Hello I need help in answering these questions for a discussion In…

Question Answered step-by-step Hello I need help in answering these questions for a discussion In… Hello I need help in answering these questions for a discussion In order to do so i will have to read chapter 3 is the one I will leave here for you to help me please and thank you .  1) What are the easiest and hardest parts of this chapter? Why?2) What do you think is the most important concept in this chapter? Why?3) Discuss a ‘real life ‘ situation where the concepts in this chapter would be used.  CHAPTER 3 CENTRAL TENDENCY  Tools You Will NeedThe following items are considered essential background material for this chapter. If you doubt your knowledge of any of these items, you should review the appropriate chapter or section before proceeding.Summation notation (Chapter 1)Frequency distributions (Chapter 2)PreviewResearch has now confirmed what you already suspected to be true—alcohol consumption increases the attractiveness of opposite-sex individuals (Jones, Jones, Thomas, and Piper, 2003). In the study, college-age participants were recruited from bars and restaurants near campus and asked to participate in a “market research” study. During the introductory conversation, they were asked to report their alcohol consumption for the day and were told that moderate consumption would not prevent them from taking part in the study. Participants were then shown a series of photographs of male and female faces and asked to rate the attractiveness of each face on a 1-7 scale. Figure 3.1 shows the general pattern of results obtained in the study. The two polygons in the figure show the distributions of attractiveness ratings for one male photograph obtained from two groups of females: those who had no alcohol and those with moderate alcohol consumption. Note that the attractiveness ratings from the alcohol group are noticeably higher than the ratings from the no-alcohol group. Incidentally, the same pattern of results was obtained for the male’s ratings of female photographs. Figure 3.1.DetailsFrequency distributions for ratings of attractiveness of a male face shown in a photograph for two groups of female participants: those who had consumed no alcohol and those who had consumed moderate amounts of alcohol.Although it seems obvious that the alcohol-based ratings are noticeably higher than the no-alcohol ratings, this conclusion is based on a general impression, or a subjective interpretation, of the figure. In fact, this conclusion is not always true. For example, there is overlap between the two groups so that some of the no-alcohol females actually rate the photograph as more attractive than some of the alcohol-consuming females. What we need is a method to summarize each group as a whole so that we can objectively describe how much difference exists between the two groups.The solution to this problem is to identify the typical or average rating as the representative for each group. Then the research results can be described by saying that the typical rating for the alcohol group is higher than the typical rating for the no-alcohol group. On average, the male in the photograph really is seen as more attractive by the alcohol-consuming females.In this chapter we introduce the statistical techniques used to identify the typical or average score for a distribution. Although there are several reasons for defining the average score, the primary advantage of an average is that it provides a single number that describes an entire distribution and can be used for comparison with other distributions.  3.1. OverviewThe general purpose of descriptive statistical methods is to organize and summarize a set of scores. Perhaps the most common method for summarizing and describing a distribution is to find a single value that defines the average score and can serve as a typical example to represent the entire distribution. In statistics, the concept of an average or representative score is called central tendency. The goal in measuring central tendency is to describe a distribution of scores by determining a single value that identifies the center of the distribution. Ideally, this central value will be the score that is the best representative value for all of the individuals in the distribution.DefinitionCentral tendency is a statistical measure to determine a single score that defines the center of a distribution. The goal of central tendency is to find the single score that is most typical or most representative of the entire group.In everyday language, central tendency attempts to identify the “average” or “typical” individual. This average value can then be used to provide a simple description of an entire population or a sample. In addition to describing an entire