Chapter 1: Statistics - Radford University

Chapter 1: Statistics - Radford University

Chapter 2 ~ Descriptive Analysis & Presentation of Single-Variable Data 9 E JT Black Bears Mean: 60.07 inches Median: 62.50 inches Range: 42 inches 20 Variance: 117.681 Standard deviation: 10.85 inches Minimum: 36 inches Frequency Maximum: 78 inches 10 First quartile: 51.63 inches Third quartile: 67.38 inches Count: 58 bears Sum: 3438.1 inches 0

30 40 50 60 70 80 Length in Inches 1 9 E JT Chapter Goals Learn how to present and describe sets of data Learn measures of central tendency, measures of dispersion (spread), measures of position, and types of distributions Learn how to interpret findings so that we know what the data is telling us about the sampled population

2 9 E JT 2.1 ~ Graphic Presentation of Data Use initial exploratory data-analysis techniques to produce a pictorial representation of the data Resulting displays reveal patterns of behavior of the variable being studied The method used is determined by the type of data and the idea to be presented No single correct answer when constructing a graphic display 3 9 E JT Circle Graphs & Bar Graphs Circle Graphs and Bar Graphs: Graphs that are used to summarize attribute data Circle graphs (pie diagrams) show the amount of data that belongs to each category as a proportional part of a circle Bar graphs show the amount of data that belongs to each

category as proportionally sized rectangular areas 4 9 E JT Example Example: The table below lists the number of automobiles sold last week by day for a local dealership. Describe the data using a circle graph and a bar graph: Day Number Sold Monday 15 Tuesday 23 Wednesday 35 Thursday 11

Friday 12 Saturday 42 5 9 E JT Circle Graph Solution Automobiles Sold Last Week 6 9 E JT Bar Graph Solution Automobiles Sold Last Week Frequency 7

9 E JT Pareto Diagram Pareto Diagram: A bar graph with the bars arranged from the most numerous category to the least numerous category. It includes a line graph displaying the cumulative percentages and counts for the bars. Notes: Used to identify the number and type of defects that happen within a product or service Separates the vital few from the trivial many The Pareto diagram is often used in quality control applications 8 9 E JT Example Example: The final daily inspection defect report for a cabinet

manufacturer is given in the table below: Defect Dent Stain Blemish Chip Scratch Others Number 5 12 43 25 40 10 1) Construct a Pareto diagram for this defect report. Management has given the cabinet production line the goal of reducing their defects by 50%. 2) What two defects should they give special attention to in working toward this goal? 9 9 E JT Solutions Daily Defect Inspection Report 1)

140 100 120 80 100 Count 60 80 Percent 60 40 40 20 20 0 Defect: Count Percent

Cum% 0 Blemish Scratch Chip Stain Others Dent 43 31.9 31.9 40 29.6 61.5 25 18.5 80.0 12 8.9 88.9

10 7.4 96.3 5 3.7 100.0 2) The production line should try to eliminate blemishes and scratches. This would cut defects by more than 50%. 10 9 E JT Key Definitions Quantitative Data: One reason for constructing a graph of quantitative data is to examine the distribution - is the data compact, spread out, skewed, symmetric, etc. Distribution: The pattern of variability displayed by the data of a variable. The distribution displays the frequency of each value of the variable. Dotplot Display: Displays the data of a sample by representing each piece of data with a dot positioned along a scale. This scale can be either horizontal or vertical. The frequency of the values is represented along the other scale. 11

9 E JT Example Example: A random sample of the lifetime (in years) of 50 home washing machines is given below: 2.5 16.9 4.5 0.9 1.5 17.8 8.5 8.9 2.5 6.4 14.5 0.7 7.3 1.4 12.2 3.5 2.9 4.0 3.7

6.8 7.4 4.1 0.4 3.3 0.9 4.2 3.3 4.7 18.1 2.6 4.4 7.2 6.9 7.0 0.7 1.6 2.2 9.2 5.2 15.3 4.0 10.4 12.2 4.0 4.1

