Statistics in criminology

User Generated

Ynqlybir1

Mathematics

Description

Answer each question with a minimum of 300 words. Use provided material (PDF book) to answer questions. Must also use two other sources to complete assignment. Here is the reference for the PDF in APA format: Walker, J. (2009). Statistics in criminology and criminal justice (4th ed.). Sudbury, MA: Jones and Bartlett.

All of the questions refer to the results of the SPSS analysis presented on pages 161-165 of the textbook.

1.) For the variable HOME, what are the modes? Is the data normally distributed?

2.) For the variable ARREST, what are the modes? Is the data normally distributed?

3.) Why are we concerned about the distribution of data?

4.) What difference does it make in the case of each of the variables (HOME and ARREST) if the data is not normally distributed?

Unformatted Attachment Preview

© Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. What do you want to do? Make inferences Describe How many variables? Univariate Bivariate What level of data? Nominal Ordinal Central tendency Central tendency Mode Dispersion Form Form Skew 16304_CH06_Walker.indd 146 T I F F A NMedian Y Dispersion Range, Index of dispersion Kurtosis L I D D E L L , Kurtosis Multivariate Interval/ Ratio Central tendency Mean Dispersion 1Average Absolute 5deviation 6 8 T Skew S Variance, Standard deviation Form Skew Kurtosis 7/12/12 10:02:10 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. Chapter 6 The Form of a Distribution Learning Objectives L Understand the number of modes,I skewness, and kurtosis as they relate to ­explaining a data set. D ■■ Explain the difference between the mode and the number of modes. D ■■ Interpret the values of skewness and kurtosis as they relate to univariate E ­analysis. ■■ Discuss the importance of the normal L curve in statistics. ■■ Describe the properties of the normal curve. L The final univariate descriptive statistic,,the form of the distribution, ties together the ■■ central tendency and dispersion of the data. Three characteristics make up the form of the distribution: the number of modes, the symmetry, and the kurtosis. In addressing the form of a distribution, a polygon canTgenerally be used to represent these characteristics visually. I F 6-1 Moments of a Distribution F In some statistics books and other places,A distributions and the form of distributions are referred to in terms of the moments of the distribution. There are four moments that are N considered important to a distribution. Moments are calculated as follows: Y S1X 2 X2i 1N where 1 X 2 X 2 represents the deviations5from the mean (as has been the case in Chapters 4 and 5), N is the total number of cases in the distribution, and i is the moment 6 being calculated. 8 the mean is always zero, the first moment Since the sum of the deviations around is always zero. If X is taken to the second T power in the formula above, you can see that this is the formula for the variance—thus, the second moment is the variance. The third moment is usually associatedSwith the skew of the distribution, although the exact formula is to divide the formula for the third moment by the variance to the power of 1.5. Similarly, the kurtosis of a distribution is associated with the fourth 147 16304_CH06_Walker.indd 147 7/12/12 10:02:10 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 148    Chapter 6 n The Form of a Distribution moment, although the exact formula is to divide the formula for the fourth moment by the variance squared. The mean and variance were discussed in “Measures of Central Tendency” and “Measures of Dispersion” (Chapters 4 and 5). The skew and kurtosis of a distribution are discussed in this chapter, together with the third measure of the form of a distribution: the number of modes. 6-2 Number of Modes L The first measure of the form of a distribution is the number of modes. The number I of modes is important to higher-order analyses because it is indicative of the normalD multivariate statistical procedures, a ity of the distribution. To use many bivariate and unimodal distribution is preferred. D In determining the number of modes, a slight deviation from determining the E mode may be necessary. Recall from the discussion of central tendency that it is common to count only the highest frequency in aLdistribution as the mode, even though some people argue that all peaks of a distribution L should be considered. For determining the number of modes in an analysis of form, it may be more beneficial to look at , peaks rather than to find the one, highest value. Consider, for example, the distribution in Figure 6-1. Even though there is only one highest value, there are three peaks in the distribution. These peaks may make the data T unsuitable for certain statistical procedures unless transformations are made. In this distribution, all three modes should I probably be counted in evaluating the form of the distribution even though the mode F is actually only 4. F A N Y 45 40 35 30 25 20 15 10 5 0 1 2 3 4 Figure 6-1 16304_CH06_Walker.indd 148 5 1 5 6 8 T S6 7 8 9 10 Polymodal Distribution 7/12/12 10:02:10 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 6-3 6-3 Skewness 149 Skewness The next characteristic of the form of the distribution is the degree of symmetry (skewness) of the distribution. This measure of the form of a distribution has three categories: symmetrical, positively skewed, and negatively skewed. A fully symmetrical distribution has mirror-image sides such that the distribution could be split at the mean and the sides folded over each other for a perfect match. In Figure 6-2, it is easy to see the symmetry in the distribution. This is the histogram from Figure 4-7. The frequenL cies displayed in this distribution are very balanced: categories 1 and 7 have the same I frequency, as do 2 and 6, and 3 and 5. Category 4 has the highest frequency level. It is easy to see that this distribution could be folded in half and the two sides would D match perfectly. This distribution is therefore, a perfectly symmetrical distribution. In D actual research, however, it is not common to see a perfectly symmetrical distribution. E be only close to symmetrical or not at all More typically, the distribution will either symmetrical. L It should be noted here that the number of modes does not necessarily affect the L skew of the distribution. A distribution that is bimodal could still be cut in half where the distribution mirrors itself. The only,difference in this case is that the mode and other measures of central tendency would not be the same. T I F F A N Y 1 2 3 1 5 6 8 T S4 5 6 7 Figure 6-2 Histogram and Normal Curve for a Symmetrical Distribution 16304_CH06_Walker.indd 149 7/12/12 10:02:10 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 150    Chapter 6 n The Form of a Distribution Analysis of Skew If a distribution is such that one side is different from the other, it is said to be skewed. In skewed distributions, there is no point that can