distribution, measures of central tendency are also useful for making comparisons between groups of individuals or between sets of data. For example, weather data indicate that for Seattle, Washington, the average yearly temperature is 53° and the average annual precipitation is 34 inches. By comparison, the average temperature in Phoenix, Arizona, is 71° and the average precipitation is 7.4 inches. The point of these examples is to demonstrate the great advantage of being able to describe a large set of data with a single, representative number. Central tendency characterizes what is typical for a large population and in doing so makes large amounts of data more digestible. Statisticians sometimes use the expression “number crunching” to illustrate this aspect of data description. That is, we take a distribution consisting of many scores and “crunch” them down to a single value that describes them all.Unfortunately, there is no single, standard procedure for determining central tendency. The problem is that no single measure produces a central, representative value in every situation. The three distributions shown in Figure 3.2 should help demonstrate this fact. Before we discuss the three distributions, take a moment to look at the figure and try to identify the “center” or the “most representative score” for each distribution.The first distribution (Figure 3.2(a)) is symmetrical, with the scores forming a distinct pile centered around X=5 . For this type of distribution, it is easy to identify the “center,” and most people would agree that the value X=5 is an appropriate measure of central tendency.In the second distribution (Figure 3.2(b)), however, problems begin to appear. Now the scores form a negatively skewed distribution, piling up at the high end of the scale around X=8 , but tapering off to the left all the way down to X=1 . Where is the “center” in this case? Some people might select X=8 as the center because more individuals had this score than any other single value. However, X=8 is clearly not in the middle of the distribution. In fact, the majority of the scores (10 out of 16) have values less than 8, so it seems reasonable that the “center” should be defined by a value that is less than 8.Now consider the third distribution (Figure 3.2(c)). Again, the distribution is symmetrical, but now there are two distinct piles of scores. Because the distribution is symmetrical with X=5 as the midpoint, you may choose X=5 as the “center.” However, none of the scores is located at X=5 (or even close), so this value is not particularly good as a representative score. On the other hand, because there are two separate piles of scores with one group centered at X=2 and the other centered at X=8 , it is tempting to say that this distribution has two centers. But can one distribution have two centers? Three distributions demonstrating the difficulty of defining central tendency. In each case, try to locate the “center” of the distribution.Clearly, there can be problems defining the “center” of a distribution. Occasionally, you will find a nice, neat distribution like the one shown in Figure 3.2(a), for which everyone will agree on the center. But you should realize that other distributions are possible and that there may be different opinions concerning the definition of the center. To deal with these problems, statisticians have developed three different methods for measuring central tendency: the mean, the median, and the mode. They are computed differently and have different characteristics. To decide which of the three measures is best for any particular distribution, you should keep in mind that the general purpose of central tendency is to find the single most representative score. Each of the three measures we present has been developed to work best in a specific situation. We examine this issue in more detail after we introduce the three measures. 3.2. The MeanLearning ObjectivesDefine the mean and calculate both the population mean and the sample mean.Explain the alternative definitions of the mean as the amount each individual receives when the total is divided equally and as a balancing point.Calculate a weighted mean.Find n, ΣX , and M using scores in a frequency distribution table.Describe how the mean is affected by each of the following: changing a score, adding or removing a score, adding or subtracting a constant from each score, and multiplying or dividing each score by a constant.The mean, also known as the arithmetic average, is computed by adding all the scores in the distribution and dividing by the number of scores. The mean for a population is identified by the Greek letter mu, μ (pronounced “mew”), and the mean for a sample is identified by Mor X¯ (read “x-bar”).The convention in many statistics textbooks is