2. What we are going to learn? Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. Grouped Data a. Mean b. Mode c. Median
3. Review Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Ratio Nominal Ordinal Interval What is the level of measurement for these items related to the newspaper business? The number of papers sold each Sunday during 2006. The departments, such as editorial, advertising , sports, etc. A summary of the number of papers sold by county. The number of years with the paper for each employee. Ratio Nominal Ratio Ratio P14. N.2 Ch.1
4. Review Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Sample Population For the follow questions, would you collect information using a sample or a population? Statistics 201 is a course taught at a university. Professor A has taught nearly 1,500 students in the course over the past 5 years. You would like to know the average grade for the course You are looking forward to graduation project and your first job as a salesperson for one of five large corporations. Planning for your interviews, you will need to know about each company’s mission, profitability, products, and markets. Sample Population P16. N.8 Ch.1
5. Review-Qualitative Data Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Pie Chart Bar Chart
6. Review-Quantitative Data Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Cumulative Frequency Distribution Histogram Polygon
7. Review-Quantitative Data Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B.GroupedData a. Mean b. Mode c. Median A Cumulative Frequency Distribution A (21, 30) Around 43% of the vehicleswereseldbelow $21,000.
8. Review Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median A set of data contains53observations. The lowestvalue is 43 and the largest is 129. The data are to beorganizedinto a frequency distribution. a. Howmanyclasseswouldyousuggest? 130 - 43 6 i > ≈ 15 25 = 32, 26 = 64, suggests 6 classes b. Whatwouldyousuggest as class interval & the lower limit of the firstclass? Use interval of 15 And start first class at 40 P34. N.10 Ch.2
9. Central Tendency Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Parameter: a numerical characteristic of a population. Example:The fraction of U. S. voters who support Sen. McCain for President is a parameter. Statistic: A statistic is a numerical characteristic of a sample. Example: If we select a simple random sample of n = 1067 voters from the population of all U. S. voters, the fraction of people in the sample who support Sen. McCain is a statistic.
10. Central Tendency Parameter & Statistics Review Chapter 3-A: Central Tendency A. Grouped Data a. Mean b. Mode c. Median B. Ungrouped Data a. Mean b. Mode c. Median
11. Central Tendency: Mean Sum of all the values in the population Population mean = Number of values in the population Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Example:
12. Central Tendency: Mean Sum of all the values in the sample Sample mean = Number of values in the sample Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median
13. Central Tendency: Mean Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Example: A sample of five executives received the following bonus last year ($000): 14.0, 15.0, 17.0, 16.0, 15.0 $ 15,400 Every set of interval- or ratio-level data has a mean All the values are included in computing the mean The mean is unique.
14. Central Tendency: Mean Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Example: Consider the set of values: 3, 8, and 4. The mean is 5. 4. The sum of the deviations of each value from the mean is zero.
15. Central Tendency: WeightedMean Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Weighted Mean: a set of numbers X1, X2, ..., Xn, with corresponding weights w1, w2, ...,wn, is computed from the following formula:
16. Central Tendency: WeightedMean Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Weighted Mean: Example: During a one hour period on a hot Saturday afternoon, Julie served fifty lemon drinks. She sold five drinks for $0.50, fifteen for $0.75, fifteen for $0.90, and fifteen for $1.10. Compute the weighted mean of the price of the drinks.
17. Exercise Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median The Bookstallsoldbooks via internet. Paperbacks are $1.00 each, and hardcover books are $3.50. Of the 50 bookssoldon last Tuesday, 40 were paperback and the rest were hardcover. What was the weightedmeanprice of a book? 40 paperback $1.00 10 hardcover $3.50 P62. N.14 Ch.3
18. Central Tendency: Mode Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Mode: There is one situation in which the mode is the only measure of central tendency that can be used – when we have categorical, or non-numeric data. In this situation, we cannot calculate a mean or a median. The mode is the most typical value of the categorical data. Example: Suppose I have collected data on religious affiliation of citizens of the U.S. The modal, or most Typical value, is Roman Catholic, since The Roman Catholic Church is the largest religious organization in the U.S.
19. Central Tendency: Mode Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Mode: The value of the observation that appears most frequently.
20. Central Tendency: Mode Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Mode: The value of the observation that appears most frequently. Example: The exam scores for ten students are: 81, 93, 84, 75, 68, 87, 81, 75, 81, 87. Because the score of 81 occurs the most often, it is the mode.
21. Central Tendency: Median Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Median: the midpoint of the values after they have been ordered from the smallest to the largest. Example: The ages for a sample of five college students are: 21, 25, 19, 20, 22 Arranging the data in ascending order gives: 19, 20, 21, 22, 25. Thus the median is21.