be drawn in the polygon where it could be divided into two similar parts. If the point of the curve is to the left of the graph, it is said to be negatively skewed (the tail of the graph points to the negative end of the scale, smaller positive numbers). In Figure 6-3, “children” is an example of a negatively skewed distribution. Here, the point of the curve is toward category 1 or the left of L of the graph, it is said to be positively the graph. If the point of the curve is to the right skewed (the tail of the graph points toward theI positive end of the scale, larger positive numbers). In Figure 6-3, “gun-wher” is an example of a positively skewed distribution. D 12.5 or the right of the graph. Here, the point of the curve points toward category D output. A value of 0 means there SPSS provides measures of skew in frequency is no skew to the data. Skew values of zero are E almost never obtained, however, and a distribution is considered symmetrical if the skew value in SPSS is between 21 and L if it has a skew greater than +1.00 1.1 A distribution is generally considered skewed or less than (a greater negative number) 21.00. L The magnitude of the number will represent the degree of skew. When conducting , research, it is desirable to obtain a distribution that has a skew as close to zero as possible. If the skew is outside +1 to 21, the distribution may be too skewed to work with, and efforts should be made to get the distribution closer to normal. This is T done through transformations, which is addressed in the discussion on regression. I The frequency distribution that has been used with the other univariate measures F is 20.477, which means that the disis shown in Table 6-1. Here the value of skew tribution is not perfectly symmetrical but thatFit exhibits an acceptable level of skew (it is within the acceptable range of 0 to 21.00). There is some negative skew to this A distribution, as exhibited by the negative value, but it is not enough to warrant addiN value had been less than 21.00 (e.g., tional analyses or give cause for concern. If this 22.77), it might have been necessary to transform Y the distribution. 1 5 6 8 T S 30 400 300 20 200 10 100 0 Std. Dev. = 0.22 Mean = 1.95 N = 324.00 1.00 CHILDREN 1.50 2.00 0 Std. Dev. = 1.85 Mean = 2.2 N = 47.00 0.0 2.5 5.0 7.5 10.0 12.5 GUN_WHER Figure 6-3 Negatively and Positively Skewed Distributions 16304_CH06_Walker.indd 150 7/12/12 10:02:11 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 6-3 Skewness 151 What is your highest level of education? Value Label Value Frequency Percent Valid Percent Cumulative Percent Less than High School 1 16    4.6    4.8    4.8 GED 2 59 17.0 17.6 22.3 High School Graduate 3    8    2.3    2.4 24.7 Some College 4 117 33.7 34.8 59.5 College Graduate 5 Post Graduate 6    Missing    Total Valid N Missing Mean Median Mode L 72 I 64 D11 347 D E 336 L11 L4.08 , 4.00 20.7 21.4 81.0 18.4 19.0 100.0    3.2 100.0 100.00 4 Std. Deviation 1.460 Variance Skewness Std. Error of Skewness Kurtosis Std. Error of Kurtosis Table 6-1 T2.131 I2.477 F .133 F2.705 A .265 SPSS Output N of Measures of Form Y Although quantitative measures of skewness and kurtosis will almost always be available when conducting actual research, an estimate of the skew of a distribution 1 If the mean and the median are different, can be made even without a skew calculation. the distribution is at least somewhat skewed, 5 although it is not possible to tell if it is beyond +1 or 21. Additionally, the skew is in the direction of the mean. For example, 6 if the distribution is positively skewed, the mean will be larger than the median, but 8 smaller than the median. It should also be if the skew is negative, the mean will be noted that the mode is generally on the opposite side of the median from the mean in T skewed distributions. This is not always the case, however, and should not be treated S as a rule. MacGillivray (1981) discusses the conditions under which each of these examples would fall. 16304_CH06_Walker.indd 151 7/12/12 10:02:11 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 152    Chapter 6 6-4 n The Form of a Distribution Kurtosis The last characteristic of the form of a distribution is the kurtosis. For kurtosis, think again of stacking blocks (or beer cans) on top of each other to represent the frequency of the categories in a histogram. The kurtosis is the extent to which cases are piled up around the measure of central tendency or in the tails of the distribution. If most of the values in the distribution are very close to the measure of central tendency, the distribution is said to be leptokurtic (as shown on the left in Figure 6-4). If most of the L values in the distribution are out in the tails, the distribution is said to be platykurtic, I in the distribution are such that they as shown on the right in Figure 6-4. If the values represent a distribution such as that shown in D Figure 6-2, the distribution is said to be mesokurtic, as shown in the center of Figure 6-4. It is desirable to have a mesokurtic D distribution in research; otherwise, the data may have to be transformed. E L L , T I 1 2 3 4 5 6 7 1 2 3 4 5 6 7 7 1 2 3 4 5 6 F Figure 6-4 Leptokurtic, Mesokurtic, and Platykurtic Distributions F A The shape of these curves also offers anN opportunity to talk about variance and standard deviation. As discussed in “Measures of Dispersion” (Chapter 5), the variance and standard deviation dictate the shapeY of the distribution. In a leptokurtic distribution, the variance and standard deviation would be smaller than in a mesokurtic distribution. The variance and standard deviation 1 of a platykurtic distribution would be larger than either a mesokurtic or leptokurtic distribution. This is one application of 5 the variance and standard deviation. A more expanded discussion of this application is presented later in the section on the normal curve. 6 8 T as skew. A value between +1 and 21 In SPSS, kurtosis is measured in the same way represents a mesokurtic distribution. Positive S numbers greater than 1 represent leptoAnalysis of Kurtosis kurtic curves. Negative numbers less than 21 (a greater negative number) represent platykurtic curves. As with skew, it is desirable to get the kurtosis as close as possible 16304_CH06_Walker.indd 152 7/12/12 10:02:11 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 6-6 Design of the Normal Curve 153 to zero, using transformations if necessary. Examining the kurtosis value in Table 6-1 shows that the distribution is mesokurtic because the value (20.705) is between 21.00 and 0. If this value had been 21.705, the distribution would have been platykurtic. 