to use X¯ to represent the mean for a sample. However, in manuscripts and in published research reports the letter M is the standard notation for a sample mean. Because you will encounter the letter M when reading research reports and because you should use the letter M when writing research reports, we have decided to use the same notation in this text. Keep in mind that the X¯ notation is still appropriate for identifying a sample mean, and you may find it used on occasion, especially in textbooks.DefinitionThe mean for a distribution is the sum of the scores divided by the number of scores.The formula for the population mean is(3.1). μ=ΣXNFirst, add all the scores in the population, and then divide by N. For a sample, the computation exactly the same, but the formula for the sample mean uses symbols that signify sample values:(3.2). sample mean=M=ΣXnIn general, we use Greek letters to identify characteristics of a population (parameters) and letters of our own alphabet to stand for sample values (statistics). If a mean is identified with the symbol M, you should realize that we are dealing with a sample. Also note that the equation for the sample mean uses a lowercase n as the symbol for the number of scores in the sample.  Image transcription textCisco Finesse x Discussion X *frequency hw.docx X Unit 2 AS1 Central Tender X Ebooks -Webreader.io X Homework Help – Q&A fr x + X – > C Aebooks.cenreader.com/#!/reader/bfa11fb6-56da-4843-bee-17c20d06eed2/page… Show more… Show more   Alternative Definitions for the MeanAlthough the procedure of adding the scores and dividing by the number of scores provides a useful definition of the mean, there are two alternative definitions that may give you a better understanding of this important measure of central tendency.  Image transcription textCisco Finesse x Discussion X *frequency hw.docx * *Unit 2 AS1 Central Tender X Ebooks -Webreader.io X Homework Help – Q&A fr x + X – > C Aebooks.cenreader.com/#!/reader/bfa11fb6-56da-4843-bee-17c20d06eed2/page… Show more… Show more Image transcription textCisco Finesse x Discussion X *frequency hw.docx * *Unit 2 AS1 Central Tender X Ebooks -Webreader.io X Homework Help – Q&A fr x + X – > C Aebooks.cenreader.com/#!/reader/bfa11fb6-56da-4843-bee-17c20d06eed2/page… Show more… Show more Image transcription textCisco Finesse x Discussion X *frequency hw.docx * *Unit 2 AS1 Central Tender X Ebooks -Webreader.io X Homework Help – Q&A fr x + X – > C Aebooks.cenreader.com/#!/reader/bfa11fb6-56da-4843-bee-17c20d06eed2/page… Show more… Show moreThe reason the seesaw is balanced over the mean becomes clear when we measures the distance of each box (score) from the mean:ScoreDistance from the MeanX=14 points below the meanX=23 points below the meanX=61 point above the meanX=61 point above the meanX=105 points above the mean Notice that the mean balances the distances. That is, the total distance below the mean is the same as the total distance above the mean:below the mean:4+3=7⁢pointsabove the mean:1+1+5=7pointsBecause the mean serves as a balance point, the value of the mean will always be located somewhere between the highest score and the lowest score; that is, the mean can never be outside the range of scores. If the lowest score in a distribution is X=8 and the highest isX=15 , then the mean must be between 8 and 15. If you calculate a value that is outside this range, then you have made an error. The image of a seesaw with the mean at the balance point is also useful for determining how a distribution is affected if a new score is added or if an existing score is removed. For the distribution in Figure 3.3, for example, what would happen to the mean (balance point) if a new score were added at X=10 ? The Weighted MeanOften it is necessary to combine two sets of scores and then find the overall mean for the combined group. Suppose, for example, that we begin with two separate samples. The first sample has n=12 scores and a mean of M=6 . The second sample has n=8 andM=7 . If the two samples are combined, what is the mean for the total group?To calculate the overall mean, we need two values:the overall sum of the scores for the combined group (ΣX) , andthe total number of scores in the combined group (n).The total number of scores in the combined group can be found easily by adding the number of scores in the first sample (n1) and the number in the second sample (n2) . In this case, there are 12 scores in the first sample and 8 in the second, for a total of 12+8=20scores in the combined group. Similarly, the overall sum for the combined group can be found by adding the sum for the first sample (ΣX1) and the sum for the second sample (ΣX2). With these two values, we can compute the mean using the basic equationImage transcription textCisco Finesse x Discussion X *frequency hw.docx * *Unit 2 AS1 Central Tender X Ebooks -Webreader.io X Homework Help – Q&A fr x + X – > C Aebooks.cenreader.com/#!