1.8 21.8 18.3 3.6 The figure below is a. dotplot for the 50 lifetimes: : . . .:. . ..: :.::::::.. .::. ... . : . . . +---------+---------+---------+---------+---------+------- 0.0 4.0 8.0 12.0

16.0 :. . 20.0 Note: Notice how the data is bunched near the lower extreme and more spread out near the higher extreme 12 9 E JT Stem & Leaf Display Background: The stem-and-leaf display has become very popular for summarizing numerical data It is a combination of graphing and sorting The actual data is part of the graph Well-suited for computers Stem-and-Leaf Display: Pictures the data of a sample using the actual digits that make up the data values. Each numerical data is divided into two parts: The leading digit(s) becomes the stem, and the trailing digit(s) becomes the leaf. The stems are located along the main axis, and a leaf for each piece of data is located so as to display the distribution of the data.

13 9 E JT Example Example: A city police officer, using radar, checked the speed of cars as they were traveling down the main street in town. Construct a stem-and-leaf plot for this data: 41 31 33 35 36 37 39 49 33 19 26 27 24 32 40 39 16 55 38 36 Solution: All the speeds are in the 10s, 20s, 30s, 40s, and 50s. Use the first digit of each speed as the stem and the second digit as the leaf. Draw a vertical line and list the stems, in order to the left of the line. Place each leaf on its stem: place the trailing digit on the right side of the vertical line opposite its corresponding leading digit. 14 9 E JT Example 20 Speeds --------------------------------------1 | 6 9 2 | 4 6 7 3 | 1 2 3 3 5 6 6 7 8 9 9 4 | 0 1 9

5 | 5 ---------------------------------------- The speeds are centered around the 30s Note: The display could be constructed so that only five possible values (instead of ten) could fall in each stem. What would the stems look like? Would there be a difference in appearance? 15 9 E JT Remember! 1. It is fairly typical of many variables to display a distribution that is concentrated (mounded) about a central value and then in some manner be dispersed in both directions. (Why?) 2. A display that indicates two mounds may really be two overlapping distributions 3. A back-to-back stem-and-leaf display makes it possible to compare two distributions graphically 4. A side-by-side dotplot is also useful for comparing two distributions 16 9 E JT

2.2 ~ Frequency Distributions & Histograms Stem-and-leaf plots often present adequate summaries, but they can get very big, very fast Need other techniques for summarizing data Frequency distributions and histograms are used to summarize large data sets 17 9 E JT Frequency Distributions Frequency Distribution: A listing, often expressed in chart form, that pairs each value of a variable with its frequency Ungrouped Frequency Distribution: Each value of x in the distribution stands alone Grouped Frequency Distribution: Group the values into a set of classes 1. A table that summarizes data by classes, or class intervals 2. In a typical grouped frequency distribution, there are usually 5-12 classes of equal width 3. The table may contain columns for class number, class interval, tally (if constructing by hand), frequency, relative frequency, cumulative relative frequency, and class midpoint

4. In an ungrouped frequency distribution each class consists of a single value 18 9 E JT Frequency Distribution Guidelines for constructing a frequency distribution: 1. All classes should be of the same width 2. Classes should be set up so that they do not overlap and so that each piece of data belongs to exactly one class 3. For problems in the text, 5-12 classes are most desirable. The square root of n is a reasonable guideline for the number of classes if n is less than 150. 4. Use a system that takes advantage of a number pattern, to guarantee accuracy 5. If possible, an even class width is often advantageous 19 9 E JT Frequency Distributions Procedure for constructing a frequency distribution: 1. Identify the high (H) and low (L) scores. Find the range. Range = H - L

2. Select a number of classes and a class width so that the product is a bit larger than the range 3. Pick a starting point a little smaller than L. Count from L by the width to obtain the class boundaries. Observations that fall on class boundaries are placed into the class interval to the right. 20 9 E JT Example Example: The hemoglobin test, a blood test given to diabetics during their periodic checkups, indicates the level of control of blood sugar during the past two to three months. The data in the table below was obtained for 40 different diabetics at a university clinic that treats diabetic patients: 6.5 6.4 5.0 7.9 5.0 6.0 8.0 6.0