6-5 The Importance of Skew and Kurtosis It is important to know the skew and kurtosis because some statistical procedures do not work well with skewed data or data that isLnot mesokurtic. If data in a research project is found to be skewed or kurtose, it may be necessary to transform the data. Initially, you I must remember two things about transformation. First, if the data is not within acceptable tolerances for skew and kurtosis, theD data will need to be transformed prior to using some statistical procedures. Second, after D making transformations, recheck both the skew and kurtosis. Transforming the data may bring one of these within acceptable tolE If that happens, you will need to choose erances but may make the other unacceptable. another transformation. You should thenLrecheck the skew and kurtosis again. This process should continue until you reach a point where both the skew and kurtosis are L acceptable. If it is not possible to get both the skew and kurtosis in an acceptable range, , analysis procedure that is not susceptible to you may need to consider using a different nonnormal curves. 6-6 T Design of the Normal Curve I Extending the concepts of the frequencyF distribution, graphical representation of data, and measures of central tendency, dispersion, and form brings us to the point of disF cussing a key concept in statistical analysis, the normal curve. At this point, you Aof the normal curve; that is covered in more should not be concerned with applications detail in the chapters of the book on inferential analysis. The purpose of the present N discussion is to introduce the properties of the normal curve. An introduction to the normal curve Y is included in descriptive analyses rather than inferential analyses for two reasons. First, the normal curve can be used to provide an interpretation of the variance and standard 1 deviation. Second, the normal curve is important to a number of statistical procedures that will be discussed before reaching 5 information on inferential statistical procedures. An example of relatively normally 6 distributed data can be shown in grades in a course (see Figure 6-5). Say that most people 8 taking the course score a C on the first test. This would be the modal grade (the top part of the curve). There are those who receive high A’s, but there would be onlyTa few of these; they would be at the positive end of the curve. There are also those who Sreceive very low F’s, but these are also few; they would be at the negative end of the curve. Most people would be in between these two extremes, with more people making scores around C’s than other grades, and 16304_CH06_Walker.indd 153 7/12/12 10:02:11 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 154    Chapter 6 n The Form of a Distribution L I D D E L Curve Figure 6-5 Normal L more people making B’s and D’s than A’s and, F’s. For the sake of argument, though, say that no one received a 100 and there were a few people who did not take the test. Therefore, the ends of the tails will never completely touch the baseline. This type of data represents a special formTof distribution called the normal curve. This type of curve or distribution is very much like those that have been used in this I chapter and at the end of Chapter 4. A normal curve is special because it has certain F characteristics. First, a normal curve is symmetrical in that it can be folded in half, and both sides would be exactly the same (asF in Figure 6-2, where the frequencies of category 1 and 3 are the same as those of category 5 and 7, respectively). Note, though, A that this does not mean this curve cannot be kurtose. Some symmetrical distributions are leptokurtic or platykurtic. This is shown N in Figure 6-4, where each of those distributions was symmetrical even if it was kurtose. Y Also, a normal curve is unimodal; there is one, and only one, peak. This peak is at the maximum frequency of the data distribution, so that the mean, median, and mode all have the same value. The normal 1 From the peak, the tails of a normal curve shown in Figure 6-5 has only one mode. curve fall off on both ends and extend to infinity, 5 always getting closer to the baseline but never touching it. This is shown in Figure 6-5, where the bottom part of the curve 6 straightens out and runs relatively parallel to the X axis. You may say this makes no 8 the normal curve not have an end? sense; all distributions have an end, so why would The answer lies in the scientific process. TakeT the example of computers. Less than 20 years ago, scientists and engineers thought they had achieved the ultimate when they S were able to reach 640K of random access memory (RAM) in a computer. They felt that this was the maximum that could be achieved and all that anyone would ever need. To them, the distribution limits were set. We now know, of course, that 640K was only 16304_CH06_Walker.indd 154 7/12/12 10:02:11 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 6-6 Design of the Normal Curve 155 the beginning and that computers are far beyond that now. It would have been foolish, then, to have the curve touch the line at 640K; it should not touch the line because we do not know what will come in the future. A final characteristic of the normal curve that merits discussion is that the area under a normal curve is always the same, regardless of the data set. The area under the normal curve is 1.00, or 100% of all values in the distribution. This is extremely important in the section of this book concerning inferential analyses because of its importance in estimating the placement of a sample distribution within a population or another L sample. The area under the normal curve also offers the opportunity to put the variance I and standard deviation into practice. Say, for example, that a researcher was examinD before they committed another crime or ing the time prisoners were out on parole returned to prison on a technical violation. D If the time each parolee took before being reincarcerated was plotted, it might look as in Figure 6-6. There were a few people E who returned to prison right away, most of the parolees who returned to prison did it within two to four years, and some tookL longer. Some had not recidivated at the time of the research, so the end of the distribution L is open. An analysis of the central tendency would put the mean of this distribution at 36 , line. This is good information: the avermonths, which is represented by the vertical age length of time for parolees to be reincarcerated is three years. It is obvious from this distribution, however, that not all ofTthe parolees were reincarcerated at the same time. The span of time runs from a couple of months to more than five years. To get a Frequency I F F A N Y 12 24 1 5 6 8 T S 36 48 60 Months Figure 6-6 Distribution of Time to Reincarceration for Parolees 16304_CH06_Walker.indd 155 7/12/12 10:02:12 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 156    Chapter 6 n The Form of a Distribution more accurate picture of the distribution of parolees, we might want to know, on average, how far each of them is from the mean. To calculate this, the procedure would be to take each person (each dot in Figure 6-6) and determine how far from the mean it is. This could be completed by measuring the distance with a ruler, but because this is a numeric scale, it could also be completed