/reader/bfa11fb6-56da-4843-bee-17c20d06eed2/page… Show more… Show moreThe following table summarizes the calculations.First SampleSecond SampleCombined Samplen=12n=8n=20⁢(12+8)ΣX=72ΣX=56ΣX=128⁢(72+56)M=6M=7M=6.4Note that the overall mean is not halfway between the original two sample means. Because the samples are not the same size, one makes a larger contribution to the total group and therefore carries more weight in determining the overall mean. For this reason, the overall mean we have calculated is called the weighted mean. In this example, the overall mean ofM=6.4 is closer to the value of M=6 (the larger sample) than it is to M=7 (the smaller sample). An alternative method for finding the weighted mean is presented in Box 3.1. Box 3.1. An Alternative Procedure for Finding the Weighted MeanIn the text, the weighted mean was obtained by first determining the total number of scores (n) for the two combined samples and then determining the overall sum (ΣX)for the two combined samples. The following example demonstrates how the same result can be obtained using a slightly different conceptual approach.We begin with the same two samples that were used in the text: One sample hasM=6 for n=12 students, and the second sample has M=7 for n=8students. The goal is to determine the mean for the overall group when the two samples are combined. Image transcription textCisco Finesse x Discussion X *frequency hw.docx * *Unit 2 AS1 Central Tender X Ebooks -Webreader.io X Homework Help – Q&A fr x + X – > C Aebooks.cenreader.com/#!/reader/bfa11fb6-56da-4843-bee-17c20d06eed2/page… Show more… Show moreThe following example is an opportunity for you to test your understanding by computing a weighted mean yourself.Example 3.4.One sample has n=4 scores with a mean of M=8 and a second sample hasn=8 scores with a mean of M=5 . If the two samples are combined, what is the mean for the combined group? For this example, you should obtain a mean ofM=6 . Good luck and remember that you can use the example in the text as a model. Computing the Mean from a Frequency Distribution TableWhen a set of scores has been organized in a frequency distribution table, the calculation of the mean is usually easier if you first remove the individual scores from the table. Table 3.1shows a distribution of scores organized in a frequency distribution table. To compute the mean for this distribution you must be careful to use both the X values in the first column and the frequencies in the second column. The values in the table show that the distribution consists of one 10, two 9s, four 8s, and one 6, for a total of n=8 scores. Remember that you can determine the number of scores by adding the frequencies, n=Σf . To find the sum of the scores, you must add all eight scores:ΣX=10+9+9+8+8+8+8+6=66Table 3.1. Statistics quiz scores for a section of n=8 students.Quiz Score (X)ffX1011092188432700616Note that you can also find the sum of the scores by computing ΣfX as we demonstrated in Chapter 2 . Once you have found ΣX and n, you compute the mean as usual. For these data, Image transcription textCisco Finesse x Discussion *frequency hw.docx X Unit 2 AS1 Central Tender X Ebooks -Webreader.io X Homework Help – Q&A fr x + X – Cebooks.cenreader.com/#!/reader/bfa11fb6-56da-4843-bee-17c20d06eed2/page… Show more… Show moreCharacteristics of the MeanThe mean has many characteristics that will be important in future discussions. In general, these characteristics result from the fact that every score in the distribution contributes to the value of the mean. Specifically, every score adds to the total (ΣX) and every score contributes one point to the number of scores (n). These two values ( ΣX and n) determine the value of the mean. We now discuss four of the more important characteristics of the mean.CHANGING A SCOREChanging the value of any score will change the mean. For example, a sample of quiz scores for a psychology lab section consists of 9, 8, 7, 5, and 1. Note that the sample consists of n=5 scores with ΣX=30 . The mean for this sample isImage transcription textCisco Finesse x Discussion *frequency hw.docx X Unit 2 AS1 Central Tender X Ebooks -Webreader.io X Homework Help – Q&A fr x + X – Cebooks.cenreader.com/#!/reader/bfa11fb6-56da-4843-bee-17c20d06eed2/page… Show more… Show more Image transcription textCisco Finesse x Discussion *frequency hw.docx X Unit 2 AS1 Central Tender X Ebooks -Webreader.io X Homework Help – Q&A fr x + X – Cebooks.cenreader.com/#!