power of a Statistical Test The goodness of a statistical test is measured by the size of the two error rates: a, the probability of rejecting H0 when it is true, and b, the probability of accepting H0 when H0 is false and Ha is true. A “good” test is one for which both of these error rates are 9.3 A LARGE-SAMPLE TEST ABOUT A POPULATION MEAN ❍ 357 small. The experimenter begins by selecting a, the probability of a Type I error. If he or she also decides to control the value of b, the probability of accepting H0 when Ha is true, then an appropriate sample size is chosen. Another way of evaluating a test is to look at the complement of a Type II error— that is, rejecting H
0 when Ha is true—which has probability 1 b P(reject H0 when Ha is true) The quantity (1 b) is called the power of the test because it measures the probability of taking the action that we wish to have occur—that is, rejecting the null hypothesis when it is false and Ha is true. Definition The power of a statistical test, given as 1 b P(reject H0 when Ha is true) measures the ability of the test to perform as required. A graph of (1 b), the probability of rejecting H0 when in fact H0 is false, as a function of the true value of the parameter of interest is called the power curve for the statistical test. Ideally, you would like a to be small and the power (1 b) to be large. EXAMPLE 9.8 Refer to Example 9.5. Calculate b and the power of the test (1 b) when m is actually equal to 870 tons. Solution The acceptance region for the test of Example 9.5 is located in the interval [m0 1.96(s/n)]. Substituting numerical values, you get 2 1 or 874.18 to 885.82 880 1.96 0 5 The probability of accepting H0, given m 870, is equal to the area under the sampling distribution for the test statistic x in the interval from 874.18 to 885.82. Since x is normally distributed with a mean of 870 and SE 21/50 2.97, b is equal to the area under the normal curve with m 870 located between 874.18 and 885.82 (see Figure 9.8). Calculating the z-values corresponding to 874.18 and 885.82, you get F IGU RE 9.8 Calculating b in Example 9.8 ● f(x) α /2 =.025 Ha true: µ = 870 Ho true: µ = 880 β α /2 =.025 870 874.18 Rejection region µ 0 = 880 Acceptance region 885.82 x Rejection region 358 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES x 874 870 m 8.1 1.41 z1 50 n / s/ 1 m 2.8 5.33 z2 50 n / s/ 1 870 885 2 2 x Then b P(accept H0 when m
870) P(874.18 x 885.82 when m 870) P(1.41 z 5.33) You can see from Figure 9.8 that the area under the normal curve with m 870 above x 885.82 (or z 5.33) is negligible. Therefore, b P(z 1.41) From Table 3 in Appendix I you can ﬁnd b 1.9207.0793 Hence, the power of the test is 1 b 1.0793.9207 The probability of correctly rejecting H0, given that m is really equal to 870, is.9207, or approximately 92 chances in 100. You can use the Power of a z-Test applet to calculate the power for the hypothesis test in Example 9.8, and also for the same test when the sample size is changed. Refer to Figure 9.9. The applet in Figure 9.9 shows a sample size of n 50. The slider at the bottom of the applet allows you to change the true value of m; the power is recalculated as the mean changes. What is the true value of m and the power of the test shown in the applet? Compare this to the value found in Table 9.2. The slider on the left side of the applet allows you to change a, and the slider on the right allows you to change the sample size n. Remember that n must be 30 for the z-test to be appropriate. You will use these applets to explore power using the MyApplet Exercises at the end of the chapter. F IGU RE 9.9 Power of a z-Test applet ● 9.3 A LARGE-SAMPLE TEST ABOUT A POPULATION MEAN ❍ 359 Values of (1 b) can be calculated for various values of ma different from m0 880 to measure the power of the test. For example, if ma 885, b P(874.18 x 885.82 when m 885) P(3.64 z.28).6103 0.6103 and the power is (1 b).3897. Table 9.2 shows the power of the test for various values of ma, and a power curve is graphed in Figure 9.10. Note that the power of the test increases as the distance between ma and m0 increases. The result is a U-shaped curve for this two-tailed test
. TABLE 9.2 ● Value of (1 b) for Various Values of ma for Example 9.8 ma 865 870 872 875 877 880 (1 b).9990.9207.7673.3897.1726.0500 ma 883 885 888 890 895 (1 b).1726.3897.7673.9207.9990 FI GUR E 9.1 0 Power curve for Example 9.8 ● Power, 1 – β 1.0.8.6.4.2 865 870 875 880 885 890 895 µ There are many important links among the two error rates, a and b, the power, (1 b), and the sample size, n. Look at the two curves shown in Figure 9.8. • If a (the sum of the two tail areas in the curve on the right) is increased, the shaded area corresponding to b decreases, and vice versa. • The only way to decrease b for a ﬁxed a is to “buy” more information—that is, increase the sample size n. What would happen to the area b as the curve on the left is moved closer to the curve on the right (m 880)? With the rejection region in the right curve ﬁxed, the value of b will increase. What effect does this have on the power of the test? Look at Figure 9.10. 360 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES You may also want to use the Power of a z-Test applet to help you visualize the following statements: • As the distance between the true (ma) and hypothesized (m0) values of the mean increases, the power (1 b) increases. The test is better at detecting differences when the distance is large. • The closer the true value (ma) gets to the hypothesized value (m0), the less power (1 b) the test has to detect the difference. • The only way to increase the power (1 b) for a ﬁxed a is to “buy” more information—that is, increase the sample size, n. The experimenter must decide on the values of a and b—measuring the risks of the possible errors he or she can tolerate. He or she also must decide how much power is needed to detect differences that