5.6 5.6 6.5 5.6 7.6 6.0 6.1 6.0 4.8 5.7 6.4 6.2 8.0 9.2 6.6 7.7 7.5 8.1 7.2 6.7 7.9 8.0 5.9 7.7

8.0 6.5 4.0 8.2 9.2 6.6 5.7 9.0 1) Construct a grouped frequency distribution using the classes 3.7 - <4.7, 4.7 - <5.7, 5.7 - <6.7, etc. 2) Which class has the highest frequency? 21 9 E JT 1) Solutions Class Frequency Relative Cumulative Class Boundaries f Frequency Rel. Frequency Midpoint, x

--------------------------------------------------------------------------------------3.7 - <4.7 1 0.025 0.025 4.2 4.7 - <5.7 6 0.150 0.175 5.2 5.7 - <6.7 16 0.400 0.575 6.2 6.7 - <7.7 4 0.100 0.675 7.2 7.7 - <8.7 10 0.250 0.925 8.2 8.7 - <9.7 3 0.075 1.000 9.2

2) The class 5.7 - <6.7 has the highest frequency. The frequency is 16 and the relative frequency is 0.40 22 9 E JT Histogram Histogram: A bar graph representing a frequency distribution of a quantitative variable. A histogram is made up of the following components: 1. A title, which identifies the population of interest 2. A vertical scale, which identifies the frequencies in the various classes 3. A horizontal scale, which identifies the variable x. Values for the class boundaries or class midpoints may be labeled along the x-axis. Use whichever method of labeling the axis best presents the variable. Notes: The relative frequency is sometimes used on the vertical scale It is possible to create a histogram based on class midpoints 23 9

E JT Example Example: Construct a histogram for the blood test results given in the previous example The Hemoglobin Test Solution: 15 10 Frequency 5 0 4.2 5.2 6.2 7.2 8.2 9.2

Blood Test 24 9 E JT Example Example: A recent survey of Roman Catholic nuns summarized their ages in the table below. Construct a histogram for this age data: Age Frequency Class Midpoint -----------------------------------------------------------20 up to 30 34 25 30 up to 40 58 35 40 up to 50 76 45 50 up to 60 187 55 60 up to 70 254

65 70 up to 80 241 75 80 up to 90 147 85 25 9 E JT Solution Roman Catholic Nuns 200 Frequency 100 0 25 35 45 55

65 75 85 Age 26 9 E JT Terms Used to Describe Histograms Symmetrical: Both sides of the distribution are identical mirror images. There is a line of symmetry. Uniform (Rectangular): Every value appears with equal frequency Skewed: One tail is stretched out longer than the other. The direction of skewness is on the side of the longer tail. (Positively skewed vs. negatively skewed) J-Shaped: There is no tail on the side of the class with the highest frequency Bimodal: The two largest classes are separated by one or more classes. Often implies two populations are sampled. Normal: A symmetrical distribution is mounded about the mean and becomes sparse at the extremes 27

9 E JT Important Reminders The mode is the value that occurs with greatest frequency (discussed in Section 2.3) The modal class is the class with the greatest frequency A bimodal distribution has two high-frequency classes separated by classes with lower frequencies Graphical representations of data should include a descriptive, meaningful title and proper identification of the vertical and horizontal scales 28 9 E JT

Cumulative Frequency Distribution Cumulative Frequency Distribution: A frequency distribution that pairs cumulative frequencies with values of the variable The cumulative frequency for any given class is the sum of the frequency for that class and the frequencies of all classes of smaller values The cumulative relative frequency for any given class is the sum of the relative frequency for that class and the relative frequencies of all classes of smaller values 29 9 E JT Example Example: A computer science aptitude test was given to 50 students. The table below summarizes the data: Class Relative Cumulative Cumulative Boundaries Frequency Frequency Frequency Rel. Frequency ------------------------------------------------------------------------------------0 up to 4

4 0.08 4 0.08 4 up to 8 8 0.16 12 0.24 8 up to 12 8 0.16 20 0.40 12 up to 16 20 0.40 40 0.80 16 up to 20 6

