Download Section 4

Introductory Statistics for
Laboratorians dealing with High
Throughput Data sets
Centers for Disease Control
Interpreting Scores
What do the numbers mean?
Johnny came home from 4th grade and told his
mother he’d made 100 on his test.
•
•
•
•
That’s good!
But it was a 200 point test.
That’s bad!
But it was a very difficult test and Johnny’s score
was one of the highest in the district.
• That’s good!
• But Johnny wasn’t the only one who got 100, the
average score on the test was 100.
• That’s not so good.
What have we learned?
• The fact is that a raw score by itself is
meaningless.
• To interpret a persons score you must know
how everybody else scored.
• For a score to have meaning, you have to
know where that score is in the distribution.
The two main things we need to
know to interpret a score are:
• How far is is from the mean
• How spread out are the scores
The Deviation Score
• Deviation score commonly used in statistics
to make a score more interpretable.
• Deviation score: how far the score is from
the mean
Some Notation
• In statistics the raw score is symbolized by
a UPPER CASE X
• The mean of the raw scores is symbolized
by X
• The deviation score is symbolized by a
lower case
• The deviation score is computed by
subtracting the mean from the score:
x
xX X
• If someone scores at the mean, the deviation
score would be zero.
• If someone scores above average, the
deviation score will be a positive number.
• If the score is below the mean the deviation
score will be a negative number
• If Johnny had come home and told his
mother that his deviation score on the test
was 0, she would have known immediately
that he was average.
•
(Johnny’s mother is a professor of statistics at the local college)
But that is not all.
• While the distance a persons score is from
the mean is more meaningful than the raw
score, the interpretation of the distance from
the mean depends on how spread out the
scores are.
The importance of Dispersion
• For example, if Johnny tells his mother he
scored 10 points above the mean on a test,
we know right away that he is above
average.
• Question is, how much above average.
• If the average
score on the test
is 55 and Johnny
scores 65 and
that is the
highest score on
the test then
scoring 10
points above the
mean is very
good. (see
figure1)
• If on the
other hand,
the highest
score on the
test is 100,
then a 65 is
not so great.
20
15
Johnny's Score = 65
10
5
0
55
-5
Score
So?
• What we really need is a way to express a
score that takes into account both how far
the score is from the mean and how spread
out the scores are.
z-Scores
• The standard deviation is the parameter that
measures the dispersion or spread of the
distribution.
• z-scores measure the distance from the
mean in standard deviation units.
X X
z
s
z-Scores
• If a person scores 1 standard deviation (SD) above
the mean, the z-score will be +1
• If they score 1 SD below the mean the z-score will
be –1
• If they score 2 SD’s above the mean the z-score
will be +2
• If they score at the mean the z-score will be zero.
• Etc.
Areas Under the Normal Curve
• The proportion of the
area under the normal
curve can be
interpreted as the
probability that a score
appears in that area.
• Areas here are shown
for standard deviation
units.
Areas Under the Curve
• As shown here, the
percentage of the
distribution in a
standard deviation
band is the same
regardless of the shape
of the distribution
Problem 10: Compute z-Scores
Subject
Score
S1
1
S2
4
S3
4
S4
5
S5
5
S6
6
S7
7
S8
8
N=
Total =
Mean =
x = X - Mean
x2
z score
SS =
s=
SS
s
N
XX x
z

s
s
Problem 10: Compute z-Scores
Subject
Score
x = X - Mean
x2
z score
S1
1
-4
16
-2
S2
4
-1
1
-0.5
S3
4
-1
1
-0.5
S4
5
0
0
0
S5
5
0
0
0
S6
6
1
1
0.5
S7
7
2
4
1
S8
8
3
9
1.5
N= 8
Total = 40
Mean = 5
SS = 32
s=2
SS
s
N
XX x
z

s
s
Problem 11: Properties of z-Scores
Subject
z – Scores (from
Problem 10)
Deviation score of Squared
the z’s
deviations of z’s
S1
S2
S3
S4
S5
S6
S7
S8
N=
Total of z’s =
Mean of z’s =
SS of z’s =
Standard deviation
of z’s =
Problem 11: Properties of z-Scores
Subject
z – Scores (from
Problem 10)
Deviation score of Squared
the z’s
deviations of z’s
S1
-2
-2
4
S2
-0.5
-0.5
0.25
S3
-0.5
-0.5
0.25
S4
0
0
0
S5
0
0
0
S6
0.5
0.5
0.25
S7
1
1
1
S8
1.5
1.5
2.25
N= 8
Total of z’s = 0
Mean of z’s = 0
SS of z’s = 8
Standard deviation
of z’s = 1
Using the Standard Normal Distribution
• Because all Normal distributions share the same properties, we
can us the standard normal distribution (the distribution of zscores) for our computations and get the same results.
• In the distribution with mean of 64.5 and standard deviation of 2.5,
68% of the distribution is between 62 and 67 (-1 SD to +1 SD).
• In the standard normal distribution (with mean 0 and standard
deviation 1), 68% of the distribution is between -1 SD and +1 SD.
N(64.5, 2.5)
N(0,1)
=>
x
z
Standardized height
(no units)
Problem 12: Women’s Heights
• The average woman is
64.5 inches tall.
• Mean = 64.5
• Standard Deviation =
2.5
Problem 12: Women’s Heights
• Maria is 67 inches tall
(5’ 7”).
• What is Maria’s zscore?
• What percent of
women are shorter
than Maria?
• What percent are
taller?
Problem 12: Women’s Heights
• Alexis is 62 inches tall
(5’ 2”).
• What is Alexis’ zscore?
• What percent of
women are shorter
than Alexis?
• What percent are
taller?
Problem 12: Women’s Heights
• Barbie is 69.5 inches tall
(5’ 9.5”).
• What is Barbie’s z-score?
• What percent of women
are shorter than Barbie?
• What percent are between
Alexis and Barbie?
Problem 12: Women’s Heights
• Leela is 68 ¾ inches
tall (5’ 8 ¾ ”).
• What is Leela’s zscore?
• Can we compute the
percent of women who
are shorter than Leela?
• Why or why not?
Problem 12: Women’s Heights
• Leela is 68 ¾ inches tall.
• Her z-score is 1.5
• Use http://davidmlane.com/hyperstat/z_table.html to
compute the percent of women who are
shorter than Leela.
Problem 12: Women’s Heights
• How tall do you have
to be to be taller than
50% of the women?
• How tall do you have
to be to be taller than
84% of the women?
• How tall do you have
to be to be taller than
97.6% of the women?
Problem 12: Women’s Heights
• Use http://davidmlane.com/hyperstat/z_table.html for
the following problems:
• How tall do you have to be to be taller than
95% of the women?
• How tall do you have to be to be taller than
99% of the women?
• We can be sure that 95% of the women are
between what heights?
Problem 13
• Use http://davidmlane.com/hyperstat/z_table.html for the
following problem:
• You have been timing how long it takes to get to work in
the mornings. The mean is 22.6 minutes with a standard
deviation of 8.16 minutes.
• You have to be at work at 8:30 am at the latest.
• How many minutes before 8:30 do you have to leave to be
95% confident that you will get there at or before 8:30?
• When do you have to leave to be 99% sure you’ll be there
by 8:30?