are practically important in the experiment. Once these decisions are made, the sample size can be chosen by consulting the power curves corresponding to various sample sizes for the chosen test. How Do I Calculate b? 1. Find the critical value or values of x used to separate the acceptance and rejec- tion regions. 2. Using one or more values for m consistent with the alternative hypothesis Ha, calculate the probability that the sample mean x falls in the acceptance region. This produces the value b P(accept Ha when m ma). 3. Remember that the power of the test is (1 b). 2.2 EXERCISES 9.3 EXERCISE REPS These exercises refer back to the MyPersonal Trainer section on page 355. 9.1 Critical Value Approach Fill in the blanks in the table below. Test Statistic z 0.88 z 2.67 z 5.05 z 1.22 Signiﬁcance Level a.05 a.05 a.01 a.01 One or Two-Tailed Test? Two-tailed One-tailed (lower) Two-tailed One-tailed (lower) Critical Value Rejection Region Conclusion 9.2 p-value Approach Fill in the blanks in the table below. Test Statistic z 3.01 z 2.47 z 1.30 z 2.88 Signiﬁcance Level a.05 a.05 a.01 a.01 One or Two-Tailed Test? Two-tailed One-tailed (upper) Two-tailed One-tailed (lower) p-Value p-Value b a? Conclusion BASIC TECHNIQUES 9.3 Find the appropriate rejection regions for the large-sample test statistic z in these cases: a. A right-tailed test with a.01 b. A two-tailed test at the 5% signiﬁcance level c. A left-tailed test at the 1% signiﬁcance level d. A two-tailed test with a.01 9.4 Find the p-value for the following large-sample z tests: a. A right-tailed test with observed z 1.15 b. A two-tailed test with observed z 2.78 c. A left-tailed test with observed z 1.81 9.5 For the three tests given in Exercise 9.4, use the p-value to determine the signiﬁcance of the results. Explain what “statistically sign
iﬁcant” means in terms of rejecting or accepting H0 and Ha. 9.6 A random sample of n 35 observations from a quantitative population produced a mean x 2.4 and a standard deviation s.29. Suppose your research objective is to show that the population mean m exceeds 2.3. a. Give the null and alternative hypotheses for the test. b. Locate the rejection region for the test using a 5% signiﬁcance level. c. Find the standard error of the mean. d. Before you conduct the test, use your intuition to decide whether the sample mean x 2.4 is likely or unlikely, assuming that m 2.3. Now conduct the test. Do the data provide sufficient evidence to indicate that m 2.3? 9.7 Refer to Exercise 9.6. a. Calculate the p-value for the test statistic in part d. b. Use the p-value to draw a conclusion at the 5% sig- niﬁcance level. c. Compare the conclusion in part b with the conclusion reached in part d of Exercise 9.6. Are they the same? 9.8 Refer to Exercise 9.6. You want to test H0 : m 2.3 against Ha : m 2.3. a. Find the critical value of x used for rejecting H0. b. Calculate b P(accept H0 when m 2.4). c. Repeat the calculation of b for m 2.3, 2.5, and 2.6. d. Use the values of b from parts b and c to graph the power curve for the test. 9.3 A LARGE-SAMPLE TEST ABOUT A POPULATION MEAN ❍ 361 9.9 A random sample of 100 observations from a quantitative population produced a sample mean of 26.8 and a sample standard deviation of 6.5. Use the p-value approach to determine whether the population mean is different from 28. Explain your conclusions. APPLICATIONS 9.10 Airline Occupancy Rates High airline occupancy rates on scheduled ﬂights are essential to corporate proﬁtability. Suppose a scheduled ﬂight must average at least 60% occupancy in order to be profitable, and an examination of the occupancy rate for 120 10:00 A.M. ﬂights from Atlanta to Dallas showed a mean occupancy per ﬂight of 58% and a standard deviation of 11%. a. If