22. Central Tendency: Median Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median For an even set of values, the median will be the arithmetic average of the two middle numbers. Example: The heights of four basketball players, in inches, are: 76, 73, 80, 75 Arranging the data in ascending order gives: 73, 75, 76, 80. Thus the median is 75.5
23. Central Tendency: Median Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Example: Finding the median 72 68 65 70 75 79 73 65 68 70 72 73 75 79 65 68 70 72 73 75 79 79 72.5 65 68 70 72 73 75 79 79,000 72.5 Median is notinfluencedby the extreme value.
24. Central Tendency: Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median List below are the total automobile sales (in millions of dollars) for the last 14 years. What was the mediannumber of automobiles sold? What is the mode? 41 15 39 54 31 15 33 Mean= 32.57; Median=33; Mode=15 P65. N.22 Ch.3
25. Central Tendency: Review Chapter 3-A: Central Tendency A. Grouped Data a. Mean b. Mode c. Median B. Ungrouped Data a. Mean b. Mode c. Median Central Tendency Mean, Mode, Median P69. N.26 Ch.3
26. Central Tendency: Mean Review Chapter 3-A: Central Tendency A. Grouped Data a. Mean b. Mode c. Median B. Ungrouped Data a. Mean b. Mode c. Median
27. Central Tendency: Mean Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median
28. Central Tendency: Mean Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Determine the mean of the followingfrequency distribution. X=380/30=12.67 P87. N.58 Ch.3
29. Central Tendency: Mode Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Example: Finding the mode forgrouped data Step 2: Step 1: Midpoint of the modal class is the mode Modal class with the highest frequency 19.5
30. Central Tendency: Median Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Example: Finding the medianforgrouped data CumulativeFrequency Distribution Step 1:
31. Central Tendency: Median Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Determine the position of the median and the medianclass Step 2:
32. Central Tendency: Median Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Draw twolines (value & position) Step 3: A Value: 100 Median 150 B Position: 201 300.5 388 Median – 100 150 - 100 300.5 – 201 388 - 201 300.5 – 201 388 - 201 = Median = * 50 + 100 = 126.60 (dollars)
33. Exercise Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median SCCoast, an Internet provider in the Southeast, developed the following frequency distribution on the age of Internet users. Describe the central tendency: X = 2410 / 60 = 40.17 (years) Mode = 45 (years) Median = ? (years) P87 N.60 Ch.3
34. Exercise Value:40 50 Location: 28 48 Review Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. GroupedData a. Mean b. Mode c. Median Step 1: Define the location of the median Step 2: Calculate the median M Lm=(60+1)/2=30.5 30.5-28 48-28 M-40 50-40 30.5 = Median= 41.25 years P87 N.60 Ch.3
35.
36. Chapter 3-A: Central Tendency A. Ungrouped Data a. Mean b. Mode c. Median B. Grouped Data a. Mean b. Mode c. Median
37. Chapter 3: Describing Data 8. The Relative Positions of the Mean, Median, and Mode skewed
38. Chapter 3: Describing Data 8. The Relative Positions of the Mean, Median, and Mode Zero skewness mode=median=mean
39. Chapter 3: Describing Data 7. The Relative Positions of the Mean, Median, and Mode positive skewness Mode median mean < <
40. Chapter 3: Describing Data 8. The Relative Positions of the Mean, Median, and Mode negative skewness Mode median mean > >
42. Chapter 3: Describing Data 9. The GeometricMean Geometric mean (GM) : a set of n numbers is defined as the nth root of the product of the n numbers. The formula is: The geometric mean is used to average percents, indexes, and relatives. The geometric mean is not applicable when some numbers are negative.
43. Chapter 3: Describing Data 9. The GeometricMean Example: Suppose you receive a 5 percent increase in salary this year and a 15 percent increase next year. The average annual percent increase is 9.886, not 10.0. Why is this so? We begin by calculating the geometric mean. Not understand percentage? Click here
44. Chapter 3: Describing Data 9. The GeometricMean Example: The return on investment earned by Atkins construction Company for four successive years was: 30 percent, 20 percent, -40 percent, and 200 percent. What is the geometric mean rate of return on investment?
45. Chapter 3: Describing Data 9. The GeometricMean Geometric mean (GM) : Another use of the geometric mean is to determine the percent increase in sales, production or other business or economic series from one time period to another.
46. Chapter 3: Describing Data 9. The GeometricMean Example: The total number of females enrolled in American colleges increased from 755,000 in 1992 to 835,000 in 2000. That is, the geometric mean rate of increase is1.27%.
47. Chapter 3: Describing Data 9. The GeometricMean Example: A banker wants to get an annual return of 100% on its loan in credit card business. What monthly interest rate should he charge? A monthly interest rate of 5.9%.