0.12 46 0.92 20 up to 24 3 0.06 49 0.98 24 up to 28 1 0.02 50 1.00 30 9 E JT Ogive Ogive: A line graph of a cumulative frequency or cumulative relative frequency distribution. An ogive has the following components: 1. A title, which identifies the population or sample 2. A vertical scale, which identifies either the cumulative frequencies or the cumulative relative frequencies 3. A horizontal scale, which identifies the upper class boundaries. Until

the upper boundary of a class has been reached, you cannot be sure you have accumulated all the data in the class. Therefore, the horizontal scale for an ogive is always based on the upper class boundaries. Note: Every ogive starts on the left with a relative frequency of zero at the lower class boundary of the first class and ends on the right with a relative frequency of 100% at the upper class boundary of the last class. 31 9 E JT Example Example: The graph below is an ogive using cumulative relative frequencies for the computer science aptitude data: Computer Science Aptitude Test 1.0 0.9 0.8 0.7 Cumulative Relative Frequency 0.6 0.5 0.4

0.3 0.2 0.1 0.0 0 4 8 12 16 20 24 28 Test Score 32 9 E JT 2.3 ~ Measures of Central Tendency Numerical values used to locate the middle of a

set of data, or where the data is clustered The term average is often associated with all measures of central tendency 33 9 E JT Mean Mean: The type of average with which you are probably most familiar. The mean is the sum of all the values divided by the total number of values, n: 1 1 xn ) x xi ( x1 x 2 n n Notes: The population mean, , (lowercase mu, Greek alphabet), is the mean of all x values for the entire population

We usually cannot measure but would like to estimate its value A physical representation: the mean is the value that balances the weights on the number line 34 9 E JT Example Example: The following data represents the number of accidents in each of the last 8 years at a dangerous intersection. Find the mean number of accidents: 8, 9, 3, 5, 2, 6, 4, 5: Solution: 1 x (8 9 3 5 2 6 4 5) 5.25 8 In the data above, change 6 to 26: Solution:

1 x (8 9 3 5 2 26 4 5) 7.75 8 Note: The mean can be greatly influenced by outliers 35 9 E JT Median Median: The value of the data that occupies the middle position when the data are ranked in order according to size Notes: Denoted by x tilde: ~ x The population median, (uppercase mu, Greek alphabet), is the data value in the middle position of the entire population To find the median: 1. Rank the data 2. Determine the depth of the median: 3. Determine the value of the median d (~

x ) n 1 2 36 9 E JT Example Example: Find the median for the set of data: {4, 8, 3, 8, 2, 9, 2, 11, 3} Solution: 1. Rank the data: 2, 2, 3, 3, 4, 8, 8, 9, 11 2. Find the depth: ~ d ( x ) (9 1)/ 2 5 3. The median is the fifth number from either end in the ranked data: ~ x 4 Suppose the data set is {4, 8, 3, 8, 2, 9, 2, 11, 3, 15}: 1. Rank the data: 2, 2, 3, 3, 4, 8, 8, 9, 11, 15 2. Find the depth: d ( ~x ) (10 1) / 2 5.5 3. The median is halfway between the fifth and sixth ~

x (4 8)/ 2 6 observations: 37 9 E JT Mode & Midrange Mode: The mode is the value of x that occurs most frequently Note: If two or more values in a sample are tied for the highest frequency (number of occurrences), there is no mode Midrange: The number exactly midway between a lowest value data L and a highest value data H. It is found by averaging the low and the high values: LH midrange 2 38 9 E JT Example Example: Consider the data set {12.7, 27.1, 35.6, 44.2, 18.0} H 12.7 44.2

L 2845 Midrange . 2 2 Notes: When rounding off an answer, a common rule-of-thumb is to keep one more decimal place in the answer than was present in the original data To avoid round-off buildup, round off only the final answer, not intermediate steps 39 9 E JT 2.4 ~ Measures of Dispersion Measures of central tendency alone cannot completely characterize a set of data. Two very different data sets may have similar measures of central tendency.