m is the mean occupancy per ﬂight and if the company wishes to determine whether or not this scheduled ﬂight is unproﬁtable, give the alternative and the null hypotheses for the test. b. Does the alternative hypothesis in part a imply a one- or two-tailed test? Explain. c. Do the occupancy data for the 120 ﬂights suggest that this scheduled ﬂight is unproﬁtable? Test using a.05. 9.11 Hamburger Meat Exercise 8.33 involved the meat department of a local supermarket chain that packages ground beef in trays of two sizes. The smaller tray is intended to hold 1 pound of meat. A random sample of 35 packages in the smaller meat tray produced weight measurements with an average of 1.01 pounds and a standard deviation of.18 pound. a. If you were the quality control manager and wanted to make sure that the average amount of ground beef was indeed 1 pound, what hypotheses would you test? b. Find the p-value for the test and use it to perform the test in part a. c. How would you, as the quality control manager, report the results of your study to a consumer interest group? 9.12 Invasive Species In a study of the pernicious weed giant hogweed, one of the tallest herbaceous species in Europe, Jan Pergl1 and associates compared the density of these plants in both managed and unmanaged sites within the Caucasus region of Russia. In its native area, the average density was found to be 5 plants/m2. In an invaded area in the Czech Republic, 362 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES a sample of n 50 plants produced an average density of 11.17 plants/m2 with a standard deviation of 3.9 plants/m2. a. Does the invaded area in the Czech Republic have an average density of giant hogweed that is different from m 5 at the a.05 level of signiﬁcance? b. What is the p-value associated with the test in part a? Can you reject H0 at the 5% level of signiﬁcance using the p-value? 9.13 Potency of an Antibiotic A drug manufacturer claimed that the mean potency of one of its antibiotics was 80%. A random sample of n 100 capsules were tested and produced a sample mean of x 79.7%
with a standard deviation of s.8%. Do the data present sufficient evidence to refute the manufacturer’s claim? Let a.05. a. State the null hypothesis to be tested. b. State the alternative hypothesis. c. Conduct a statistical test of the null hypothesis and state your conclusion. 9.14 Flextime Many companies are becoming involved in ﬂextime, in which a worker schedules his or her own work hours or compresses work weeks. A company that was contemplating the installation of a ﬂextime schedule estimated that it needed a minimum mean of 7 hours per day per assembly worker in order to operate effectively. Each of a random sample of 80 of the company’s assemblers was asked to submit a tentative ﬂextime schedule. If the mean number of hours per day for Monday was 6.7 hours and the standard deviation was 2.7 hours, do the data provide sufﬁcient evidence to indicate that the mean number of hours worked per day on Mondays, for all of the company’s assemblers, will be less than 7 hours? Test using a.05. 9.15 Does College Pay Off? An article in Time describing various aspects of American life indicated that higher educational achievement paid off! College grads work 7.4 hours per day, fewer than those with less than a college education.2 Suppose that the average work day for a random sample of n 100 individuals who had less than a four-year college education was calculated to be x 7.9 hours with a standard deviation of s 1.9 hours. a. Use the p-value approach to test the hypothesis that the average number of hours worked by individuals having less than a college degree is greater than individuals having a college degree. At what level can you reject H0? b. If you were a college graduate, how would you state your conclusion to put yourself in the best possible light? c. If you were not a college graduate, how might you state your conclusion? 9.16 What’s Normal? What is normal, when it comes to people’s body temperatures? A random sample of 130 human body temperatures, provided by Allen Shoemaker3 in the Journal of Statistical Education, had a mean of 98.25 degrees and a standard deviation of 0.73 degrees. Does the data indicate that the average body temperature for healthy humans is different from 98.6 degrees, the usual average temperature cited by physicians and others? Test using both methods given
in this section. a. Use the p-value approach with a.05. b. Use the critical value approach with a.05. c. Compare the conclusions from parts a and b. Are they the same? d. The 98.6 standard was derived by a German doctor in 1868, who claimed to have recorded 1 million temperatures in the course of his research.4 What conclusions can you draw about his research in light of your conclusions in parts a and b? 9.17 Sports and Achilles Tendon Injuries Some sports that involve a signiﬁcant amount of running, jumping, or hopping put participants at risk for Achilles tendinopathy (AT), an inﬂammation and thickening of the Achilles tendon. A study in The American Journal of Sports Medicine looked at the diameter (in mm) of the affected tendons for patients who participated in these types of sports activities.5 Suppose that the Achilles tendon diameters in the general population have a mean of 5.97 millimeters (mm). When the diameters of the affected tendon were measured for a random sample of 31 patients, the average diameter was 9.80 with a standard deviation of 1.95 mm. Is there sufficient evidence to indicate that the average diameter of the tendon for patients with AT is greater than 5.97 mm? Test at the 5% level of signiﬁcance. 