Download Chapter 6 The Standard Deviation as a Ruler and the Normal

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
Chapter 6
The Standard Deviation as a
Ruler and the Normal Model
Math2200
Examples: How do we compare
measurements on different scales ?
SAT score 1500 versus ACT score 21
Women’s heptathlon
200-m runs, 800-m runs, 100-m high hurdles
Shot put, javelin, high jump, long jump
Make them on the same scale by
standardization
Count how many standard deviations away
from the mean
The Standard Deviation as a Ruler
 The standard deviation tells us how the whole
collection of values varies, so it’s a natural ruler
for comparing an individual to a group.
 Standardizing with z-scores
We compare individual data values to their mean,
relative to their standard deviation using the following
formula:
y  y

z
s
We call the resulting values standardized values,
denoted as z. They can also be called z-scores.
Heptathlon: Kluft versus Skujyte
Carolina Kluft
(Sweden)
Austra Skujyte
(Lithuania)
Gold Medal
Silver Medal
Kluft versus Skujyte
Long jump
shot put
mean
6.16m
13.29m
sd
0.23m
1.24m
Kluft
6.78m
14.77m
z-score
2.70 = (6.78-6.16)/0.23
1.19 = (14.77-13.29)/1.24
Skujyte
6.30m
16.40m
z-score
0.61 = (6.30-6.16)/0.23
2.51=(16.40-13.29)/1.24
Total z-score
Kluft
2.70+1.19=3.89
Skujyte
0.61+2.51=3.12
Standardizing with z-scores (cont.)
Standardized values (z-scores) have no
units.
z-scores measure the distance of each
data value from the mean in standard
deviations.
A negative z-score tells us that the data
value is below the mean, while a positive
z-score tells us that the data value is
above the mean.
Benefits of Standardizing
Standardized values have been converted
from their original units to the standard
statistical unit of “standard deviations from
the mean.” Thus we can compare values
that are measured on different scales, or
in different units.
Shifting Data
Adding (or subtracting) a constant to every
data value adds (or subtracts) the same
constant to measures of position.
Shifting the data
-Adding (or subtracting) a constant to each value will
increase (or decrease) measures of position: center,
percentiles, max or min by the same constant. Its shape
and spread, range, IQR, standard deviation - remain
unchanged.
Mean
+c
sd
unchanged
Median
+c
IQR
unchanged
Min
+c
Max
+c
Q1
+c
Q3
+c
Rescaling Data
Multiply (or divide) all the data values by
any constant
 The men’s weight data set measured weights in
kilograms. If we want to think about these weights
in pounds, we would rescale the data:
Rescaling Data (cont.)
 All measures of position (such as the mean, median, and
percentiles) and measures of spread (such as the range,
the IQR, and the standard deviation) are multiplied (or
divided) by that same constant.
Mean
*a
sd
*a
Median *a
IQR *a
Min
*a
Max
Q1
Q3
*a
*a
*a
How does the standardization changes
the distribution?
 Standardizing data into z-scores shifts the data
by subtracting the mean and rescales the values
by dividing by their standard deviation.
 Shape
No change
 Mean
0 after standardization
 Standard deviation
1 after standardization
How should we read a z-score?
 The larger a
z-score is
(negative or
positive), the
more unusual
it is.
 How do we
evaluate how
unusual it is?
Long jump
shot put
mean
6.16m
13.29m
sd
0.23m
1.24m
Kluft
6.78m
14.77m
z-score
2.70 = (6.786.16)/0.23
1.19 = (14.7713.29)/1.24
Skujyte
6.30m
16.40m
z-score
0.61 = (6.306.16)/0.23
2.51=(16.4013.29)/1.24
Total zscore
Kluft
2.70+1.19=3.89
Skujyte
0.61+2.51=3.12
Normal models
 Quantitative variables
 Unimodal & symmetric
 N(μ,σ)
 μ : mean σ: sd
Standard normal N(0,1)
 Parameters (μ,σ)
 Statistics ( ,
)
If (μ,σ) are given,
If (μ,σ) are not given,
y  y

z
s
Data, Model
Data , Statistics
Model , Parameter
A tool to describe the data with a number of
parameters
Questions related to the data can be answered
by the value of the parameters
Estimate parameters by certain statistics
Mean from the model, sample mean
SD from the model, sample standard deviation
Proportion parameter, sample proportion
When to use the Normal model?
When we use the Normal model, we are
assuming the distribution is Normal.
Nearly Normal Condition: The shape of the
data’s distribution is unimodal and symmetric.
This condition can be checked with a histogram
or a Normal probability plot.
68-95-99.7 rule
How do we measure how extreme a value
is using normal models?
68% within
95% within
99.7% within
Finding normal probability using TI-83
 2nd + VARS (DISTR)
 normalcdf(lowerbound, upperbound, μ,σ)
If the data is from N(0,1), what is the chance to see a
value between -0.5 and 1?
 normalcdf(-0.5,1,0, 1) = 0.5328072082
If the data follow the N(1,1.5) distribution, what is the
chance to see a value less or equal to 5?
 normalcdf(-1E99, 5,1,1.5) = 0.9961695749
From probability to Z-scores
 For a given probability, how do we find the
corresponding z-score or the original data value.
 Example: What is Q1 in a standard Normal
model?
 TI-83: invNorm(probability, μ,σ)
invNorm(0.25,0,1) = -0.6744897495
Example: SAT score
A college only admits people with SAT
scores among the top 10%. Assume that
SAT scores follow the N(500,100)
distribution. If you want to be admitted,
how high your SAT score needs to be?
invNorm(0.9,500,100) = 628.1551567
Finding the parameters
 While only 5% of babies have learned to walk by
the age of 10 months, 75% are walking by 13
months of age. If the age at which babies
develop the ability to walk can be described by a
Normal model, find the parameters?
Z-score corresponding to 5%:
 (10- μ)/ σ = invNorm(0.05,0,1) = -1.644853626
Z-score corresponding to 75%:
 (13- μ)/ σ = invNorm(0.75,0,1) = 0.6744897495
Solve the two equations, we have
 μ=12.12756806
 σ=1.293469536
Normal Probability Plots
 If the distribution of the data is roughly Normal,
the Normal probability plot approximates a
diagonal straight line. Deviations from a straight
line indicate that the distribution is not Normal.
 Nearly Normal data have a histogram and a
Normal probability plot that look somewhat like
this example:
Normal Probability Plots (cont.)
A skewed distribution might have a
histogram and Normal probability plot like
this:
How to make Normal Probability Plots?
 Suppose that we measured the fuel efficiency of
a car 100 times
 The smallest has a z-score of -3.16
 If the data are normally distributed, the model
tells us that we should expect the smallest zscore in a batch of 100 is -2.58.
This calculation is beyond the scope of this class.
 Plot the point whose X-axis is for the z-scores
given by the normal model and Y-axis is for that
from the data
 Keep doing this for every value in the data set
 If the data are normally distributed, the two
scores should be close, and graphically, all the
points should be roughly on a diagonal line.
What Can Go Wrong?
 Don’t use a Normal model when the distribution is
not unimodal and symmetric.
What Can Go Wrong? (cont.)
Don’t use the sample mean and sample
standard deviation when outliers are
present—the mean and standard deviation
can both be distorted by outliers.
Don’t round your results in the middle of a
calculation.