Measures of dispersion are used to describe the spread, or variability, of a distribution Common measures of dispersion: range, variance, and standard deviation 40 9 E JT Range Range: The difference in value between the highest-valued (H) and the lowestvalued (L) pieces of data: rangeH L Other measures of dispersion are based on the following quantity Deviation from the Mean: A deviation from the mean, is the difference between the value of x and the mean , x x x 41 9

E JT Example Example: Consider the sample {12, 23, 17, 15, 18}. Find 1) the range and 2) each deviation from the mean. Solutions: 1) range H L 23 12 11 1 2) x (12 2317 1518) 17 5 Data Deviation from Mean x x x _________________________ -5 12 6 23 17 0 15 -2 18 1 42

9 E JT Note: Mean Absolute Deviation (x x) 0 (Always!) Mean Absolute Deviation: The mean of the absolute values of the deviations from the mean: 1 Mean absolute deviation n | x x| For the previous example: 1 n 1 14

| x x | (5 6 0 2 1) 2.8 5 5 43 9 E JT Sample Variance & Standard Deviation Sample Variance: The sample variance, s2, is the mean of the squared deviations, calculated using n 1 as the divisor: 1 s n 1 2 ( x x)2 where n is the sample size

Note: The numerator for the sample variance is called the sum of squares for x, denoted SS(x): s2 SS( x) n 1 where SS( x ) 2 ( x x ) 2 x 1 n 2 x Standard Deviation: The standard deviation of a sample, s, is the positive square root of the variance:

s s2 44 9 E JT Example Example: Find the 1) variance and 2) standard deviation for the 7, 1, 3, 8}: data {5, Solutions: First: x 1(5 7 1 3 8) 48 . 5 Sum: 1) s 2 x x x ( x x)2

5 7 1 3 8 24 0.2 2.2 -3.8 -1.8 3.2 0 0.04 4.84 14.44 3.24 10.24 32.08 1 ( 32 . 8 ) 8 . 2 4 2) s 8 . 2 2 . 86 45

9 E JT Notes The shortcut formula for the sample variance: s2 x 2 x 2 n n 1 The unit of measure for the standard deviation is the same as the unit of measure for the data

46 9 E JT 2.5 ~ Mean & Standard Deviation of Frequency Distribution If the data is given in the form of a frequency distribution, we need to make a few changes to the formulas for the mean, variance, and standard deviation Complete the extension table in order to find these summary statistics 47 9 E JT To Calculate In order to calculate the mean, variance, and standard deviation for data: 1. In an ungrouped frequency distribution, use the frequency of occurrence, f, of each observation 2. In a grouped frequency distribution, we use the frequency of occurrence associated with each class midpoint:

xf x f 2 s x 2 xf f 2 f f 1 48 9 E JT

Example Example: A survey of students in the first grade at a local school asked for the number of brothers and/or sisters for each child. The results are summarized in the table below. Find 1) the mean, 2) the variance, and 3) the standard deviation: Solutions: First: x f xf x2 f 0 1 2 4 5 15 17 23 5 2 62

0 17 46 20 10 93 0 17 92 80 50 239 Sum: 1) x 93/ 62 15 . 2 ( 93 ) 239 . 2) s2 62 62 1 163 . 128

. 3) s 163 49 9 E JT TI-83 Calculations When dealing with a grouped frequency distribution, use the following technique: Input the class midpoints or data values into L1 and the frequencies into L2; then continue with Highlight: Enter: Highlight: Enter: Highlight: L3 L3 = L1*L2 L4 L4 = L1*L3 L5(1) (first position in L5 column) 50 9 E JT

TI-83 Calculations Enter: L5(1) = sum(L2) (f) To find sum use 2nd List>Math>5:sum( L5(2) = sum(L3) (xf) L5(3) = sum(L4) (x2f) L5(4) = L5(2)/L5(1) to find the mean L5(5) = (L5(3)-(L5(2))2/L5(1))/(L5(1)-1) to find the variance L5(6) = 2nd (L5(5) to find the standard deviation Lets work problem 2.108 as an example! 51 9 E JT Problem 2.108 Find the mean and the variance for this grouped frequency distribution: Class Boundaries

f 26 7 6 10 15 10 14 22 14 18 14 18 22 2 Step 1: Enter the midpoints into L1 Step 2: Enter the frequencies into L2 52 9 E JT Problem 2.108 Highlight L3 and enter L1 * L2 Highlight L4 and enter L1 * L3 Highlight L5(1) and enter Sum(L2) 53 9 E