9.4 A LARGE-SAMPLE TEST OF HYPOTHESIS FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS ❍ 363 A LARGE-SAMPLE TEST OF HYPOTHESIS FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS 9.4 In many situations, the statistical question to be answered involves a comparison of two population means. For example, the U.S. Postal Service is interested in reducing its massive 350 million gallons/year gasoline bill by replacing gasoline-powered trucks with electric-powered trucks. To determine whether signiﬁcant savings in operating costs are achieved by changing to electric-powered trucks, a pilot study should be undertaken using, say, 100 conventional gasoline-powered mail trucks and 100 electricpowered mail trucks operated under similar conditions. The statistic that summarizes the sample information regarding the difference in population means (m1 m2) is the difference in sample means (x1 x2). Therefore, in testing whether the difference in sample means indicates that the true difference in population means differs from a speci�
��ed value, (m1 m2) D0, you can use the standard error of (x1 x2), 1 2 estimated by SE s s 2 2 2 ns2 s2 1 n 1 2 n1 n2 in the form of a z-statistic to measure how many standard deviations the difference (x1 x2) lies from the hypothesized difference D0. The formal testing procedure is described next. LARGE-SAMPLE STATISTICAL TEST FOR (m1 m2) 1. Null hypothesis: H0 : (m1 m2) D0, where D0 is some speciﬁed difference that you wish to test. For many tests, you will hypothesize that there is no difference between m1 and m2; that is, D0 0. 2. Alternative hypothesis: One-Tailed Test Two-Tailed Test Ha : (m1 m2) D0 Ha : (m1 m2) D0 [or Ha : (m1 m2) D0] 3. Test statistic: z (x1 ) D0 x2 S E (x1 x2) D0 n s2 1 n 1 s2 2 2 4. Rejection region: Reject H0 when One-Tailed Test Two-Tailed Test z za/2 or z za/2 z za [or z za when the alternative hypothesis is Ha : (m1 m2) D0] or when p-value a 364 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES EXAMPLE 9.9 α α/2 0 zα –zα/2 0 α/2 zα/2 Assumptions: The samples are randomly and independently selected from the two populations and n1 30 and n2 30. To determine whether car ownership affects a student’s academic achievement, two random samples of 100 male students were each drawn from the student body. The grade point average for the n1 100 non-owners of cars had an average and variance equal to x1 2.70 and s2 2.40 for the n2 100 car 1.36, while x2 2.54 and s2 owners. Do the data present sufficient evidence to indicate a difference in the mean achievements between car owners and nonowners of cars? Test using a.05. Solution To detect a difference, if it exists, between the mean academic achievements for non-owners of
cars m1 and car owners m2, you will test the null hypothesis that there is no difference between the means against the alternative hypothesis that (m1 m2 ) 0; that is, H0 : (m1 m2) D0 0 versus Ha : (m1 m2) 0 Substituting into the formula for the test statistic, you get z (x1 x2) D0 ns2 s2 1 n 1 2 2 2.70 2.54 0.4 0 0 1 6.3 0 0 1 1.84 test statistic critical value ⇔ reject H0 • The critical value approach: Using a two-tailed test with signiﬁcance level a.05, you place a/2.025 in each tail of the z distribution and reject H0 if z 1.96 or z 1.96. Since z 1.84 does not exceed 1.96 and is not less than 1.96, H0 cannot be rejected (see Figure 9.11). That is, there is insufficient evidence to declare a difference in the average academic achievements for the two groups. Remember that you should not be willing to accept H0—declare the two means to be the same—until b is evaluated for some meaningful values of (m1 m2). F IG URE 9. 11 Rejection region and p-value for Example 9.9 ● f(z) 1 2 p-value 1 2 p-value –1.84 0 1.84 z Reject Ho (z < –1.96) Reject Ho (z > 1.96) 9.4 A LARGE-SAMPLE TEST OF HYPOTHESIS FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS ❍ 365 • The p-value approach: Calculate the p-value, the probability that z is greater than z 1.84 plus the probability that z is less than z 1.84, as shown in Figure 9.11: p-value P(z 1.84) P(z 1.84) (1.9671).0329.0658 The p-value lies between.10 and.05, so you can reject H0 at the.10 level but not at the.05 level of signiﬁcance. Since the p-value of.0658 exceeds the speciﬁed signiﬁcance level a.05, H0 cannot be
rejected. Again, you should not be willing to accept H0 until b is evaluated for some meaningful values of (m1 m2). Hypothesis Testing and Confidence Intervals Whether you use the critical value or the p-value approach for testing hypotheses about (m1 m2), you will always reach the same conclusion because the calculated value of the test statistic and the critical value are related exactly in the same way that the p-value and the signiﬁcance level a are related. You might remember that the conﬁdence intervals constructed in Chapter 8 could also be used to answer questions about the difference between two population means. In fact, for a two-tailed test, the (1 a)100% conﬁdence interval for the parameter of interest can be used to test its value, just as you did informally in Chapter 8. The value of a indicated by the conﬁdence coefficient in the conﬁdence interval is equivalent to the signiﬁcance level a in the statistical test. For a one-tailed test, the equivalent conﬁdence interval approach would use the one-sided conﬁdence bounds in Section 8.8 with conﬁdence coefficient a. In addition, by using the conﬁdence