IBS Statistics Year 1
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
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median
Review
What is the level of measurement for these items related to the newspaper business?
a. The number of papers sold each Sunday during 2006.
b. The departments, such as editorial, advertising , sports, etc.
c. A summary of the number of papers sold by county.
d. The number of years with the paper for each employee.
RatioRatio
RatioRatio
NominalNominal
RatioRatio
P14. N.2 Ch.1
RatioRatioNominalNominal OrdinalOrdinal IntervalInterval
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median
Review
For the follow questions, would you collect information using a sample or a population?
a. 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
b. 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.
SampleSample
PopulationPopulation P16. N.8 Ch.1
SampleSample PopulationPopulation
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median
Review-Qualitative DataBar Chart
Pie Chart
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median HistogramPolygon
Cumulative Frequency
Distribution
Review-Quantitative Data
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B.Grouped Data
a. Mean
b. Mode
c. Median
Cumulative Frequency
Distribution
Review-Quantitative Data
A (21, 30) Around 43% of the vehicles were seld below $21,000.
A
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median
Review
P34. N.10 Ch.2
A set of data contains 53 observations. The lowest value is 43 and the largest is 129. The data are to be organized into a frequency distribution.
a. How many classes would you suggest?
25 = 32, 26 = 64, suggests 6 classes
i > ≈ 15130 - 43
6
Use interval of 15And start first class at 40
b. What would you suggest as class interval & the lower limit of the first class?
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median
Central Tendency
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.
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
Parameter & Statistics
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median
Central Tendency: Mean
N
X=μ
∑
Population mean = Sum of all the values in the population
Number of values in the population
Example:
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median
Sample mean = Sum of all the values in the sample
Number of values in the sample
nΣX
=X
Central Tendency: Mean
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
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
15.4 =577
=5
15.0+...+14.0=
nΣX
=X $ 15,400
1. Every set of interval- or ratio-level data has a mean
2. All the values are included in computing the mean
3. The mean is unique.
Central Tendency: Mean
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median
4. The sum of the deviations of each value from the mean is zero.
Central Tendency: Mean
Example: Consider the set of values: 3, 8, and 4. The mean is 5.
0=5)(4+5)(8+5)(3=)X -Σ(X ---
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
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:
)n21
nn2211w ...w+w+(w
)Xw+...+Xw+X(w=X
Central Tendency: Weighted Mean
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median
Weighted Mean:
)n21
nn2211w ...w+w+(w
)Xw+...+Xw+X(w=X
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.
$0.89=50
$44.50=
15+15+15+515($1.10)+15($0.90)+15($0.75)+5($0.50)
=Xw
Central Tendency: Weighted Mean
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median
Exercise
P62. N.14 Ch.3
The Bookstall sold books via internet. Paperbacks are $1.00 each, and hardcover books are $3.50. Of the 50 books sold on last Tuesday, 40 were paperback and the rest were hardcover. What was the weighted mean price of a book?
)n21
nn2211w ...w+w+(w
)Xw+...+Xw+X(w=X
50.1$50
$3.50)*10+$1.00*(40=Xw
40 paperback$1.00
40 paperback$1.00
10 hardcover$3.50
10 hardcover$3.50
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median
Central Tendency: Mode
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.
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median
Central Tendency: Mode
Mode:
The value of the observation that appears most frequently.
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median
Central Tendency: Mode
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.
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median
Central Tendency: 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 is 21.
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median
Central Tendency: 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
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median
Central Tendency: Median
72 68 65 70 75 79 73
Example:
Finding the median
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 not influenced by the extreme value.
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median
Central Tendency:
P65. N.22 Ch.3
Mean= 32.57; Median=33; Mode=15
List below are the total automobile sales (in millions of dollars) for the last 14 years. What was the median number of automobiles sold? What is the mode?
41 15 39 54 31 15 33
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:
P69. N.26 Ch.3
Mean Mode Median
City - - -
Wind direction
- Southwest -
Temperature
91 o F 92 o F 92 o F
Pavement - Wet & Dry TraceCentral Tendency
Mean, Mode, M
edian
Review
Chapter 3-A:
Central Tendency
A. Grouped Data
a. Mean
b. Mode
c. Median
B. Ungrouped Dataa. Mean
b. Mode
c. Median
Central Tendency: Mean
32
Recall in Chapter 2, we constructed a frequency distribution for the vehicle selling prices. The information is repeated below. Determine the arithmetic mean vehicle selling price.
The Arithmetic Mean of Grouped Data -Example
32
Recall in Chapter 2, we constructed a frequency distribution for the vehicle selling prices. The information is repeated below. Determine the arithmetic mean vehicle selling price.
The Arithmetic Mean of Grouped Data -Example
33
The Arithmetic Mean of Grouped Data -Example
33
The Arithmetic Mean of Grouped Data -Example
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Dataa. Mean
b. Mode
c. Median
Central Tendency: Mean
33
The Arithmetic Mean of Grouped Data -Example
31
The Arithmetic Mean of Grouped Data
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Dataa. Mean
b. Mode
c. Median
Central Tendency: Mean
P87. N.58 Ch.3
Determine the mean of the following frequency distribution.
31
The Arithmetic Mean of Grouped Data
X=380/30=12.67
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Dataa. Mean
b. Mode
c. Median
Central Tendency: ModeExample:
Finding the mode for grouped data
Step 1:
Modal class with the highest frequency
32
Recall in Chapter 2, we constructed a frequency distribution for the vehicle selling prices. The information is repeated below. Determine the arithmetic mean vehicle selling price.