JT Problem 2.108 Highlight L5(2) and enter Sum(L3) Highlight L5(3) and enter Sum(L4) Highlight L5(4) and enter L5(2)/L5(1) Highlight L5(5) and enter (L5(3)-(L5(2)) /L5(1))/(L5(1)-1) Finally highlight L5(6) and enter 2 nd (L5(5)) 2 54 9 E JT Problem 2.73 Runs At Home Runs Away Difference Mean

Median 9.77 9.65 9.80 9.78 -0.03 -0.06 Maximum Minimum Midrange 13.65 7.64 10.65 11.06 8.67 9.87 4.89 -1.74 1.58 55 9

E JT Problem 2.73 cont. Teams playing the Rockies at Coors Field generated the maximum number of runs scored (13.65). On the other hand, while playing their opponents away, the Rockies and their opponents, were able to generate only 8.76 runs, which ranked second from the bottom. Collectively, these two performances produced the greatest spread (4.89) by a considerable margin between runs scored at home and runs scored away by any stadium/team combination in the major leagues. This unusually large value inflates the midrange difference. It appears the playing conditions at Coors Field, therefore, are more responsible for producing the higher combined scores than the strength of either the Rockies or their opponents bats or any weakness of the pitching staffs. 56 9 E JT a. b. c. d. e.

f. g. h. Problem 2.75 x need to be 500, therefore need any three numbers that total 330. Need two numbers smaller than 70 and one larger Need multiple 87s Need any two numbers that total 140 for the extreme values where one is 100 or larger Need two numbers smaller than 70 and one larger than 70 so their total is 330 Need two numbers of 87 and a third number large enough so that the total of all five is 500. Mean equal to 100 requires the five data to total 500 and the midrange of 70 requires the total of L and H to be 140; 40, __, 70, ___, 100; that is a sum of 210, meaning the other two data must total 290. One of the last two numbers must be larger than 145, which would then become H and change the midrange. Impossible. There must be two 87s in order to have a mode, and there can only be two data larger than 70 in order for 70 to be the median. , 70, 87, 87, 100; Impossible 57 9 E JT

Problem 2.83 a. Range = H L = 9 2 = 7 b. 1st: find the mean: 6 30 x x xbar (x xbar)2 2 -4 16 4 -2 4 7 1

1 8 2 4 9 3 9 0 34 s2 = (x-xbar)2/(n-1) = 34/4 = 8.5 s = s2 = 8.5 = 2.9 58 9 E JT 2.6 ~ Measures of Position Measures of position are used to describe the

relative location of an observation Quartiles and percentiles are two of the most popular measures of position An additional measure of central tendency, the midquartile, is defined using quartiles Quartiles are part of the 5-number summary 59 9 E JT Quartiles Quartiles: Values of the variable that divide the ranked data into quarters; each set of data has three quartiles 1. The first quartile, Q1, is a number such that at most 25% of the data are smaller in value than Q1 and at most 75% are larger 2. The second quartile, Q2, is the median 3. The third quartile, Q3, is a number such that at most 75% of the data are smaller in value than Q3 and at most 25% are larger Ranked data, increasing order 25% L

25% Q1 25% Q2 25% Q3 H 60 9 E JT Percentiles Percentiles: Values of the variable that divide a set of ranked data into 100 equal subsets; each set of data has 99 percentiles. The kth percentile, Pk, is a value such that at most k% of the data is smaller in value than Pk and at most (100 k)% of the data is larger. at most k % L at most (100 - k )% Pk H