interval approach, you gain a range of possible values for the parameter of interest, regardless of the outcome of the test of hypothesis. • • If the conﬁdence interval you construct contains the value of the parameter speciﬁed by H0, then that value is one of the likely or possible values of the parameter and H0 should not be rejected. If the hypothesized value lies outside of the conﬁdence limits, the null hypothesis is rejected at the a level of signiﬁcance. Construct a 95% conﬁdence interval for the difference in average academic achievements between car owners and non-owners. Using the conﬁdence interval, can you conclude that there is a difference in the population means for the two groups of students? Solution For the large-sample statistics discussed in Chapter 8, the 95% conﬁdence interval is given as Point estimator 1.96 (Standard error of the estimator) For the difference in two population means, the conﬁdence interval is approximated as EXAMPLE
9.10 (x1 x2) 1.96 (2.70 2.54) 1.96 2 ns2 s2 1 n 1 6.3 0 0 1 2 0.4 0 0 1.16.17 366 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES or.01 (m1 m2).33. This interval gives you a range of possible values for the difference in the population means. Since the hypothesized difference, (m1 m2) 0, is contained in the conﬁdence interval, you should not reject H0. Look at the signs of the possible values in the conﬁdence interval. You cannot tell from the interval whether the difference in the means is negative (), positive (), or zero (0)—the latter of the three would indicate that the two means are the same. Hence, you can really reach no conclusion in terms of the question posed. There is not enough evidence to indicate that there is a difference in the average achievements for car owners versus non-owners. The conclusion is the same one reached in Example 9.9. 9.4 EXERCISES BASIC TECHNIQUES 9.18 Independent random samples of 80 measurements were drawn from two quantitative populations, 1 and 2. Here is a summary of the sample data: Sample 1 Sample 2 80 9.7 38.4 80 11.6 27.9 Sample Size Sample Mean Sample Variance a. If your research objective is to show that m1 is larger than m2, state the alternative and the null hypotheses that you would choose for a statistical test. b. Is the test in part a one- or two-tailed? c. Calculate the test statistic that you would use for the test in part a. Based on your knowledge of the standard normal distribution, is this a likely or unlikely observation, assuming that H0 is true and the two population means are the same? d. p-value approach: Find the p-value for the test. Test for a signiﬁcant difference in the population means at the 1% signiﬁcance level. e. Critical value approach: Find the rejection region when a.01. Do the data provide sufficient evidence to indicate a difference in the population means? 9.19 Independent random samples of 36 and 45 observations are drawn from two quantitative populations, 1 and 2, respectively. The sample data summary is shown here: Sample 1
Sample 2 Sample Size Sample Mean Sample Variance 36 1.24.0560 45 1.31.0540 Do the data present sufficient evidence to indicate that the mean for population 1 is smaller than the mean for population 2? Use one of the two methods of testing presented in this section, and explain your conclusions. 9.20 Suppose you wish to detect a difference between m1 and m2 (either m1 m2 or m1 m2) and, instead of running a two-tailed test using a.05, you use the following test procedure. You wait until you have collected the sample data and have calculated x1 and x2. If x1 is larger than x2, you choose the alternative hypothesis Ha : m1 m2 and run a one-tailed test placing a1.05 in the upper tail of the z distribution. If, on the other hand, x2 is larger than x1, you reverse the procedure and run a one-tailed test, placing a2.05 in the lower tail of the z distribution. If you use this procedure and if m1 actually equals m2, what is the probability a that you will conclude that m1 is not equal to m2 (i.e., what is the probability a that you will incorrectly reject H0 when H0 is true)? This exercise demonstrates why statistical tests should be formulated prior to observing the data. APPLICATIONS 9.21 Cure for the Common Cold? An experiment was planned to compare the mean time (in days) required to recover from a common cold for persons given a daily dose of 4 milligrams (mg) of vitamin C versus those who were not given a vitamin supplement. Suppose that 35 adults were randomly selected for each treatment category and that the mean recovery times and standard deviations for the two groups were as follows: 9.4 A LARGE-SAMPLE TEST OF HYPOTHESIS FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS ❍ 367 No Vitamin Supplement 4 mg Vitamin C Sample Size Sample Mean Sample Standard Deviation 35 6.9 2.9 35 5.8 1.2 a. Suppose your research objective is to show that the use of vitamin C reduces the mean time required to recover from a common cold and its complications. Give the null and alternative hypotheses for the test. Is this a one- or a two-tailed test? b. Conduct the statistical test of the null hypothesis in part a and state your conclusion. Test using a.05. 9.