by subtracting each value from the mean. This operation would be noted as X 2 X. Summing all of these values, represented by S 1 X 2 X 2 , would provide us with the total of the distance from each value to the mean. L But remember that the sum of each value subtracted from the mean is zero, so that does not help us. A solution to this is I to square each value before summing it. This calculation, S 1 X 2 X 2 2, will give a positive value. Knowing the total distance betweenDthe mean and each value is good, but a simpler value would be the average of the distance D from each value to the mean. The calculation for this procedure is: E S 1 X 2 XL2 2 N L The only problem with this calculation is that the value that would be obtained is not , on the same scale as the original data. To return this value to the original scale, take the square root of the value. Of course, you recognize that this is the procedure for calculating the variance and, ultimately, the T standard deviation for the distribution. Although this procedure can be completed for any distribution, it is particularly imporI tant for a normal curve because of what the standard deviation represents. Because the area under the normal curveFis always 100% and because standard deviations represent standard distances in relation F to the mean of the distribution, standard deviations can be used to calculate the area under the normal curve for particular values. For example, between the mean and 1 A standard deviation under a normal curve lies 34.13% of all the data. Between the mean N and 2 standard deviations is 47.72% of all the data. Between the mean and 3 standard Y deviations is 49.87% of all the data. That covers 49.87% of the possible 50% of one half of the curve. Since the normal curve is symmetrical, the same values can be obtained whether counting from the left or right of the mean. This also means that the1values could be doubled from one side and the area under the whole normal curve could be determined for values plus or 5 minus a certain number of standard deviations from the mean. If the figures above are 6 between 21 and 1 standard deviation, doubled, the result would be 68.26% of the data 95.44% of the data between 22 and 2 standard 8 deviations, and 99.74% of the data between 23 and 3 standard deviations. T This is very useful to know because it is now possible to determine what percentSstandard deviation within the distribuage of the scores lie between the mean and any tion. The problem is that researchers are not always interested in determining values (or determining the area under a normal curve) only for values directly on a particular 16304_CH06_Walker.indd 156 7/12/12 10:02:12 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 6-6 Design of the Normal Curve 157 standard deviation. Often, calculations need to be made that allow any value in the distribution to be converted to standard deviation units such that the area under the normal curve can be calculated for that value. The procedure for doing this is to calculate a Z score for that value. Z scores convert any value in the distribution to standard deviation units. The formula for calculating a Z score is Z5 X2X s L where X is the number to be converted to a Z score, X the mean of the distribution, and I s is the standard deviation of the distribution. The process of converting values to D Z scores is simply a method of deriving standard scores. To be able to make comparisons D between things, it is necessary to set a standard on which both items may be measured. If measuring distance, we could use E feet or meters; weight could be in pounds or kilograms. But what about social data L in terms of crime rates. Calculating crime such as crime? Crime is generally discussed rates is simply a process of standardizingLcrime in terms of the population of a city— taking numbers that may be difficult to compare between different things and making , them comparable. The same can be done with the mean, standard deviation, and normal curve through Z scores. This standardizes scores based on the normal curve. What this means is that if a researcher T knows the mean and standard deviation of a population, Z scores can be used to calculate the distance between any value and I the mean. Using the area under the normal curve allows inferential analyses (see the F to be used to determine the chances that chapters in this book on inferential analyses) any number in a distribution will be in aF sample drawn from that population. The process of calculating a Z score, determining the area under the normal curve A between that value and the mean, and examining how that score relates to the mean, is Nfrom the mean. This determines whether the rather simple. First, subtract the raw score number is above or below the mean: negative Y numbers are below the mean, positive numbers above the mean. Then divide the number by the standard deviation to determine how many standard deviation units the number is above or below the mean. The answer obtained from this calculation is1the score’s deviation from the mean in standard units (the Z score in standard deviation 5 units). Note that you may get a negative number; all this means is that the Z score is to the left of the mean. It does not affect the calculations at all and can be dropped6in the remainder of the procedure. The next step is to take this number8and turn to Table B-1 in Appendix B, “Statistical Tables.” The row to use corresponds to this number in column a of the table. T The columns to use within this row depend on what area under the normal curve is S area between the score and the mean, look being examined. If you are looking for the to column b in the table; if examining the area beyond that score, which is farther into the tail of the distribution, use column c. 16304_CH06_Walker.indd 157 7/12/12 10:02:12 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 158    Chapter 6 n The Form of a Distribution Suppose that the number of complaints against a police department averaged 12 complaints a month with a standard deviation of 1.5. Then suppose that the chief wants you to determine what percentage of the scores fell between the mean and the current month’s score of 14. This would require determining the area between the mean and 14. To do this, first calculate a Z score, as shown below: Z5 5 X2X s L 14 2I 12 1.5 D 2 1.5 D E 5 1.33 L L Then look at column a in Table B-1 in Appendix B, “Statistical Tables,” and find 1.33. , 5 You are looking for the area between the score and the mean, so you would use column b to get the percentage of the area under the normal curve that falls between the mean and 14. This value is 0.4082, so the areaT in this distribution between 12 and 14 is 40.82%. What does this mean? It means that almost 41% of the months in the current I year had fewer complaints recorded than those in the current month. F many months had more complaints But what if the chief wanted to know how recorded than the current month? To determine F the percentage