Notes: The 1st quartile and the 25th percentile are the same: Q1 = P25 The median, the 2nd quartile, and the 50th percentile are ~ all the same: x Q2 P50 61 9 E JT Finding Pk (and Quartiles) Procedure for finding Pk (and quartiles): 1. Rank the n observations, lowest to highest 2. Compute A = (nk)/100 3. If A is an integer: d(Pk) = A.5 (depth) Pk is halfway between the value of the data in the Ath position and the value of the next data If A is a fraction: d(Pk) = B, the next larger integer Pk is the value of the data in the Bth position

62 9 E JT Example Example: The following data represents the pH levels of a random sample of swimming pools in a California town. Find: 1) the first quartile, 2) the third quartile, and 3) the 37th percentile: 5.6 6.0 6.7 7.0 5.6 6.1 6.8 7.3 5.8 6.2 6.8 7.4 5.9 6.3 6.8

7.4 6.0 6.4 6.9 7.5 Solutions: 1) k = 25: (20) (25) / 100 = 5, depth = 5.5, Q1 = 6 2) k = 75: (20) (75) / 100 = 15, depth = 15.5, Q3 = 6.95 3) k = 37: (20) (37) / 100 = 7.4, depth = 8, P37 = 6.2 63 9 E JT Midquartile Midquartile: The numerical value midway between the first and third quartile:

Q Q midquartile 1 2 3 Example: Find the midquartile for the 20 pH values in the previous example: Q3 6 6.95 12.95 Q 1 6.475 midquartil e 2 2 2 Note: The mean, median, midrange, and midquartile are all measures of central tendency. They are not necessarily equal. Can you think of an example when they would be the same value? 64 9 E JT 5-Number Summary 5-Number Summary: The 5-number summary is composed of: 1. L, the smallest value in the data set 2. Q1, the first quartile (also P25)

~ 3. x , the median (also P50 and 2nd quartile) 4. Q3, the third quartile (also P75) 5. H, the largest value in the data set Notes: The 5-number summary indicates how much the data is spread out in each quarter The interquartile range is the difference between the first and third quartiles. It is the range of the middle 50% of the data 65 9 E JT Box-and-Whisker Display Box-and-Whisker Display: A graphic representation of the 5-number summary: The five numerical values (smallest, first quartile, median, third quartile, and largest) are located on a scale, either vertical or horizontal The box is used to depict the middle half of the data that lies between the two quartiles

The whiskers are line segments used to depict the other half of the data One line segment represents the quarter of the data that is smaller in value than the first quartile The second line segment represents the quarter of the data that is larger in value that the third quartile 66 9 E JT Example Example: A random sample of students in a sixth grade class was selected. Their weights are given in the table below. Find the 5-number summary for this data and construct a boxplot: 63 64 88 89 93 109 94 112

76 90 97 76 91 99 81 92 99 83 85 93 99 86 93 101 108 Solution: 63 L

85 Q1 92 ~ x 99 Q3 112 H 67 9 E JT Boxplot for Weight Data Weights from Sixth Grade Class 60 70 80 90

100 110 Weight L Q1 ~ x Q3 H 68 9 E JT z-Score z-Score: The position a particular value of x has relative to the mean, measured in standard deviations. The z-score is found by the formula: z

value mean x x st.dev. s Notes: Typically, the calculated value of z is rounded to the nearest hundredth The z-score measures the number of standard deviations above/below, or away from, the mean z-scores typically range from -3.00 to +3.00 z-scores may be used to make comparisons of raw scores 69 9 E JT Example Example: A certain data set has mean 35.6 and standard

7.1. Find the z-scores for 46 and 33: deviation Solutions: z x s x 46 35.6 1.46 7.1 46 is 1.46 standard deviations above the mean x x 33 35.6 z 0.37 s 7.1 33 is 0.37 standard deviations below the mean. 70 9 E JT 2.7 ~ Interpreting & Understanding Standard Deviation Standard deviation is a measure of variability, or spread Two rules for describing data rely on the standard deviation: Empirical rule: applies to a variable that is normally distributed Chebyshevs theorem: applies to any distribution