22 Healthy Eating Americans are becoming more conscious about the importance of good nutrition, and some researchers believe we may be altering our diets to include less red meat and more fruits and vegetables. To test the theory that the consumption of red meat has decreased over the last 10 years, a researcher decides to select hospital nutrition records for 400 subjects surveyed 10 years ago and to compare their average amount of beef consumed per year to amounts consumed by an equal number of subjects interviewed this year. The data are given in the table. Ten Years Ago This Year Sample Mean Sample Standard Deviation 73 25 63 28 a. Do the data present sufficient evidence to indicate that per-capita beef consumption has decreased in the last 10 years? Test at the 1% level of signiﬁcance. b. Find a 99% lower conﬁdence bound for the difference in the average per-capita beef consumptions for the two groups. (This calculation was done as part of Exercise 8.76.) Does your conﬁdence bound conﬁrm your conclusions in part a? Explain. What additional information does the conﬁdence bound give you? 9.23 Lead Levels in Drinking Water Analyses of drinking water samples for 100 homes in each of two different sections of a city gave the following means and standard deviations of lead levels (in parts per million): Section 1 Section 2 Sample Size Mean Standard Deviation 100 34.1 5.9 100 36.0 6.0 a. Calculate the test statistic and its p-value (observed signiﬁcance level) to test for a difference in the two population means. Use the p-value to evaluate the statistical signiﬁcance of the results at the 5% level. b. Use a 95% conﬁdence interval to estimate the difference in the mean lead levels for the two sections of the city. c. Suppose that the city environmental engineers will be concerned only if they detect a difference of more than 5 parts per million in the two sections of the city. Based on your conﬁdence interval in part b, is the statistical signiﬁcance in part a of practical signiﬁcance to the city engineers? Explain. 9.24 Starting Salaries, again In an attempt to compare the starting salaries for college graduates who majored in chemical engineering and computer science (see Exercise 8.45), random samples of 50 recent college graduates
in each major were selected and the following information obtained. Mean Major SD Chemical Engineering Computer Science $53,659 51,042 2225 2375 a. Do the data provide sufficient evidence to indicate a difference in average starting salaries for college graduates who majored in chemical engineering and computer science? Test using a.05. b. Compare your conclusions in part a with the results of part b in Exercise 8.45. Are they the same? Explain. 9.25 Hotel Costs In Exercise 8.18, we explored the average cost of lodging at three different hotel chains.6 We randomly select 50 billing statements from the computer databases of the Marriott, Radisson, and Wyndham hotel chains, and record the nightly room rates. A portion of the sample data is shown in the table. Marriott Radisson Sample Average Sample Standard Deviation $170 17.5 $145 10 a. Before looking at the data, would you have any preconceived idea about the direction of the difference between the average room rates for these two hotels? If not, what null and alternative hypotheses should you test? b. Use the critical value approach to determine if there is a signiﬁcant difference in the average room rates for the Marriott and the Radisson hotel chains. Use a.01. c. Find the p-value for this test. Does this p-value con- ﬁrm the results of part b? 368 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES 9.26 Hotel Costs II Refer to Exercise 9.25. The table below shows the sample data collected to compare the average room rates at the Wyndham and Radisson hotel chains.6 Wyndham Radisson Sample Average Sample Standard Deviation $150 16.5 $145 10 a. Do the data provide sufficient evidence to indicate a difference in the average room rates for the Wyndham and the Radisson hotel chains? Use a.05. b. Construct a 95% conﬁdence interval for the difference in the average room rates for the two chains. Does this interval conﬁrm your conclusions in part a? 9.27 MMT in Gasoline The addition of MMT, a compound containing manganese (Mn), to gasoline as an octane enhancer has caused concern about human exposure to Mn because high intakes have been linked to serious health effects. In a study of ambient air concentrations of ﬁne Mn, Wallace and Slonecker
