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GG313 Lecture 8
9/15/05
Parametric Tests
Cruise Meeting
1:30 PM tomorrow, POST 703
Surf’s Up
“Peak Oil and the Future of Civilization”
12:30 PM tomorrow POST 723(?)
Parametric Tests
What are parametric tests? Parametric tests make
certain assumptions about the data. First, there is the
assumption that the data are drawn from a normal
distribution, second, the data are measured on an
interval scale. Thirdly, parametric tests make use of
parameters such as the mean and standard deviation.
TWO-SAMPLE T-Test: For example, we have taken
data in two samples and the samples have different
means. What is the probability that the samples come
from the same population?
This is a job for the Student’s T-test again. We will
assume that the populations have the same variance
but possibly different means.
We want the probability distribution of x1-x2, the
difference between the means of our samples. The
distribution should be normal if the samples are
independent and random, with a mean of µ1-µ2. The

standard
deviation will be:
 e   p 1 n1  1 n 2 
p 
{2.6} where  2p is called the pooled variance.
n1 112 + n 2 1 22
n1  n 2  2
{2.7}
We get the t - statistic using :
 
x1  x 2
t 1 2 
e
n1 1s12 + n 2 1s22 1 n  1 n
 1
2
n1  n 2  2
{2.8}
Our hypothesis is that the means are different, so our null
hypothesis (Ho) is µ1=µ2. If n1 and n2 are large, then we
can use z statistics:

z
s
2
1
x1  x 2
n1  s n2 
2
2
{2.9}
Example: We get two random samples of magnetite from
separate outcrops. The measured magnetizations (Am2/kg)
are: 
Outcrop 1 : 87.4, 93.4, 96.8, 86.1, 96.4 n1  5
Outcrop 2 : 106.2, 102.2, 105.7, 93.4, 95.0, 97.0 n 2  6
We’ll arbitrarily used the common 95% confidence
interval, so =0.05. The degrees of freedom are =5+62=9, and tinv(0.95,9)=2.262 gives the t-value. We reject
Ho if the t=value for the data is greater than 2.262. Our
data yield: x1=92.0, s1=5.0, and x2=99.9, s1=5.5.
Thus our samples look like they come from the
distributions shown below:
0.09
0.08
0.07
sigma=92, mean=5
0.06
sigma=5.5, mean=99.9
0.05
0.04
0.03
0.02
0.01
0
80
85
90
95
100
105
110
115
120
Could they come from the same population? By using the
distribution of x1-x2 we can think of this as asking whether
this distribution is statistically different from a distribution
with µ=0.
Using eqn. 2.8, we get:
t=

99.9 - 92.0
 2.5
2
2
4.5  5  5.5 1 1 
  
5 6 
9
{2.12}
Since t>2.262, we can reject H0, that the means are the
same with 95% confidence, and can conclude that the

magnetizations
of the two outcrops are different.
Try another one:
We have age determinations from two different rocks
dredged from an Emperor seamount. They are:
Sample 1: 42.2, 43.2, 41.9, 43.0 MY
And
Sample 2: 39.9, 41.5, 41.0, 42.0 MY
At 95% confidence, do these rocks have different ages?
Tests on the Standard deviation:
We want to test our sample standard deviations, s, to see
how well they likely reflect the population standard
deviation ,. We may, for example, want to test whether
the standard deviation of our sample is different from the
variance of a known population. The statistic used is 2,
pronounced Chi-squared..
2 
(n 1)s2

2
{2.13}

Notice that the distribution is not symmetric, so the
probabilities in our tails will need to be computed separately.
The value of 2 depends on the degrees of freedom,
(pronounced: nu) = n-1, the standard deviation of our
sample, s, and the standard deviation of our reference
population, .
Like the z-statistic for the normalized normal distribution
and the t-statistic for small sample sizes are used to
evaluate means, the 2 statistic is used to evaluate
standard deviations for small sample sizes.
The 2 distribution shows how the standard deviation
would vary over a large number of samples with n items.
Since the terms are squared, it is always positive. Small
values of 2 imply nearly identical values of xi, and large
values imply a large spread in xi.
Chi^2 Distribution
0.3
Probability Density
0.25
nu=3
nu=4
nu=5
nu=7
nu=10
nu=14
0.2
0.15
0.1
0.05
0
0
5
10
15
20
25
30
x
This plot was generated using Excel:
=-(CHIDIST($A7+0.01,B$3)-CHIDIST($A7,B$3))*100
Where A$7 is the deg of freedom and B$3 is x.
We use the 2 statistic to get the confidence interval on the
variance:  2  n 1s2   2 {2.14}
1