/reader/bfa11fb6-56da-4843-bee-17c20d06eed2/page… Show more… Show moreNow imagine what would happen if we removed the score (the box) at X=9 . This time the seesaw would tip to the left and, once again, we would need to change the mean to restore balance.Finally, consider what would happen if we added a new score of X=7 , exactly equal to the mean. It should be clear that the seesaw would not tilt in either direction, so the mean would stay in exactly the same place. Also note that if we remove the new score at X=7 , the seesaw will remain balanced and the mean will not change. In general, adding a newscore or removing an existing score will cause the mean to change unless the new score (or existing score) is located exactly at the mean.The following example demonstrates exactly how the new mean is computed when a new score is added to an existing sample. Example 3.5.Adding a score (or removing a score) has the same effect on the mean whether the original set of scores is a sample or a population. To demonstrate the calculation of the new mean, we will use the set of scores that is shown in Figure 3.4. This time, however, we will treat the scores as a sample with n=5 and M=7 . Note that this sample must have ΣX=35 . What will happen to the mean if a new score ofX=13 is added to the sample?To find the new sample mean, we must determine how the values for n and ΣX will be changed by a new score. We begin with the original sample and then consider the effect of adding the new score. The original sample had n=5 scores, so adding one new score will produce n=6 . Similarly, the original sample had ΣX=35 . Adding a score of X=13 will increase the sum by 13 points, producing a new sum of ΣX=35+13=48 . Finally, the new mean is computed using the new values for n and ΣX .Image transcription textCisco Finesse x Discussion *frequency hw.docx X Unit 2 AS1 Central Tender X Ebooks -Webreader.io X Homework Help – Q&A fr x + X – Cebooks.cenreader.com/#!/reader/bfa11fb6-56da-4843-bee-17c20d06eed2/page… Show more… Show moreExample 3.6.We begin with a sample of n=5 scores with ΣX=35 and M=7 . If one score with a value of X=11 is removed from the sample, what is the mean for the remaining scores? You should obtain a mean of M=6 . Good luck and remember that you can use Example 3.5 as a model.ADDING OR SUBTRACTING A CONSTANT FROM EACH SCOREIf a constant value is added to every score in a distribution, the same constant will be added to the mean. Similarly, if you subtract a constant from every score, the same constant will be subtracted from the mean.In the Preview for this Chapter, we described a study that examined the effect of alcohol consumption on the perceived attractiveness of opposite-sex individuals shown in photographs (Jones et al., 2003). The results showed that alcohol significantly increased ratings of attractiveness. Image transcription textCisco Finesse x Discussion *frequency hw.docx X Unit 2 AS1 Central Tender X Ebooks -Webreader.io X Homework Help – Q&A fr x + X – Cebooks.cenreader.com/#!/reader/bfa11fb6-56da-4843-bee-17c20d06eed2/page… Show more… Show moreTable 3.2.Attractiveness ratings of a male face for a sample of n=6 females.ParticipantNo AlcoholModerate AlcoholA45B23C34D34E23F34 ΣX=17M=2.83ΣX=23M=3.83MULTIPLYING OR DIVIDING EACH SCORE BY A CONSTANTIf every score in a distribution is multiplied by (or divided by) a constant value, the mean will change in the same way.Multiplying (or dividing) each score by a constant value is a common method for changing the unit of measurement. To change a set of measurements from minutes to seconds, for example, you multiply by 60; to change from inches to feet, you divide by 12. One common task for researchers is converting measurements into metric units to conform to international standards. For example, publication guidelines of the American Psychological Association call for metric equivalents to be reported in parentheses when most nonmetric units are used.Table 3.3 shows how a sample of n=5 scores measured in inches would be transformed to a set of scores measured in centimeters. (Note that 1 inch equals 2.54 centimeters.) The first column shows the original scores that total ΣX=50 with M=10 inches. In the second column, each of the original scores has been multiplied by 2.54 (to convert from inches to centimeters) and the resulting values total ΣX=127 , with M=25.4 . Multiplying each score by 2.54 has also caused the mean to be multiplied by 2.54. You should realize, however, that although the numerical values for the individual scores and the sample mean have changed, the actual measurements are not changed.Image transcription textCisco Finesse x Discussion *frequency hw.docx X Unit 2 AS1 Central Tender X Ebooks -Webreader.io X Homework Help – Q&A fr x + X – C ebooks.cenreader.com/#!/reader/bfa11fb6-56da-4843-bee-17c20d06eed2/page/f28ageedd320399d78564d7f1 1925… Show more… Show more  Math Statistics and Probability MATH MA-135 Share QuestionEmailCopy link Comments (0)