(Journal of the Air and Waste Management Association) presented the accompanying summary information about the amounts of ﬁne Mn (in nanograms per cubic meter) in mostly rural national park sites and in mostly urban California sites.7 National Parks California Mean Standard Deviation Number of Sites.94 1.2 36 2.8 2.8 26 a. Is there sufficient evidence to indicate that the mean concentrations differ for these two types of sites at the a.05 level of signiﬁcance? Use the largesample z-test. What is the p-value of this test? b. Construct a 95% conﬁdence interval for (m1 m2). Does this interval conﬁrm your conclusions in part a? 9.28 Noise and Stress In Exercise 8.48, you compared the effect of stress in the form of noise on the ability to perform a simple task. Seventy subjects were divided into two groups; the ﬁrst group of 30 subjects acted as a control, while the second group of 40 was the experimental group. Although each subject performed the task in the same control room, each of the experimental group subjects had to perform the task while loud rock music was played. The time to ﬁnish the task was recorded for each subject and the following summary was obtained: Control Experimental n x s 30 15 minutes 4 minutes 40 23 minutes 10 minutes a. Is there sufficient evidence to indicate that the average time to complete the task was longer for the experimental “rock music” group? Test at the 1% level of signiﬁcance. b. Construct a 99% one-sided upper bound for the difference (control experimental) in average times for the two groups. Does this interval conﬁrm your conclusions in part a? 9.29 What’s Normal II Of the 130 people in Exercise 9.16, 65 were female and 65 were male.3 The means and standard deviations of their temperatures are shown below. Men Women Sample Mean Standard Deviation 98.11 0.70 98.39 0.74 a. Use the p-value approach to test for a signiﬁcant difference in the average temperatures for males versus females. b. Are the results signiﬁcant at the 5% level? At the 1% level? A LARGE-SAMPLE TEST OF HYPOTHESIS FOR A BINOMIAL PROPORTION 9.
5 When a random sample of n identical trials is drawn from a binomial population, the sample proportion pˆ has an approximately normal distribution when n is large, with mean p and standard error q SE p n 9.5 A LARGE-SAMPLE TEST OF HYPOTHESIS FOR A BINOMIAL PROPORTION ❍ 369 When you test a hypothesis about p, the proportion in the population possessing a certain attribute, the test follows the same general form as the large-sample tests in Sections 9.3 and 9.4. To test a hypothesis of the form H0 : p p0 versus a one- or two-tailed alternative Ha : p p0 or Ha : p p0 or Ha : p p0 the test statistic is constructed using pˆ, the best estimator of the true population proportion p. The sample proportion pˆ is standardized, using the hypothesized mean and standard error, to form a test statistic z, which has a standard normal distribution if H0 is true. This large-sample test is summarized next. LARGE-SAMPLE STATISTICAL TEST FOR p 1. Null hypothesis: H0 : p p0 2. Alternative hypothesis: One-Tailed Test Two-Tailed Test Ha : p p0 Ha : p p0 (or, Ha : p p0) 3. Test statistic: z pˆ p0 SE pˆ p0 q0 p0 n x with pˆ n where x is the number of successes in n binomial trials.† 4. Rejection region: Reject H0 when One-Tailed Test Two-Tailed Test z za/2 or z za/2 z za (or z za when the alternative hypothesis is Ha : p p0) or when p-value a α α/2 0 zα –zα/2 0 α/2 zα/2 Assumption: The sampling satisﬁes the assumptions of a binomial experiment (see Section 5.2), and n is large enough so that the sampling distribution of pˆ can be approximated by a normal distribution (np0 5 and nq0 5). †An equivalent test statistic can be found by multiplying the numerator and denominator by z by n to obtain p n x 0 z 0q np 0 370 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES EXAMPLE 9.11 Regardless of age
, about 20% of American adults participate in ﬁtness activities at least twice a week. However, these ﬁtness activities change as the people get older, and occasionally participants become nonparticipants as they age. In a local survey of n 100 adults over 40 years old, a total of 15 people indicated that they participated in a ﬁtness activity at least twice a week. Do these data indicate that the participation rate for adults over 40 years of age is signiﬁcantly less than the 20% ﬁgure? Calculate the p-value and use it to draw the appropriate conclusions. Solution Assuming that the sampling procedure satisﬁes the requirements of a binomial experiment, you can answer the question posed using a one-tailed test of hypothesis: H0 : p.2 versus Ha : p.2 Begin by assuming that H0 is true—that is, the true value of p is p0.2. Then pˆ x/n will have an approximate normal distribution with mean p0 and standard error p0q0/n. (NOTE: This is different from the estimation procedure in which the unknown standard error is estimated by pˆqˆ/n.) The observed value of pˆ is 15/100.15 and the test statistic is pˆ p0 q0 p0.15.20.80) (.20 ( ) 0 0 1 1.25 z n p-value a ⇔ reject H0 p-value a ⇔ do not reject H0 The p-value associated with this test is found as the area under the standard normal curve to the left of z 1.25 as shown in Figure 9.12. Therefore, p-value P(z 1.25).1056 F IG URE 9. 12 p-value for Example 9.11 ● f(z) p-value =.1056 –1.25 0 z If you use the guidelines for evaluating p-values, then.1056 is greater than.10, and you would not reject H0. There is insufficient evidence to conclude that the percentage of adults over age 40 who participate in ﬁtness activities twice a week is less than 20%. Statistical Significance and Practical Importance It is important to understand the difference between results that are “signiﬁcant” and results that are practically “important.” In statistical language
