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# Chapter 9 - PowerPoint PPT Presentation

Chapter 9. Statistics. Frequency Distributions; Measures of Central Tendency. Frequency Distributions Three types of frequency distributions: Categorical – primarily for nominal, ordinal level data (FYI) Grouped – range of data is large

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### Chapter 9

Statistics

• Frequency Distributions

• Three types of frequency distributions:

• Categorical – primarily for nominal, ordinal level data (FYI)

• Grouped – range of data is large

• Ungrouped – range of data is small, single data values for each class (FYI)

• Grouped Frequency Distributions

• Step 1: Order data from smallest to largest

• Step 2: Determine the number of classes (e.g. class intervals) using Sturges’ Rule k=1+3.322(log10n) where n is the number of observations (data values). *Always round up

• Class intervals are contiguous, nonoverlapping intervals selected in such a way that they are mutually exclusive and exhaustive. That is, each and every value in the set of data can be placed in one, and only one, of the intervals.

• Grouped Frequency Distributions

• Step 3: Determine width of class intervals

• Width (W) = Range (R)

k

where Range= largest value-smallest value

k represents Sturges’ Rule

• Grouped Frequency Distributions

• Step 4: Assign observations to class intervals

• The count in each class interval represents the frequency for that interval.

• The smallest observation serves as the first lower class limit (LCL). Add the ‘width minus one’ to the LCL to get UCL (upper class limit)

• NOTE: Technically, class limits (i.e., 0-5, 6-11, 12-17 and so on) are not adjacent.

However, class boundaries account for the space between the class limit intervals (i.e., 0.5 – 5.5, 5.5-11.5, 11.5-17.5 and so on). Boundaries are written for convenience but understood to mean all values up to but not including the upper boundary.

• Grouped Frequency Distributions

• Step 5: Calculate cumulative & relative frequencies

• Cumulative Frequency-Add number of observations from the first interval through the preceding interval, inclusive.

• Relative Frequency – Divide number of observations in each class interval by the total number of observations

• Cumulative Relative Frequency-Same calculation as cum-ulative frequency, but using the relative frequencies

• A Frequency Distribution Table

Class Int. Freq. Cum. Freq. Rel. Freq. Cum. Rel. Freq.

LCL - UCL

• Measures of Central Tendency – the value(s) the data tends to center around

• Arithmetic mean (average)

• Mode

• Median

• Measures of Central Tendency

• Arithmetic mean (sample mean or sample average) --“x-bar”

• Ungrouped data (individual data such as 5, 6, 10, 14, etc.

_

x =  xi

n

_

x = x1 + x2 + x3 +… + xn

n

• where xi is each data value (observation) in the data set.

• where n is the number of observations in the data set

• Calculate the sample mean for ungrouped data:

• Step 1: add all values in a data set

• Step 2: divide the total by the number of values summed.

• Example

• 7.0 6.2 7.7 8.0 6.4 6.2 7.2 5.4 6.4 6.5 7.2 5.4

• n = 12 *This is ungrouped data

_

• x = 7.0+6.2+7.7+8.0+6.4+6.2+7.2+5.4+6.4+6.5+7.2+5.4

12

• = 79.6

12

• = 6.63

• Grouped data (assumes each value (observation) falling within a given class interval is equal to the value of the midpoint of that interval

_

x =  fi xi

n

• where xi represents each class interval midpoint (class mark)*

*an easy way to determine the class mark is to simply add the upper class limit (boundary) to the lower class limit (boundary) then divide by 2.

• Calculate the sample mean for grouped data:

• Step 1: multiply each class mark by its corresponding frequency

• Step 2: add the resulting products

• Step 3: divide the total by the number of observations

• Example

• Class Limits Frequency Class Mark xI fI

90 – 98 6 (see note below) 94 564

99-107 22 103 2266

108-116 43 112 4816

117-125 28 121 3388

126-134 9 130 1170

108 12204

_

• x = 12204 = 113

108

Note: Where did the number 6 come from? There are 6 data values

(observations) in the data set that fall between the range 90-98

(inclusive)

• Measures of Central Tendency

• Mode – value that occurs most frequently

• Ungrouped data

• Step 1: identify the data value that occurs most frequently

• Bi-modal -two values occurring at the same frequency

• No mode – all values different (not same as mode=0)

• Grouped data

• Step 1: specify the modal class (i.e., the class interval containing the largest number of observations

• For ungrouped data <mode>

• 7.0 6.2 7.7 8.0 6.4 6.2 7.2 5.4 6.4 6.5 7.2 5.4

• There are four numbers that appear two times each:

• 5.4 6.2 6.4 7.2 Therefore there are four modes.