of scores greater than 14, you would calculate the Z score exactly as before. That number, 1.33, is then found A in column a of Table B-1 in Appendix B, “Statistical Tables.” Since you are trying to N column c in the table would be used. determine the number of scores greater than 14, The value in column c that corresponds to a Y Z score of 1.33 is 0.0918, so 9.18% of the scores in the distribution are greater than 14, which means that almost 10% of the months had more complaints recorded than the current month. Actually, because the 1 not necessary to complete the second area between the mean and 14 was known, it was procedure. Because it was already determined 5that the percentage of scores between the mean and 14 was 40.82%, and because the area under the normal curve is always 6 100% and the area under half of the normal curve is 50%, the 40.08 could have been 8 subtracted from 50 to arrive at the score of 9.18%—the area beyond that value. These procedures work whether using scores above the mean or scores below the T mean. As stated previously, 21 standard deviation is the same as 1 standard deviation; S looking for the area under the normal thus, it would be the same percentage if we were curve representing less than 10. But what if you wanted to know how many scores were higher than one score and lower than another: for example, how many people 16304_CH06_Walker.indd 158 7/12/12 10:02:12 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 6-6 Design of the Normal Curve 159 scored higher than 90 and lower than 62 on a test. This case would require determining the percentage greater than the score of 90 and the percentage less than the score of 62. The final pieces of information needed here are that the mean of the distribution is 76 and the standard deviation is 10.5. The areas beyond these two scores can be determined simply by adding together the percentages in Table B-1 in Appendix B, “Statistical Tables.” The Z score calculations for these two values would be as follows: Z5 X2X L X2X        Z 5 s s I 90 2 76 D 62 2 76 5        5 10.5 10.5 D 2 22 E 5        5 1.5 1.5 L 5 1.33        5 21.33 L The results of these calculations would ,be taken to Table B-1 in Appendix B, “Statistical Tables.” Finding 1.33 in column a and wanting to determine the area under the normal curve beyond this value, we would look in column c. The value found in T column c is 0.0918. This means that 9.18% of the scores are higher than 90. Since 62 has the same value, but negative, it also Ihas a value in column c of 0.0918, so 9.18% of the scores are lower than 62. Adding these F two values together would establish what percentage of people scored greater than 90 and less than 62: F A = 0.1836 0.0918 + 0.0918 N So 18.36% of the people in the class scored higher than 90 (making an A) or lower Y than 62 (making an F). It is also possible to determine the percentage of scores that fall between two scores. In the example we have been using, the percentage of scores that fall between 90 and 62 can be determined 1 by adding together the areas between them and the mean. Looking at column b in Table B-1 in Appendix B, “Statistical Tables,” 5 the percentage of scores between the mean of 76 and a score of 90 (the Z score of 6 the same Z score value, the percentage of 1.33) is 40.82%. Since the score of 62 has scores between 62 and 76 is also 40.82%.8Adding those percentages together produces a percentage of 81.64% as shown below: T S = 0.8164 0.4082 + 0.4082 So 81.64% of the class scored a B, C, or D. 16304_CH06_Walker.indd 159 7/12/12 10:02:12 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 160    Chapter 6 n The Form of a Distribution Points to Remember about the Normal Curve Two Z scores that will become important later are 1.96 and 2.58. These correspond to 95% and 99% of the area under the normal curve, respectively. These are important because researchers often want to know where 95% or 99% of the values in the distribution fall, or they may want to compare two values and see if they fall within 95% or 99% of the values in the distribution. Another point you should remember is that the normal curve is a theoretical ideal. Normal distributions do not really exist, and L the best we can do is to gather data that is close to a normal curve but not exact. The conclusions drawn when using the theory I of the normal curve, then, will only be estimates, not hard facts. Finally, you should realize that not all D data sets even come close to a normal Dtendency (see Chapter 4), some districurve. In the discussion on measures of central butions are multimodal and almost jagged looking. E There are many distributions that are so skewed they are J shaped and do not conform at all to a normal distribution. L There are even a few distributions that are absolutely flat. Each of these distributions requires special treatment, which is discussedLin the chapters on bivariate and multivariate analysis. , 6-7 Conclusion T In this chapter we completed the description ofIone variable at a time from a distribution (univariate descriptive analysis) and you have learned how to describe a variable such F that it can be relayed to another person accurately, and in a form that allows the person F to get a mental image of the distribution. Describing single variables can take many forms: frequency distributions and graphs, measures of central tendency, measures A of dispersion, and the form of the distribution. These all have the goal of describing N the attributes of the distribution and determining if a variable is suitable for further Y analyses. In the next chapters, we put together two variables—bivariate descriptive analyses—and describe what relationship might exist between them. The success of those analyses depends on the successful univariate1description of data. 6-8 Key Terms form kurtosis leptokurtic mesokurtic moments negatively skewed normal curve 16304_CH06_Walker.indd 160 5 6 8 number of modes platykurtic T positively skewed S skew symmetry Z score 7/12/12 10:02:12 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 6-10 6-9 Exercises 161 Summary of Equations Moments of a Distribution S1X 2 X2i N Z Score LX 2 X I s D 6-10 Exercises D (from the gang database): 1. Use the following frequency tables a. To discuss the number of modes. E b. To determine the skew and kurtosis and discuss whether the distribution L is positively skewed or negatively skewed and whether it is leptokurtic, mesokurtic, or platykurtic. L , HOME: What type of house do you live in? Z5 Value Label House Duplex Trailer Apartment Other    Missing    Total N Mean Std. Error of Mean Median Mode Std. Deviation Variance Skewness Std. Error of Skewness Kurtosis Std. Error of Kurtosis Range 16304_CH06_Walker.indd 161 Value 1 2 3 4 5 Frequency T 280 I   3 F34 F21    2 A    3 343 N ValidY Missing 1 5 6 8 T S Percent 81.6    .9    9.9    6.1    .6    .9 100.0 Valid Percent 82.4    .9 10.0    6.2 .6 Cumulative Percent 82.4 83.2 93.2 99.4 100.0 100.0 340 3 1.41 .051 1 1 0.945 .892 2.001 .132 