Transcript PowerPoint

Jan. 19 Statistic for the day:
Number of Wisconsin’s 33
Senators who voted in favor of a
1988 bill that allows the blind to
hunt: 27
Source: Legislative Hotline, Madison, Wisc.
Assignment: Read Chapter 8 and do
exercises 1, 2, 3, 8, 10
These slides were created by Tom Hettmansperger
and in some cases modified by David Hunter
Age at Death of English Rulers
Revisited
60, 50, 47, 53, 48, 33, 71, 43, 65, 34,
56, 59, 49, 81, 67, 68, 49, 16, 86, 67
Turning this data into information.
Shape: Histogram
Age at death of a sample of 20 rulers of England
5
Frequency
4
3
2
1
0
10
20
30
40
50
age
60
70
80
90
Histogram but different shape
(The number of intervals has
been changed from 8 to 10.)
Sample of Rulers of England n = 20 The number
of intervals has been changed to 10.
6
Frequency
5
4
3
2
1
0
11
19
27
35
43
51
age
59
67
75
83
91
Alternatives to median and IQR:
Mean or average of the data.
 Standard deviation of the data.

The mean is easy. Just add up the numbers
and divide by the sample size.
The standard deviation is a pain. Generally you will use a
calculator or computer.
Rough way to approximate the
standard deviation:
Look at the histogram and estimate the
range of the middle 95% of the data.
The standard deviation is about
¼ of this range
Age at death of a sample of 20 rulers of England
5
Frequency
4
3
2
1
0
10
20
30
40
50
60
70
80
90
age
Take 95% range = 85 – 15 = 70
Estimate of standard deviation = .25x70 = 17.5
Using the calculation formula std dev = 16.7
Percent
10
5
0
15
20
25
HandSpan
Range of middle 95% is roughly 24 – 16 = 8
Standard deviation is roughly .25x8 = 2
Standard deviation from formula 1.927
In Summary:
The standard deviation is roughly .25 times the range of
the middle 95% of the data. Look at the histogram or
stem and leaf. The mean is the arithmetic average.
If you want to visualize a histogram and you
only know the mean and the standard deviation:
1. Put the center at the mean or the median.
2. Go out 2 standard deviations on either side of the center.
3. Draw a histogram humped up at the mean and dropping
off on either side within part 2.
Smoothing the histogram:
The Normal Curve (Chapt 8)
The histogram is rough and noisy.
Replace it with a bell shaped curve.
Center the bell at the mean.
The middle 95% of the bell should be 4 standard deviations.
Makes more accurate predictions provided the bell shape is
appropriate for the underlying population.
Histogram of HandSpan, with Normal Curve
Frequency
20
10
0
15
20
HandSpan
Mean = 20.86
Standard deviation = 1.927
25
Histogram of HandSpan, with Normal Curve
Frequency
20
10
0
15
20
HandSpan
Mean = 20.86
Standard deviation = 1.927
25
Histogram of Height, with Normal Curve
Frequency
30
20
10
0
60
70
Height
Mean = 68 inches or 5 feet 8 inches
Standard deviation = 4 inches
80
Research Question 1: How high
should I build my doorways so
that 99% of the people will not
have to duck?
Secondary Question 2: If I built my
doors 75 inches (6 feet 3 inches) high,
what percent of the people would
have to duck?
Histogram of Height, with Normal Curve
Frequency
30
20
10
0
60
70
80
Height
Question 2
Question 1
Find the value at Question 1 so that 99% of the distribution
is below it.
The value at Question 2 is 75; find the amount of distribution
above it.
Z-Scores: Measurement in
Standard Deviations
Given the mean (68), the standard deviation
(4), and a value (height say 75) compute
75  mean 75  68
Z

 1.75
SD
4
This says that 75 is 1.75 standard deviations
above the mean.
Compute your Z-score.
1. How many standard deviations are you
above or below the mean.
Use:
Mean = 68 inches
Standard deviation = 4 inches
2. Now use the table from the book to determine
what percentile you are.
Answer to Question 2: What percent of people would
have to duck if I built my doors 75 inches high?
Recall: 75 has a Z-score of 1.75
From the standard normal table in the book: .96 or
96% of the distribution is below 1.75. Hence, .04
or 4% is above 1.75.
So 4% of the distribution is above 75 inches.
Histogram of Height, with Normal Curve
Frequency
30
20
4% in here
10
0
60
70
75
80
Height
Question 2
The value at Question 2 is 75; find the amount of distribution
above it. Convert 75 to Z = 1.75 and use Table 8.1 in book.
Question 1: What is the value so that 99% of the
distribution is below it? Called the 99th percentile.
1. Look up the Z-score that corresponds to the 99th
percentile. From the table: Z = 2.33.
2. Now convert it over to inches:
h99  68
2.33 
4
h99  68  2.33x 4  77.3
Since 77 inches is 6 feet 5 inches, 99% of the distribution
is shorter than 77 inches and they will not have to duck.
Histogram of Height, with Normal Curve
Frequency
30
20
10
99% in here
0
60
70
80
Height
Question 1
77.3 inches is the 99th percentile
Find the value at Question 1 so that 99% of the distribution
is below it. Look up Z-score for 99th percentile and convert
it back to inches.
Stat 100 students Sp01
Height
80
70
60
Female
Male
Sex
Heights in Inches ( red circle is my doorway 77 inches)
85
78
75
(6-3)
65
Lakers
n=14
PSU BB
n=12
Steelers
n=30
Stat 100
Males
n=78