71 9 E JT Empirical Rule Empirical Rule: If a variable is normally distributed, then: 1. Approximately 68% of the observations lie within 1 standard deviation of the mean 2. Approximately 95% of the observations lie within 2 standard deviations of the mean 3. Approximately 99.7% of the observations lie within 3 standard deviations of the mean Notes: The empirical rule is more informative than Chebyshevs theorem since we know more about the distribution (normally distributed) Also applies to populations Can be used to determine if a distribution is normally distributed 72

9 E JT Illustration of the Empirical Rule 99.7% 95% 68% x 3s x 2s x s x x s x 2s x 3s 73 9 E JT Example

Example: A random sample of plum tomatoes was selected recorded. The mean weight was 6.5 ounces with a standard normally distributed: from a local grocery store and their weights deviation of 0.4 ounces. If the weights are 1) What percentage of weights fall between 5.7 and 7.3? 2) What percentage of weights fall above 7.7? Solutions: 1) ( x 2s, x 2s)(65 . 2(0.4), 65 . 2(0.4))(57 . , 7.3) Approximately 95% of the weights fall between 5.7 and 7.3 2) ( x 3s, x 3s)(65 . 3(0.4), 65 . 3(0.4))(53 . , 7.7) Approximately 99.7% of the weights fall between 5.3 and 7.7 Approximately 0.3% of the weights fall outside (5.3, 7.7) Approximately (0.3/2)=0.15% of the weights fall above 7.7 74 9 E JT A Note about the Empirical Rule

Note: The empirical rule may be used to determine whether or not a set of data is approximately normally distributed 1. Find the mean and standard deviation for the data 2. Compute the actual proportion of data within 1, 2, and 3 standard deviations from the mean 3. Compare these actual proportions with those given by the empirical rule 4. If the proportions found are reasonably close to those of the empirical rule, then the data is approximately normally distributed 75 9 E JT Chebyshevs Theorem Chebyshevs Theorem: The proportion of any distribution that lies within k standard deviations of the mean is at least 1 (1/k2), where k is any positive number larger than 1. This theorem applies to all distributions of data. Illustration: at least 1 12 k x ks

x x ks 76 9 E JT Important Reminders! Chebyshevs theorem is very conservative and holds for any distribution of data Chebyshevs theorem also applies to any population The two most common values used to describe a distribution of data are k = 2, 3 The table below lists some values for k and 1 - (1/k2): k

1 (1/ k 2 ) 1.7 0.65 2 0.75 2.5 0.84 3 0.89 77 9 E JT Example Example: At the close of trading, a random sample of 35 technology stocks was selected. The mean selling price was 67.75 and the standard deviation was 12.3. Use Chebyshevs theorem (with k = 2, 3) to describe the distribution.

Solutions: Using k=2: At least 75% of the observations lie within 2 standard deviations of the mean: ( x 2s, x 2 s ) (67.75 2(12.3), 67.75 2(12.3) (43.15, 92.35) Using k=3: At least 89% of the observations lie within 3 standard deviations of the mean: ( x 3s, x 3s) (67.75 3(12.3), 67.75 3(12.3) (30.85, 104.65) 78 9 E JT 2.8 ~ The Art of Statistical Deception Good Arithmetic, Bad Statistics Misleading Graphs Insufficient Information 79 9 E JT Good Arithmetic, Bad Statistics

The mean can be greatly influenced by outliers Example: The mean salary for all NBA players is $15.5 million Misleading graphs: 1. The frequency scale should start at zero to present a complete picture. Graphs that do not start at zero are used to save space. 2. Graphs that start at zero emphasize the size of the numbers involved 3. Graphs that are chopped off emphasize variation 80 9 E JT Flight Cancellations 35 30 25 Number of Cancellations 20 15 10 5 0 1996

1998 2000 2002 Year 81 9 E JT Flight Cancellations 35 34 33 Number of Cancellations 32 31 30 29 28 27 1996

1998 2000 2002 Year 82 9 E JT Insufficient Information Example: An admissions officer from a state school explains that the average tuition at a nearby private university is $13,000 and only $4500 at his school. This makes the state school look more attractive. If most students pay the full tuition, then the state school appears to be a better choice However, if most students at the private university receive substantial financial aid, then the actual tuition cost could be much lower! 83

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