2.613 .264 5 7/12/12 10:02:12 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 162    Chapter 6 n The Form of a Distribution ARREST: How many times have you been arrested? Value Frequency 0 243 1 23 2 10 3    3 Valid Percent Cumulative Percent 70.8 86.2 86.2    6.7    8.2 94.3    2.9    3.5 97.9     .9    1.1 98.9 Percent L    2     .6 I    1     .3 61 D 17.8 343 D 100.0 E282 Valid Missing L 61 L .30 , .093 5 24    Missing    Total N Mean Std. Error of Mean Median     .7 99.6     .4 100.0 100.0 0 Mode 0 T 1.567 I 2.455 F 12.692 F .145 A187.898 .289 N 24 Y Std. Deviation Variance Skewness Std. Error of Skewness Kurtosis Std. Error of Kurtosis Range TENURE: How long have you lived at your current address (months)? Value    1    2    3    4 16304_CH06_Walker.indd 162 1 Frequency 5 14 6    6 8    4    4 T S Percent Valid Percent Cumulative Percent    4.1    4.3    4.3    1.7    1.8    6.1    1.2    1.2    7.3    1.2    1.2    8.6 7/12/12 10:02:13 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 6-10 Exercises 163 TENURE: How long have you lived at your current address (months)? Cumulative Percent Frequency Percent    5    6    1.7    1.8 10.4    6    6    1.7    1.8 12.2    7    1     .3     .3 12.5    8 L   3 I   2    1 D   1 D11 E   1 L   5 L   1 , 30     .9     .9 13.5     .6     .6 14.1     .3     .3 14.4     .3     .3 14.7    3.2    3.4 18.0     .3     .3 18.3    1.5    1.5 19.9     .3     .3 20.2    8.7    9.2 29.4    1     .3     .3 29.7    1     .3     .3 30.0     .3     .3 30.3    6.4    6.7 37.0     .3     .3 37.3    3.5    3.7 41.0    7.0    7.3 48.3    4.1    4.3 52.6     .3     .3 52.9    2.3    2.4 55.4    9 10 11 12 14 18 21 24 30 31 84 T   1 I 22 F   1 F12 A24 14 N   1 Y   8 96 18    5.2    5.5 60.9 108 1   4 5   9 6 11 21 813 T11 S    1.2    1.2 62.1    2.6    2.8 64.8    3.2    3.4 68.2    6.1    6.4 74.6    3.8    4.0 78.6    3.2    3.4 82.0 32 36 42 48 60 72 76 120 132 144 156 168 16304_CH06_Walker.indd 163 Valid Percent Value 7/12/12 10:02:13 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 164    Chapter 6 n The Form of a Distribution TENURE: How long have you lived at your current address (months)? Value Frequency Percent Valid Percent Cumulative Percent 170    5    1.5    1.5 83.5 180    7    2.0    2.1 85.6 182    2     .6     .6 86.2 186    1     .3     .3 86.5    4.1    4.3 90.8     .3     .3 91.1    7.0    7.3 98.5     .9     .9 99.4     .6     .6 100.0 192 198 204 216 240    Missing    Total N L 14 I    1 D 24    3 D    2 E 16 L 343 L Valid , Missing Mean Std. Error of Mean Median Mode Std. Deviation Variance Skewness Std. Error of Skewness Kurtosis Std. Error of Kurtosis Range 16304_CH06_Walker.indd 164    4.7 100.0 100.0 327 16 T 88.77 3.880 I 72 F 24 F 70.164 A 4923.055 N .365 Y .135 21.284 1 5 6 8 T S .269 239 7/12/12 10:02:13 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 6-10 Exercises 165 SIBS: How many brothers and sisters do you have? Value 39 11.4 11.5 11.5 39.9 40.5 52.1 2 79 23.0 23.4 75.4 3 L39 I 17 13 D    6 D    4 E   1 L   1 L   1 ,   1 11.4 11.5 87.0 7 9 10 12 15 Mean Std. Error of Mean Median Mode Std. Deviation Variance Skewness Std. Error of Skewness Kurtosis Std. Error of Kurtosis Range 16304_CH06_Walker.indd 165 Cumulative Percent 137 6 N Valid Percent 1 4    Missing Percent 0 5    Total Frequency    5.0    5.0 92.0    3.8    3.8 95.9    1.7    1.8 97.6    1.2    1.2 98.8     .3     .3 99.1     .3     .3 99.4     .3     .3 99.7     .3     .3 100.0    5    1.5 343 100.0 T I Missing F F A N Y Valid 1 5 6 8 T S 338 5 1.94 .098 1 1 1.801 3.245 2.664 .133 12.027 .265 15 7/12/12 10:02:13 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 166    Chapter 6 n The Form of a Distribution 2. For a distribution with a mean of 50 and a standard deviation of 10: a. Calculate the Z score for a score of 40. b. Determine the area under the normal curve between the mean and this value. c. Determine the area under the normal curve beyond this value. d. Calculate a Z score for a score of 65. e. Determine the area under the normal curve between the mean and this value. f. Determine the area under the normal curve beyond this value. g. Determine the area under the normal L curve between 40 and 60. h. Determine the area under the normal curve outside 40 and 60. I 3. Say a parole board has a policy that it will only release prisoners who meet D a minimum number of good-time a minimum amount of time served, have points, and have made an acceptableD score on tests in their drug awareness class. If the mean of this distribution is 90, the minimum acceptable score on E these criteria is 70, and the standard deviation of scores is 15: a. Calculate the Z score for a score ofL90. b. Calculate the Z score for a score ofL50. c. Determine the area under the normal curve between the mean and the val, ues in parts a and b. d. Determine the area under the normal curve beyond these values. e. Calculate a Z score for a score of 70. T f. Determine the area under the normal curve between the mean and this valI ue. g. Determine the area under the normal F curve beyond this value. F 6-11 References A N mode inequality and skew for a class MacGillivray, H. L. (1981). The mean, median, of densities. Australian Journal of Statistics, Y23, 247. 6-12 For Further Reading1 Pearson, K. (1894). On the dissection of asymmetrical frequency-curves: General 5 theory. Philosophical Transactions of the Royal Society of London (Series A, Vol. 6 185). London: Cambridge University Press. 8 frequency curves in general: Types Pearson, K. (1895). Classification of asymmetrical actually occurring. Philosophical Transactions T of the Royal Society of London (Series A, Vol. 186). London: Cambridge University Press. S 16304_CH06_Walker.indd 166 7/12/12 10:02:13 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 6-13 Note 167 6-13 Note 1. SPSS uses the convention of +1 to 21 as the measure of acceptable skew. This is based on a particular formula that “standardizes” skew and kurtosis scores. There is a similar (and popular) formula that calculates skew and kurtosis where the acceptable range is +3 to 23. Programs such as SAS use this formula. It is important, then, that you understand which formula is being used (what the acceptable range is) for these values before making a judgment about L them. I D Criminal Justice on the Web Dto make full use of today’s teaching and techVisit http://criminaljustice.jbpub.com/Stats4e nology! Our interactive Companion Website Ehas been designed to specifically complement Statistics in Criminology and Criminal Justice: Analysis and Interpretation, 4th Edition. The L resources available include a Glossary, Flashcards, Crossword Puzzles, Practice Quizzes, Web­links, and Student Data Sets. Test yourself L today! , T I F F A N Y 1 5 6 8 T S 16304_CH06_Walker.indd 167 7/12/12 10:02:13 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. Appendix B Statistical Tables L I D D E L Table B-1 Area Under the Normal Curve L a b c b a c Area Area Area Area , between beyond between