• The data set is quad-modal

• For grouped data <modal class>

• The modal class: 108-116 or 3rd class (The class with the largest number of data values)

• Measures of Central Tendency

• Median – The value above which half the values in a data set lie and below which the other half lie. (The middle value)

• Ungrouped Data

• Step 1: arrange the values in order of magnitude (smallest to largest)

• Step 2: locate the middle value

• For ungrouped data <median>

• 5.4 5.4 6.2 6.2 6.4 6.4 6.5 7.0 7.2 7.2 7.7 8.0

• Even number of values therefore we must get an average of the middle two values

• 6.4 + 6.5 = 6.45

2

• Range (R) (for ungrouped data only)

• Ungrouped data

• Step 1: Take the difference between the largest and smallest values in a data set. For example, a data set such as 5, 6, 10, 14 has a range of 9 because 14 (the largest value) minus 5 (the smallest value) is 9.

• Deviations from the Mean

• Differences found by subtracting the mean from each number in a sample

• Given 3, 5, 2, 6

• The mean ( ) is 4

• The deviations from the mean would be -1, 1, -2, 2

• Variance (s2) - an average of the squares of the deviations of the individual values from their mean.

• Ungrouped data

s2 =  (xi – )2

n-1

• Standard deviation (s)

• Step 1: Calculate the sample standard deviation for grouped or ungrouped data by:

• taking the square root of the variance

• Example

• 8 6 3 0 0 5 9

2 1 3 7 10 0 3 6

_ *This is ungrouped data

• x = 4.2

n = 15

• (a) Range (R) = 10 – 0 = 10

• (b) variance (s2) = (8-4.2)2 + (6-4.2)2 + (3-4.2)2 + (0-4.2)2 + (0-4.2)2 + (5-4.2)2 + (9-4.2)2 + (2-4.2)2 + (1-4.2)2 + (3-4.2)2 + (7-4.2)2 + (10-4.2)2+ (0-4.2)2 +(3-4.2)2 + (6-4.2)2 _________

15-1

= 158.40__

14

= 11.31

• (c) standard deviation (s) = the square root of 11.31 = 3.36

• Grouped data

s2 = n ( xi2 fi) - (xi fi)2

n(n-1)

• where xi represents each class boundary (or limit) midpoint (class mark)*

• where fi represents each class frequency

*an easy way to determine the class mark is to simply add the upper class limit (boundary) to the lower class limit

(boundary) then divide by 2.

• Calculate the sample variance for grouped data:

• Step 1: multiply each squared class mark by its corresponding frequency

• Step 2: add the resulting products

• Step 3: multiply the sum by n [A]

• Step 4: multiply each class mark by its corresponding frequency

• Step 5: add the resulting products

• Step 6 :square the sum [B]

• Step 7: perform subtraction [C] = [A] – [B]

• Step 8: divide [C] by n(n-1)

• Example

• Class limits freq(fi) xi xifi xi2fi

90 – 98 6 94 564 (946) 53,016 [(942)6]

99-107 22 103 2266 233,398

108-116 43 112 4816 539,392

117-125 28 121 3388 409,948

126-134 9 130 1170152,100

108 12204 1,387,854

• Refer to the formula for variance of grouped data below and see if you can fill in the formula using values from the table on the previous slide.

s2 = n ( xi2 fi) - (xi fi)2

n(n-1)

• s2 = 108(1,387,854) – (12,204)2

108(107)

• = 149,888,232.0 - 148,937,616.0

11,556

• = 950,616

11,556

• = 82.26

• Therefore s = 9.07

• The Normal Distribution

• Also known as the “bell-shaped” curve

• Some statisticians say it is the most important distribution in statistics

• Most popular distribution in statistics

• The normal density function is given by

• where ∏≈ 3.142 and ex ≈ 2.718

• Properties of the Normal Distribution

- mean = median = mode

- area under the curve = 1

- each different and specifies different normal distribution, thus the normal distribution is really a family of distributions

- a very important member of the family is the standard normal distribution

• The Standard Normal Distribution

• has mean (μ) = 0

• has standard deviation (σ) = 1

• the normal density function reduces to

• The probability that z lies between any two points on the z-axis is determined by the area bounded by perpendiculars erected at each of the points, the curve, and the horizontal axis.