The Arithmetic Mean of Grouped Data -Example
Step 2:
Midpoint of the modal class is the mode
19.5
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Dataa. Mean
b. Mode
c. Median
Central Tendency: MedianExample:
Finding the median for grouped data
Step 1: Cumulative Frequency Distribution
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Dataa. Mean
b. Mode
c. Median
Central Tendency: MedianStep 2: Determine the position of the median and the median
class
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Dataa. Mean
b. Mode
c. Median
Central Tendency: Median
Step 3: Draw two lines (value & position)
Median – 100 150 - 100
300.5 – 201388 - 201=
A
B
Value: 100 Median 150
Position: 201 300.5 388
Median = 300.5 – 201388 - 201 * 50 + 100 = 126.60 (dollars)
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Dataa. Mean
b. Mode
c. Median
P87 N.60 Ch.3
SCCoast, an Internet provider in the Southeast, developed the following frequency distribution on the age of Internet users. Describe the central tendency:
Mode = 45 (years)
Median = ? (years)
Exercise
X = 2410 / 60 = 40.17 (years)
Review
Chapter 3-A:
Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Dataa. Mean
b. Mode
c. Median
P87 N.60 Ch.3
Lm=(60+1)/2=30.5 Value:40 50
Location: 28 48
30.5
30.5-2848-28 =
M-4050-40
Median= 41.25 years
Step 1: Define the location of the median Step 2: Calculate the median
M
Exercise
What we have learnt?
• Review
• Chapter 3-A: Central Tendency
A. Ungrouped Data
a. Mean
b. Mode
c. Median
B. Grouped Data
a. Mean
b. Mode
c. Median
8. The Relative Positions of the Mean, Median, and Mode
Chapter 3: Describing Data
skewed
8. The Relative Positions of the Mean, Median, and Mode
Chapter 3: Describing Data
Zero skewness
mode=median=mean
Zero skewness
mode=median=mean
7. The Relative Positions of the Mean, Median, and Mode
Chapter 3: Describing Data
positive skewness
Mode median mean
positive skewness
Mode median mean< <
8. The Relative Positions of the Mean, Median, and Mode
Chapter 3: Describing Data
negative skewness
Mode median mean
negative skewness
Mode median mean> >
8. The Relative Positions of the Mean, Median, and Mode
Chapter 3: Describing Data
9. The Geometric Mean
Chapter 3: Describing Data
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.
n n321 ))...(X)(X)(X(X=GM
9. The Geometric Mean
Chapter 3: Describing Data
n n321 ))...(X)(X)(X(X=GM
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.
098861151051 . ).)(.(GM
Not understand percentage?Click here
9. The Geometric Mean
Chapter 3: Describing Data
n n321 ))...(X)(X)(X(X=GM
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?
..).)(.)(.)(.(GM 2941808203602131 44
9. The Geometric Mean
Chapter 3: Describing Data
1period) of beginningat (Value
period) of endat (Value=GM n -
Geometric mean (GM) :Another use of the geometric mean is to determine the percent increase insales, production or other business or economic series from one time periodto another.
9. The Geometric Mean
Chapter 3: Describing Data
1period) of beginningat (Value
period) of endat (Value=GM n -
Example: The total number of females enrolled in American colleges increased from755,000 in 1992 to 835,000 in 2000.
.0127=1 - 755,000
835,000=GM 8
That is, the geometric mean rate of increase is 1.27%.
9. The Geometric Mean
Chapter 3: Describing Data
Example: A banker wants to get an annual return of 100% on its loan in credit cardbusiness. What monthly interest rate should he charge?
.059=1 - 100
200=GM 12
A monthly interest rate of 5.9%.
9. The Geometric Mean
Chapter 3: Describing Data
Example: The Chinese government claimed in 1990 that their GDP will double in 20years. What must the annual GDP growth rate be for this dream to come true?
A annual GDP growth of 3.5%.
.035=1 - 100
200=GM 20
9. The Geometric Mean
Chapter 3: Describing Data
Example: The 2006 population size of Duval County was 837,964. The population grew by7.6% between 2000 and 2006. We want to project the size of the population in2030, 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. Theaverage growth rate over the 24 years is found by calculating the geometric mean:
The average growth rate is just what we expect.
076.1076.1076.1076.1076.14 GM
Exercise
P71. N.32 Ch.3
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?
Chapter 3: Describing Data
5.11% found by -1292.570.605
Review
Chapter 2: Describing Data
P27. N.4 Ch.2
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 Indianapolis best, 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.
Review
Chapter 2: Describing Data
P34. N.12 Ch.2
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?
24 = 16, 25 = 32, suggests 5 classes
i > ≈ 1099 - 51
5
Use interval of 10
b. What class interval would you suggest?
Review
Chapter 2: Describing Data
P34. N.12 Ch.2
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
Review
Chapter 2: Describing Data
P34. N.12 Ch.2
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.
Review
Chapter 2: Describing Data
P34. N.12 Ch.2
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.
Exercise
P65. N.20 Ch.3
Chapter 3: Describing Data
Determine the mean, median, mode12 8 17 6 11 14 8 17 10 8
Mean=11.10; Median=10.50; Mode=8
Exercise
P60. N.2 Ch.3
a. Compute the mean of the following population values: 7, 5, 7, 3, 7, 4
Chapter 3: Describing Data
μ = 5.5 found by (7+5+7+3+7+4)/6
Exercise
P60. N.4 Ch.3
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
Chapter 3: Describing Data
(1.3-4.2)+(7.0-4.2)+(3.6-4.2)+(4.1-4.2)+(5.0-4.2)=0
X = 4.2 found by 21/5
More Information
Source: Keller, Statistics for Management and Economics, 2005
More InformationPercentage
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.