beyond Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.20 0.21 0.22 0.23 0.24 – X and Z 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753 0.0793 0.0832 0.0871 0.0910 0.0948 Z 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247 0.4207 0.4168 0.4129 0.4090 0.4052 Z 0.25 0.26 0.27 0.28 0.29 0.30 0.31 0.32 0.33 0.34 0.35 0.36 0.37 0.38 0.39 0.40 0.41 0.42 0.43 0.44 0.45 0.46 0.47 0.48 0.49 – X and Z 0.0987 0.1026 0.1064 0.1103 0.1141 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879 Z 0.4013 0.3974 0.3936 0.3897 0.3859 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.3483 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121 a Z 0.50 0.51 0.52 0.53 0.54 0.55 0.56 0.57 0.58 0.59 0.60 0.61 0.62 0.63 0.64 0.65 0.66 0.67 0.68 0.69 0.70 0.71 0.72 0.73 0.74 T I F F A N Y 1 5 6 8 T S b Area between – X and Z 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549 0.2580 0.2611 0.2642 0.2673 0.2704 c Area beyond Z 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451 0.2420 0.2389 0.2358 0.2327 0.2296 a Z 0.75 0.76 0.77 0.78 0.79 0.80 0.81 0.82 0.83 0.84 0.85 0.86 0.87 0.88 0.89 0.90 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 b Area between – X and Z c Area beyond Z 0.2734 0.2266 0.2764 0.2236 0.2794 0.2206 0.2823 0.2177 0.2852 0.2148 0.2881 0.2119 0.2910 0.2090 0.2939 0.2061 0.2967 0.2033 0.2995 0.2005 0.3023 0.1977 0.3051 0.1949 0.3078 0.1922 0.3106 0.1894 0.3133 0.1867 0.3159 0.1841 0.3186 0.1814 0.3212 0.1788 0.3238 0.1762 0.3264 0.1736 0.3289 0.1711 0.3315 0.1685 0.3340 0.1660 0.3365 0.1635 0.3389 0.1611 (continues) 501 16304_APPB_Walker.indd 501 7/7/12 9:59:19 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 502    Appendix B n Statistical Tables Table B-1 Area Under the Normal Curve (continued) a Z 1.00 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20 1.21 1.22 1.23 1.24 1.25 1.26 1.27 1.28 1.29 1.30 1.31 1.32 1.33 1.34 1.35 1.36 1.37 1.38 1.39 1.40 1.41 1.42 1.43 1.44 b Area between – X and Z 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177 0.4192 0.4207 0.4222 0.4236 0.4251 c Area beyond Z 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823 0.0808 0.0793 0.0778 0.0764 0.0749 a Z 1.45 1.46 1.47 1.48 1.49 1.50 1.51 1.52 1.53 1.54 1.55 1.56 1.57 1.58 1.59 1.60 1.61 1.62 1.63 1.64 1.65 1.66 1.67 1.68 1.69 1.70 1.71 1.72 1.73 1.74 1.75 1.76 1.77 1.78 1.79 1.80 1.81 1.82 1.83 1.84 1.85 1.86 1.87 1.88 1.89 b Area between – X and Z 0.4265 0.4279 0.4292 0.4306 0.4319 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706 c Area beyond Z 0.0735 0.0721 0.0708 0.0694 0.0681 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.0455 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.0367 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294 b Area between – X and Z a Z 1.90 1.91 1.92 1.93 1.94 1.95 1.96 1.97 1.98 1.99 2.00 2.01 2.02 2.03 2.04 2.05 2.06 2.07 2.08 2.09 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 2.20 2.21 2.22 2.23 2.24 2.25 2.26 2.27 2.28 2.29 2.30 2.31 2.32 2.33 2.34 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890 0.4893 0.4896 0.4898 0.4901 0.4904 L I D D E L L , T I F F A N Y 1 5 6 8 T S c Area beyond Z 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183 0.0179 0.0174 0.0170 0.0166 0.0162 0.0158 0.0154 0.0150 0.0146 0.0143 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.0110 0.0107 0.0104 0.0102 0.0099 0.0096 a Z 2.35 2.36 2.37 2.38 2.39 2.40 2.41 2.42 2.43 2.44 2.45 2.46 2.47 2.48 2.49 2.50 2.51 2.52 2.53 2.54 2.55 2.56 2.57 2.58 2.59 2.60 2.61 2.62 2.63 2.64 2.65 2.66 2.67 2.68 2.69 2.70 2.71 2.72 2.73 2.74 2.75 2.76 2.77 2.78 2.79 b Area between – X and Z 0.4906 0.4909 0.4911 0.4913 0.4916 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974 c Area beyond Z 0.0094 0.0091 0.0089 0.0087 0.0084 0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.0048 0.0047 0.0045 0.0044 0.0043 0.0041 0.0040 0.0039 0.0038 0.0037 0.0036 0.0035 0.0034 0.0033 0.0032 0.0031 0.0030 0.0029 0.0028 0.0027 0.0026 (continues) 16304_APPB_Walker.indd 502 7/7/12 9:59:19 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. Statistical Tables    503 Table B-1 Area Under the Normal Curve (continued) a b c Area Area between beyond – X and Z Z Z 2.80 2.81 2.82 2.83 2.84 2.85 2.86 2.87 2.88 2.89 2.90 2.91 2.92 2.93 2.94 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981 0.4981 0.4982 0.4982 0.4983 0.4984 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.0020 0.0019 0.0019 0.0018 0.0018 0.0017 0.0016 a Z 2.95 2.96 2.97 2.98 2.99 3.00 3.01 3.02 3.03 3.04 3.05 3.06 3.07 3.08 3.09 b c Area Area between beyond – X and Z Z 0.4984 0.4985 0.4985 0.4986 0.4986 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990 0.0016 0.0015 0.0015 0.0014 0.0014 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.0010 0.0010 Table B-2 Values of Chi-Square a Z 3.10 3.11 3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19 3.20 3.21 3.22 3.23 3.24 L I D D E L L , b c Area Area between beyond – X and Z Z 0.4990 0.4991 0.4991 0.4991 0.4992 0.4992 0.4992 0.4992 0.4993 0.4993 0.4993 0.4993 0.4994 0.4994 0.4994 0.0010 0.0009 0.0009 0.0009 0.0008 0.0008 0.0008 0.0008 0.0007 0.0007 0.0007 0.0007 0.0006 0.0006 0.0006 a Z b Area between – X and Z c Area beyond Z 3.25 3.30 3.35 3.40 3.45 3.50 3.60 3.70 3.80 3.90 4.00 4.50 5.00 5.50 0.4994 0.4995 0.4996 0.4997 0.4997 0.4998 0.4998 0.4999 0.4999 0.49995 0.49997 0.4999966 0.4999997 0.4999999 0.0006 0.0005 0.0004 0.0003 0.0003 0.0002 0.0002 0.0001 0.0001 0.00005 0.00003 0.0000034 0.0000003 0.0000001 Probability (Top Row) and Significance (Bottom Row) df    1 0.999 0.99 0.0001 0.01 10.827 6.635    2 13.815 9.210    3 16.268 11.345    4 18.465 13.277    5 20.517 15.086    6 22.457 16.812    7 24.322 18.475    8 26.125 20.090    9 27.877 21.666 10 29.588 23.209 11 31.264 24.725 12 32.909 26.217 13 34.528 27.688 14 36.123 29.141 15 37.697 30.578 T0.95 I0.05 F3.841 5.991 F7.815 A 9.488 11.070 N 12.592 Y 14.067 0.90 0.80 0.70 0.10 0.20 0.30 2.706 1.642 1.074 4.605 3.219 2.408 6.251 4.624 3.665 7.779 5.989 4.878 9.236 7.289 6.064 10.645 8.558 7.231 12.017 9.803 8.383 15.507 13.362 11.030 9.524 16.919 14.684 12.242 10.656 1 5 18.307 19.675 6 21.026 8 22.362 T 23.685 S 24.996 15.987 13.442 11.781 17.275 14.631 12.899 18.549 15.812 14.011 19.812 16.985 15.119 21.064 18.151 16.222 22.307 19.311 17.322 (continues) 16304_APPB_Walker.indd 503 7/7/12 9:59:20 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. 