P(a <z< b)

• Generally we find the area under the curve for a continuous distribution via calculus by integrating the function between a & b.

dz

• However, we don't have to integrate because we have a table that has calculated this area

• See TABLE 1 of Appendix A-2

• Exercises 6-3 #7 p. 282

• Find the area under the normal distribution curve between z = 0 and z = 0.56

• So, we want P (0 < z < 0.56)

• From the standard normal table we find that

P (0 < z < 0.56) = 0.2123

where a = 0 and b = 0.56

• Exercises 6-3 #16 p. 283

• Find the area under the normal distribution curve between z = -0.87 and z = -0.21

• So we want P(-0.87 < z < -0.21)

a b 0

where a = -0.87 and b =-0.21

• Exercises 6-3 #16 p. 283 con’t

• The table gives a probability of 0.3078 at z = 0.87 (note area same for negative or positive z since distribution is symmetrical). This area covers values of z from 0 out to -.87. Since we don’t want that entire area we subtract the area from 0 out to -.21. That is , we subtract .0832 which is the area under the curve at z = 0.21

• So 0.3078 – 0.0832 = 0.2246

The Normal Distribution

• Exercises 6-3 #25 p. 283

• Find the area under the normal distribution curve to the right of z = 1.92 and to the left of

z = -0.44

• So we want P(z >1.92)  P(z < -0.44) = 0.3574

where a = -0.44 and b = 1.92

• Exercises 6-3 #25 p. 283 Con’t

• Since the area at z = .44 is 0.1700 which is the area under the curve from 0 out to 0.44, the remaining area of interest has to be 0.5 – 0.1700 = 0.3300.

AND

Since the area at z = 1.92 is .4726 which is the area under the curve from 0 out to 1.92, the remaining area of interest has to be 0.5 – 0.4726 = 0.0274. So the combined areas of interest are

0.3300 + 0.0274 = 0.3574

z 0

The Normal Distribution

• Exercises 6-3 #45 z = ?

• Given that the shaded area is 0.8962, what would be the value of z?

• z has to be equal to -1.26. Since the area from 0 out to z is equal to 0.3962 (0.8962 - 0.5000) Recall that one-half of the area under the curve is .5. If we look in the body of the standard normal table for an area of 0.3962 we find that value at the intersection of the 13th row and 7th column which corresponds to a z value of 1.26. Since z is located to the left of 0 it has to be negative, hence – 1.26.

• Section 6-4

• Applications of the Normal Distribution

• To solve problems for a normally distributed variable with a  0 or  1 we MUST transform the variable to a standard normal variable, that is

P(x1 < X < x2) becomes P(z1 < Z < z2) which

allows us to use the standard normal table.

• Using z = value – mean = x - 

standard dev.

The Normal Distribution

• Example

• A survey found that people keep their television sets an average of 4.8 years. The standard deviation is 0.89 year. If a person decides to buy a new TV set, find the probability that he or she has owned the set for the following amount of time. Assume the variable is normally distributed.

• Less than 2.5 years

• Between 3 and 4 years

• More than 4.2 years

•  = 4.8  = 0.89

• (a) P(x < 2.5) becomes P(z<-2.58) because z = (2.5 – 4.8)/ 0.89 = -2.58

The area under the curve at Z=2.58 is 0.4951 therefore the P(z<-2.58) = 0.5 – 0.4951 = 0.0049

-2.02-.9 0

The Normal Distribution

• (b) P(3 < X < 4) becomes P(-2.02 < z < -0.9) because z = (3-4.8)/ .89 = -2.02 and z=(4-4.8)/.89 = -0.90

• from the standard normal table at a z of 2.02 we get .4783 and at a z of .9 we get .3159 so the P(-2.02 < z < -0.9) = .4783 - .3159 = .1624

-.67 0

The Normal Distribution

• (c) P (x > 4.2) becomes P(z > -0.67) because z = (4.2-4.8)/.89 = -0.67

• from the standard normal table at z of .67 we get .2486 so the P(z > -0.67) = 0.2486 + 0.5 = 0.7486

-.67 0 .67

The Normal Distribution

• Review Exercises #9

• Area (%age) = .5

• = 100  = 15

• We can find the X values that correspond to the z values by using the same transformation equation.

• -0.67 = (x – 100)/15 and 0.67 = (x -100)/15

15(-.67) = x – 100 15(.67) = x - 100

x = 89.95 x = 110.05

therefore the highest and lowest scores are in the range (89.95 < x < 110.05)