DESCRIPTIVE STATISTICS: A BRIEF INTRODUCTION
BASIC LABORATORY METHODS IN A REGULATED ENVIRONMENT
LECTURE OVERVIEW
LECTURE OVERVIEW
WHY LEARN ABOUT STATISTICS?
LATER IN THIS COURSE
LECTURE OVERVIEW
VARIATION
VARIATION
EXAMPLE
VARIABILITY IS THE ISSUE
STATISTICS
STATISTICS
DESCRIPTIVE STATISTICS
SOME IMPORTANT VOCABULARY
POPULATIONS
SAMPLE
POPULATION AND SAMPLE
MORE ABOUT SAMPLES
TWO VOCABULARY WORDS
SAMPLING
EXAMPLE
EXAMPLE CONT.
VARIABLES
VARIABLES
VARIABLES
DATA
ALWAYS UNCERTAINTY
SAMPLE SIZE
INFERENTIAL STATISTICS
EXAMPLES
ANSWERS
EXAMPLE PROBLEM
ANSWER
Many abuses of statistics relate to poor sampling. The population of interest is all doctors. No way to know what the sample is. The sample could have included only relatives of employees at Brand X headquarters, or only doctors in a certain area. Therefore, the statement does not ensure that most doctors recommend Brand X. It certainly does not ensure that Brand X is best.
DESCRIBING DATA SETS
DATA SETS
DATA SETS
DESCRIPTIVE STATISTICS
LECTURE OVERVIEW
AVERAGE OR MEAN
MEDIAN AND MODE
MEASURES OF CENTRAL TENDENCY
HYPOTHETICAL DATA SET
2 5 6 7 8 3 9 3 10 4 7 4 6 11 9
Simplest way to organize them is to put in order:
2 3 3 4 4 5 6 6 7 7 8 9 9 10 11
By inspection they center around 6 or 7
MEAN
2 3 3 4 4 5 6 6 7 7 8 9 9 10 11
What is the mean for this data set?
NOMENCLATURE
EXAMPLE
2 3 3 4 5 6 7 8 9
MEAN OF A POPULATION VERSUS THE MEAN OF A SAMPLE
LECTURE OVERVIEW
DISPERSION
A 4 5 5 5 6 6
B 1 2 4 7 8 9
MEASURES OF DISPERSION
MEASURES OF DISPERSION
CALCULATIONS OF DISPERSION
EXAMPLE
2 3 3 4 4 5 6 6 7 7 8 9 9 10 11
CALCULATING VARIANCE AND STANDARD DEVIATION
4 cm 5 cm 6 cm 7 cm
7 cm 7 cm 9 cm 11 cm
DEVIATION
4 cm 5 cm 6 cm 7 cm 7 cm 7 cm 9 cm 11 cm
CALCULATION OF DEVIATIONS FROM MEAN
4 cm 5 cm 6 cm 7 cm 7 cm 7 cm 9 cm 11 cm
Value-Mean (in cm) Deviation (in cm)
(4-7) - 3
(5-7) - 2
(6-7) - 1
(7-7) 0
(7-7) 0
(7-7) 0
(9-7) +2
(11-7) +4
SUM OF DEVIATIONS
Value-Mean Deviation
(in cm)
(4-7) - 3
(5-7) - 2
(6-7) - 1
(7-7) 0
(7-7) 0
(7-7) 0
(9-7) +2
(11-7) +4
Sum of deviations = 0
SUM OF DEVIATIONS IS ZERO
SUM OF SQUARED DEVIATIONS
Value-Mean Deviation Squared Deviation
(in cm)
(4-7) - 3 9 cm2
(5-7) - 2 4 cm2
(6-7) - 1 1 cm2
(7-7) 0 0
(7-7) 0 0
(7-7) 0 0
(9-7) +2 4 cm2
(11-7) +4 16 cm2
total squared deviation = sum of squares = 34 cm2
VARIANCE
34 cm2 = 4.25 cm2
8
STANDARD DEVIATION (SD)
VARIANCE OF POPULATION VS SAMPLE
SD OF POPULATION VS SAMPLE
STANDARD DEVIATION OF A SAMPLE
EXAMPLE PROBLEM
A biotechnology company sells cultures of E. coli. The bacteria are grown in batches that are freeze dried and packaged into vials. Each vial is expected to have 200 mg of bacteria. A QC technician tests a sample of vials from each batch and reports the mean weight and SD.
EXAMPLE CONT.
Batch Q-21 has a mean weight of 200 mg and a SD of 12 mg. Batch P-34 has a mean weight of 200 mg and as SD of 4 mg. Which lot appears to have been packaged in a more controlled fashion?
ANSWER
The SD can be interpreted as an indication of consistency. The SD of the weights of Batch P-34 is lower than of Batch Q-21. Therefore, the weights for vials for Batch P-34 are less dispersed than those for Batch Q-21 and Batch P-34 appears to have been better controlled.
LECTURE OVERVIEW
FREQUENCY DISTRIBUTIONS
�THE WEIGHTS OF 175 FIELD MICE
(in grams)
19 22 20 24 22 19 27 20 21 22 20 22 24 24 21 25 19 21 20 23 25 22 19 17 20 20 21 25 21 22 27 22 19 22 23 22 25 22 24 23 20 21 22 23 21 24 19 21 22 22 25 22 23 20 23 22 22 26 21 24 23 21 25 20 23 20 21 24 23 18 20 23 21 22 22 25 21 23 22 24 20 21 23 21 19 21 24 20 22 23 20 22 19 22 24 20 25 21 22 22 24 21 22 23 25 21 19 19 21 23 22 22 24 21 23 22 23 28 20 23 26 21 22 24 20 21 23 20 22 23 21 19 20 26 22 20 21 22 23 24 20 21 23 22 24 21 23 22 24 21 22 24 20 22 21 23 26 21 22 23 24 21 23 20 20 21 25 22 20 22 21 21 23 22
FREQUENCY DISTRIBUTION TABLE OF THE WEIGHTS OF FIELD MICE
Weight Frequency
(in grams)
17 1
18 1
19 11
20 25
21 34
22 40
23 27
24 19
25 10
26 4
27 2 28 1
�
FREQUENCY TABLE
FREQUENCY DISTRIBUTIONS
FREQUENCY HISTOGRAMS
FREQUENCY HISTOGRAMS
FIRST FOUR BARS
WEIGHTS IN GRAMS
17 18 19 20
F
R
E
Q
U
E
N
C
Y
CONSTRUCTING A FREQUENCY HISTOGRAM
FREQUENCY HISTOGRAMS
FREQUENCY HISTOGRAMS
EXAMPLE PROBLEM
Multiple Choice:
What frequency distribution is illustrated in this histogram?
a. All the mice are of the same weight.
b. There are the same number of mice in each weight class.
c. Neither of the above.
ANSWER
b is correct. The histogram illustrates a situation where there are four mice in each weight class.
NORMAL FREQUENCY DISTRIBUTION
NORMAL DISTRIBUTION
WEIGHT
F
R
E
Q
U
E
N
C
Y
NORMAL DISTRIBTION
CALCULATIONS AND GRAPHICAL METHODS
CALCULATIONS AND GRAPHICAL METHODS
ASSIGNMENT
TO DELVE DEEPER INTO THE TOPICS IN THIS LECTURE