, the word signiﬁcant does not necessarily mean “important,” but only that the results could not have occurred by chance. For example, suppose that in Example 9.11, the researcher had used 9.5 A LARGE-SAMPLE TEST OF HYPOTHESIS FOR A BINOMIAL PROPORTION ❍ 371 n 400 adults in her experiment and had observed the same sample proportion. The test statistic is now pˆ p0 q0 p0.15.20.80) (.20 ( ) 0 0 4 2.50 z n with p-value P(z 2.50).0062 Now the results are highly signiﬁcant: H0 is rejected, and there is sufficient evidence to indicate that the percentage of adults over age 40 who participate in physical ﬁtness activities is less than 20%. However, is this drop in activity really important? Suppose that physicians would be concerned only about a drop in physical activity of more than 10%. If there had been a drop of more than 10% in physical activity, this would imply that the true value of p was less than.10. What is the largest possible value of p? Using a 95% upper one-sided conﬁdence bound, you have qˆ pˆ 1.645pˆ n.15 1.645(.15.85) ( ) 0 0 4.15.029 or p.179. The physical activity for adults aged 40 and older has dropped from 20%, but you cannot say that it has dropped below 10%. So, the results, although statistically signiﬁcant, are not practically important. In this book, you will learn how to determine whether results are statistically signiﬁcant. When you use these procedures in a practical situation, however, you must also make sure the results are practically important. 9.5 EXERCISES BASIC TECHNIQUES 9.30 A random sample of n 1000 observations from a binomial population produced x 279. a. If your research hypothesis is that p is less than.3, what should you choose for your alternative hypothesis? Your null hypothesis? b. What is the critical value that determines the rejec- tion region for your test with a.05? c. Do the data provide sufficient evidence to indicate that p is less than.3? Use a 5% signiﬁcance
level. 9.31 A random sample of n 1400 observations from a binomial population produced x 529. a. If your research hypothesis is that p differs from.4, what hypotheses should you test? b. Calculate the test statistic and its p-value. Use the p-value to evaluate the statistical signiﬁcance of the results at the 1% level. c. Do the data provide sufficient evidence to indicate that p is different from.4? 9.32 A random sample of 120 observations was selected from a binomial population, and 72 successes were observed. Do the data provide sufficient evidence to indicate that p is greater than.5? Use one of the two methods of testing presented in this section, and explain your conclusions. APPLICATIONS 9.33 Childhood Obesity According to PARADE magazine’s “What America Eats” survey involving 372 ❍ CHAPTER 9 LARGE-SAMPLE TESTS OF HYPOTHESES n 1015 adults, almost half of parents say their children’s weight is ﬁne.8 Only 9% of parents describe their children as overweight. However, the American Obesity Association says the number of overweight children and teens is at least 15%. Suppose that the number of parents in the sample is n 750 and the number of parents who describe their children as overweight is x 68. a. How would you proceed to test the hypothesis that the proportion of parents who describe their children as overweight is less than the actual proportion reported by the American Obesity Association? b. What conclusion are you able to draw from these data at the a.05 level of signiﬁcance? c. What is the p-value associated with this test? 9.34 Plant Genetics A peony plant with red petals was crossed with another plant having streaky petals. A geneticist states that 75% of the offspring resulting from this cross will have red ﬂowers. To test this claim, 100 seeds from this cross were collected and germinated, and 58 plants had red petals. a. What hypothesis should you use to test the geneti- cist’s claim? b. Calculate the test statistic and its p-value. Use the p-value to evaluate the statistical signiﬁcance of the results at the 1% level. 9.35 Early Detection of Breast Cancer Of those women who are diagnosed to have early-stage breast cancer, one-third eventually die of

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