504    Appendix B n Statistical Tables Table B-2 Values of Chi-Square (continued) Probability (Top Row) and Significance (Bottom Row) df 16 0.999 0.99 0.95 0.90 0.80 0.70 0.0001 0.01 0.05 0.10 0.20 0.30 32.000 26.296 23.542 20.465 18.418 39.252 17 40.790 33.409 27.587 24.769 21.615 19.511 18 42.312 34.805 28.869 25.989 22.760 20.601 19 43.820 36.191 30.144 27.204 23.900 21.689 27 55.476 46.963 L I 31.410 D 32.671 33.924D 35.172E 36.415L 37.652 L 38.885 , 40.113 28 56.893 48.278 41.337 20 45.315 37.566 21 46.797 38.932 22 48.268 40.289 23 49.728 41.638 24 51.179 42.980 25 52.620 44.314 26 54.052 45.642 29 58.302 49.588 42.557 30 59.703 50.892 43.773 T I F F A N Y 28.412 25.038 22.775 29.615 26.171 23.858 30.813 27.301 24.939 32.007 28.429 26.018 33.196 29.553 27.096 34.382 30.675 28.172 35.563 31.795 29.246 36.741 32.912 30.319 37.916 34.027 31.391 39.087 35.139 32.461 40.256 36.250 33.530 1 5 6 8 T S 16304_APPB_Walker.indd 504 7/7/12 9:59:20 AM © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. Statistical Tables    505 Table B-3 Student’s t Distribution Level of Significance for One-Tailed Test 0.10 0.05 0.025 0.01 0.005 0.0005 Level of Significance for Two-Tailed Test df 0.20 0.10 0.05 0.02 0.01 0.001 636.620    1 3.078 6.314 12.706 31.821 63.657    2 1.886 2.920 4.303 6.965 9.925 31.598    3 1.638 2.353 4.541 5.841 12.941 13 1.350 1.771 L3.182 I2.776 2.571 D 2.447 D 2.365 2.306 E 2.262 L2.228 L2.201 2.179 ,2.160 14 1.345 1.761 2.145 2.131 T2.120 I2.110 F2.101 2.093 F2.086 2.080 A 2.074 N 2.069 2.064 Y    4 1.533 2.132    5 1.476 2.015    6 1.440 1.943    7 1.415 1.895    8 1.397 1.860    9 1.383 1.833 10 1.372 1.812 11 1.363 1.796 12 1.356 1.782 15 1.341 1.753 16 1.337 1.746 17 1.333 1.740 18 1.330 1.734 19 1.328 1.729 20 1.325 1.725 21 1.323 1.721 22 1.321 1.717 23 1.319 1.714 24 1.318 1.711 3.747 4.604 8.610 3.365 4.032 6.859 3.143 3.707 5.959 2.998 3.499 5.405 2.896 3.355 5.041 2.821 3.250 4.781 2.764 3.169 4.587 2.718 3.106 4.437 2.681 3.055 4.318 2.650 3.012 4.221 2.624 2.977 4.140 2.602 2.947 4.073 2.583 2.921 4.015 2.567 2.898 3.965 2.552 2.878 3.922 2.539 2.861 3.883 2.528 2.845 3.850 2.518 2.831 3.819 2.508 2.819 3.792 2.500 2.807 3.767 2.492 2.797 3.745 25 1.316 1.708 2.060 2.485 2.787 3.725 26 1.315 1.706 2.056 2.479 2.779 3.707 27 1.314 1.703 2.473 2.771 3.690 28 1.313 1.701 29 1.311 1.699 30 1.310 1.697 40 1.303 1.684 60 1.296 1.671 120 1.289 1.658 ∞ 1.282 1.645 12.052 52.048 2.045 62.042 82.021 2.000 T1.980 1.960 S 2.467 2.763 3.674 2.462 2.756 3.659 2.457 2.750 3.646 2.423 2.704 3.551 2.390 2.660 3.460 2.358 2.617 3.373 2.326 2.576 3.291 (continues) 16304_APPB_Walker.indd 505 7/7/12 9:59:20 AM 16304_APPB_Walker.indd 506 5.9874 5.5914 5.3177 5.1174 4.9646 4.8443 4.7472 4.6672 4.6001 4.5431 4.494 4.4513 4.4139 4.3807 4.3512 4.3248 4.3009 4.2793 4.2597 4.2417 4.2252 4.21 4.196 4.183 4.1709 4.0847 4.0012 3.9201 3.8415 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 40 60 120 ∞ df2=1 2 3 4 5 Distribution of f; p = .05 2.8524 2.81 2.7729 2.7401 2.7109 3.0069 2.9647 2.9277 2.8951 2.8661 1 5 6 8 T S 3.2317 3.1504 3.0718 2.9957 3.369 3.3541 3.3404 3.3277 3.3158 2.8387 2.7581 2.6802 2.6049 2.9752 2.9604 2.9467 2.934 2.9223 2.606 2.5252 2.4472 2.3719 2.7426 2.7278 2.7141 2.7014 2.6896 2.8401 2.8167 2.7955 2.7763 2.7587 2.4495 2.3683 2.2899 2.2141 2.5868 2.5719 2.5581 2.5454 2.5336 2.6848 2.6613 2.64 2.6207 2.603 2.3359 2.2541 2.175 2.0986 2.4741 2.4591 2.4453 2.4324 2.4205 2.5727 2.5491 2.5277 2.5082 2.4904 2.7413 2.6987 2.6613 2.6283 2.599 3.0946 2.9961 2.9153 2.8477 2.7905 2.249 2.1665 2.0868 2.0096 2.3883 2.3732 2.3593 2.3463 2.3343 2.4876 2.4638 2.4422 2.4226 2.4047 2.6572 2.6143 2.5767 2.5435 2.514 3.0123 2.9134 2.8321 2.7642 2.7066 4.2067 3.787 3.5005 3.2927 3.1355 2.1802 2.097 2.0164 1.9384 2.3205 2.3053 2.2913 2.2783 2.2662 2.4205 2.3965 2.3748 2.3551 2.3371 2.5911 2.548 2.5102 2.4768 2.4471 2.948 2.8486 2.7669 2.6987 2.6408 4.1468 3.7257 3.4381 3.2296 3.0717 2.124 2.0401 1.9588 1.8799 2.2655 2.2501 2.236 2.2229 2.2107 2.366 2.3419 2.3201 2.3002 2.2821 2.5377 2.4943 2.4563 2.4227 2.3928 2.8962 2.7964 2.7144 2.6458 2.5876 4.099 3.6767 3.3881 3.1789 3.0204 3.9999 3.5747 3.2839 3.0729 2.913 2.7876 2.6866 2.6037 2.5342 2.4753 2.4247 2.3807 2.3421 2.308 2.2776 4.06 3.6365 3.3472 3.1373 2.9782 2.8536 2.7534 2.671 2.6022 2.5437 2.4935 2.4499 2.4117 2.3779 2.3479 2.0772 1.9926 1.9105 1.8307 2.2197 2.2043 2.19 2.1768 2.1646 2.321 2.2967 2.2747 2.2547 2.2365 T I F F A N Y 3.0725 3.0491 3.028 3.0088 2.9912 3.2389 3.1968 3.1599 3.1274 3.0984 3.2039 3.1059 3.0254 2.9582 2.9013 3.3567 3.2592 3.1791 3.1122 3.0556 4.2839 3.866 3.5806 3.3738 3.2172 2.0035 1.9174 1.8337 1.7522 2.1479 2.1323 2.1179 2.1045 2.0921 2.2504 2.2258 2.2036 2.1834 2.1649 1.9245 1.8364 1.7505 1.6664 2.0716 2.0558 2.0411 2.0275 2.0148 2.1757 2.1508 2.1282 2.1077 2.0889 2.3522 2.3077 2.2686 2.2341 2.2033 2.7186 2.6169 2.5331 2.463 2.4034 3.9381 3.5107 3.2184 3.0061 2.845 1.8389 1.748 1.6587 1.5705 1.9898 1.9736 1.9586 1.9446 1.9317 2.096 2.0707 2.0476 2.0267 2.0075 2.2756 2.2304 2.1906 2.1555 2.1242 2.6464 2.5436 2.4589 2.3879 2.3275 3.8742 3.4445 3.1503 2.9365 2.774 1.7929 1.7001 1.6084 1.5173 1.9464 1.9299 1.9147 1.9005 1.8874 2.054 2.0283 2.005 1.9838 1.9643 2.2354 2.1898 2.1497 2.1141 2.0825 2.609 2.5055 2.4202 2.3487 2.2878 3.8415 3.4105 3.1152 2.9005 2.7372 1.7444 1.6491 1.5543 1.4591 1.901 1.8842 1.8687 1.8543 1.8409 2.0102 1.9842 1.9605 1.939 1.9192 2.1938 2.1477 2.1071 2.0712 2.0391 2.5705 2.4663 2.3803 2.3082 2.2468 3.8082 3.3758 3.0794 2.8637 2.6996 2.4901 2.3842 2.2966 2.2229 2.1601 2.1058 2.0584 2.0166 1.9795 1.9464 2.5309 2.4259 2.3392 2.2664 2.2043 2.1507 2.104 2.0629 2.0264 1.9938 1.6928 1.5943 1.4952 1.394 1.8533 1.8361 1.8203 1.8055 1.7918 1.9645 1.938 1.9139 1.892 1.8718 1.6373 1.5343 1.429 1.318 1.8027 1.7851 1.7689 1.7537 1.7396 1.9165 1.8894 1.8648 1.8424 1.8217 3.7398 3.3043 3.0053 2.7872 2.6211 3.7743 3.3404 3.0428 2.8259 2.6609 L I D D E L L , 3.4668 3.4434 3.4221 3.4028 3.3852 3.6337 3.5915 3.5546 3.5219 3.4928 4.3874 3.9715 3.6875 3.4817 3.3258 4.5337 4.1203 3.8379 3.6331 3.478 1.5766 1.4673 1.3519 1.2214 1.7488 1.7306 1.7138 1.6981 1.6835 1.8657 1.838 1.8128 1.7896 1.7684 2.0589 2.0107 1.9681 1.9302 1.8963 2.448 2.341 2.2524 2.1778 2.1141 3.7047 3.2674 2.9669 2.7475 2.5801 1.5089 1.3893 1.2539 1 1.6906 1.6717 1.6541 1.6376 1.6223 1.8117 1.7831 1.757 1.733 1.711 2.0096 1.9604 1.9168 1.878 1.8432 2.4045 2.2962 2.2064 2.1307 2.0658 3.6689 3.2298 2.9276 2.7067 2.5379 n 3.5874 3.4903 3.4105 3.3439 3.2874 4.7571 4.3468 4.0662 3.8625 3.7083 506    Appendix B 3.9823 3.8853 3.8056 3.7389 3.6823 5.1433 4.7374 4.459 4.2565 4.1028 df1=1 2 3 4 5 6 7 8 9 10 12 15 20 24 30 40 60 120 ∞ 161.4476 199.5 215.7073 224.5832 230.1619 233.986 236.7684 238.8827 240.5433 241.8817 243.906 245.9499 248.0131 249.0518 250.0951 251.1432 252.1957 253.2529 254.3144 18.5128 19 19.1643 19.2468 19.2964 19.3295 19.3532 19.371 19.3848 19.3959 19.4125 19.4291 19.4458 19.4541 19.4624 19.4707 19.4791 19.4874 19.4957 10.128 9.5521 9.2766 9.1172 9.0135 8.9406 8.8867 8.8452 8.8123 8.7855 8.7446 8.7029 8.6602 8.6385 8.6166 8.5944 8.572 8.5494 8.5264 7.7086 6.9443 6.5914 6.3882 6.2561 6.1631 6.0942 6.041 5.9988 5.9644 5.9117 5.8578 5.8025 5.7744 5.7459 5.717 5.6877 5.6581 5.6281 6.6079 5.7861 5.4095 5.1922 5.0503 4.9503 4.8759 4.8183 4.7725 4.7351 4.6777 4.6188 4.5581 4.5272 4.4957 4.4638 4.4314 4.3985 4.365 Table B-4 © Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION. Statistical Tables 7/7/12 9:59:20 AM
User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.

This question has not been answered.

Create a free account to get help with this and any other question!

Related Tags