48. Chapter 3: Describing Data 9. The GeometricMean Example: The Chinese government claimed in 1990 that their GDP will double in 20 years. What must the annual GDP growth rate be for this dream to come true? A annual GDP growth of 3.5%.
49. Chapter 3: Describing Data 9. The GeometricMean Example: The 2006 population size of Duval County was 837,964. The population grew by 7.6% between 2000 and 2006. We want to project the size of the population in 2030, assuming that the growth rate remains the same; i.e., 7.6% every 6 years. The Projected population size in 2030 is (1.0764 X 837,964) = 1123245. The average growth rate over the 24 years is found by calculating the geometric mean: The average growth rate is just what we expect.
50. Exercise Chapter 3: Describing Data In 1976 the nationwide average price of a gallon of unleaded gasoline at a self-serve pump was $0.605. By 2005 the average price had increased to $2.57. What was the geometric mean annual increase for the period? 5.11% found by -1 2.57 0.605 29 P71. N.32 Ch.3
51. Chapter 2: Describing Data Review Qualitative Data Two thousand frequent mIdwestern business travelers are asked which Midwest city they prefer: Indianapolis, Saint Louis, Chicago, or Milwaukee. The results were 100 liked Indianapolisbest, 450 liked Saint Louis, 1,300 liked Chicago, and the remainder preferred Milwaukee. Develop a frequency table and a relative frequency table to summarize this information. P27. N.4 Ch.2
52. Chapter 2: Describing Data Review 99 - 51 5 The daily number of oil changes at the Oak Streek outlet in the past 20 days are: The data are to be organzied into a frequency distribution. a. How many classes would you recommend? i > ≈ 10 24 = 16, 25 = 32, suggests 5 classes b. What class interval would you suggest? Use interval of 10 P34. N.12 Ch.2
53. Chapter 2: Describing Data Review The daily number of oil changes at the Oak Streek outlet in the past 20 days are: The data are to be organzied into a frequency distribution. c. What lower limit would you recommend for the first class? start first class at 50 P34. N.12 Ch.2
54. Chapter 2: Describing Data Review The daily number of oil changes at the Oak Streek outlet in the past 20 days are: d. Organize the number of oil changes into a frequency distribution. P34. N.12 Ch.2
55. Chapter 2: Describing Data Review The daily number of oil changes at the Oak Streek outlet in the past 20 days are: e. Comment on the shape of the frequency distribution. Also determine the relative frequency distribution. The fewest number is about 50, the highest about 100. The greatest concentration is in classes 60 up to 70 and 70 up to 80. P34. N.12 Ch.2
57. Chapter 3: Describing Data Exercise a. Compute the mean of the following population values: 7, 5, 7, 3, 7, 4 μ = 5.5 found by (7+5+7+3+7+4)/6 P60. N.2 Ch.3
58. Chapter 3: Describing Data Exercise Compute the mean of the following sample values: 1.3 7.0 3.6 4.1 5.0 b. Show that Σ(X - X)=0 X = 4.2 found by 21/5 (1.3-4.2)+(7.0-4.2)+(3.6-4.2)+(4.1-4.2)+(5.0-4.2)=0 P60. N.4 Ch.3
59. More Information Source: Keller, Statistics for Management and Economics, 2005
60. More Information Percentage We added memory to our computer system. We had 96 MB of main memory and now with our new addition, we have 256 MB of main memory. I would like to figure out what percent increase this represents. If you go from 100 MB of memory to 200 MB then you've increased it by 100 percent, because the amount of the increase (100 MB) is 100% of the original amount (100 MB). That is... if you double your memory then you've increased it by 100 percent. If you add another 100 MB, you're adding another 100% of the original amount, so you have a 200% increase, from 100 MB to 300 MB. In this case, you have gone from about 100 to about 250. Since 250 is halfway between 200 MB and 300 MB, you could guess that the answer is about 150 percent. Does this make sense? Now let's find the actual value. I'm going to do a simple example first so you see how percentages work. If I go from 100 MB to 105 MB, what is the percent increase? In this case, the numbers are straightforward: the increase (5 MB) is 5 percent of the original amount (100 MB). But we can use a method that will work even when the numbers aren't this tidy: I ask: 100 times what number will give me 105? 100 * x = 105 x = 105 / 100 x = 1.05 Then I ask: What increase is that over 100%? x - 1 = 1.05 - 1 = 0.05 = 5/100 = 5% So I have an increase of 5%. Now let's do the same thing with your numbers: 1) 96 * x = 256 x = 256 / 96 x = 2.67 2) x - 1 = 2.67 - 1 = 1.67 = 167/100 = 167% which is pretty close to the original estimate of 150%. That gives us some confidence that we have the right answer.