2
2

2
n 1s2   2  n 1s2
2
2


1
2
{2.15}

2
Like the other statistics, if n is large, we can use zstatistics. The 2 statistic approaches the normal
distribution,
and the limits become:

s
1 z
2
2n
 
s
1 z 
2
2n
{2.16}
One test we might need is to determine whether the
standard deviation of our sample, s, is different from
that of a given population, . The null hypothesis is that
they are the same, s= .
EXAMPLE: We have 15 values of temperature with
s=1.3°C. Previous results have given us =1.5°C. Are
these results statistically different? The H0 hypothesis is s=
, and plugging =0.05,  =14 into our 2 table yielding
20.025=26.119 and 20.975=5.63. We to be sure 2 is not
greater than either side. From our data:
2
14
1.3
2 
 10.5
2
1.5
{2.17}
Since 10.5 is not greater than 26.119 and 5.63, we cannot
reject the idea that the standard deviations are the same.
Again, for large sample sizes the 2 probability approaches
a normal distribution, so we can use z statistics:
z
s 

{2.18}
2n
Test: Are two standard deviations different?

When using the t-test to see whether two sample means
were the same, we assumed that the variances of the
samples were the same. This may not be the case, and we
can test to see if they are. An appropriate statistic is the Fstatistic:
s 12 s22 , s1  s2
F   2 2
s2 s1 , s2  s1
{2.19}
The F-statistic is the ratio of the variances of the two
samples. For normal distributions, it is continuous, with a
probability distribution called the F-distribution. There
are two degrees of freedom, one for each sample, 1=n11, and 2=n2-1.
F Distributions
50,50
1,1
10,10
50,1
This plot was generated with Matlab fpdf. The F-distribution
is strongly dependent on the degrees of freedom of the two
samples, 1 and 2.
Example: In a previous example concerning
magnetization, (n1=6 and n2=5, s1=5.0 and s2=5.5 ) we
assumed that the ’s for the two populations were the
same. Is that true? We calculate the F statistic for the
data using Eqn. 2.19:
5.5 2
F
 1.21
2
5.0
And find the F value for a confidence interval of 95%
using the Matlab
finv function, finv(0.975,5,4)=9.36. Since
9.36 is larger than 1.21, the standard deviations are not
significantly different at the 95% confidence interval.
How large would s2 need to be to say that they WERE
different? In this example, s2 would need to be greater
than 15!
The 2 Test
The 2 Test is used to determine if a particular probability
distribution matches one of the theoretical distributions.
This is the last parametric test we will look at. The 2
Test usually uses “binned” data that are normalized,
squared, and summed:
2


x i 
2
2
   zi  

  
i1
i1
n
n
{2.21}
If we take all possible samples with n members from a
2
normal
 population and plot the above, we get a 
distribution. If our data are binned into k bins, we can
compare the number of elements in the jth bin with the
expected number of elements in the known distribution:
k
 
2
i1
O
 E j
2
j
Ej
{2.22}
Where Oj and Ej are the observed and expected
numbers in the jth bin. This is non-dimensional, since we

are using “counts” as the units.
From earlier, the probability that m out of n counts will
fall in the jth bin is given by the binomial distribution:
 = E j  np j
{2.23}
  np j 1 p j   np j  E j

{2.24}
How we bin the distribution to find pj depends on what
distribution we want to compare our data to.

Plugging into the equations above,
2
2
n


Oj  E j 
Oj  E j
x   

2
  
  
  

   i1  E j 
Ej
i1
 i1
n
2
n
{2.24}
This equation gives us a 2 value for our data to
compare with a value for our confidence interval.
EXAMPLE: We have 48 values of salinity from
Whitewater Bay, Florida. Do these observations
come from a normal distribution? The answer could
have implications for models of mixing salt and
fresh water.
1) Get the mean and standard deviation of the data
(Table 2.2). µ=49.59 and s=9.27.
2) Normalize the data: zi=(xi-49.54)/9.27
3) Choose bins: we want each bin to contain some data,
and we choose 5 bins with boundaries set such that the
expected value in each bin is the same: 1/n=0.2. The
expected value in each bin will then be the same for all
bins: Ej=48/5=9.6.
4) Find the bin boundaries in the distribution curve that
divide the curve into 5 bins with equal probabilities. For
this problem using the normal curve, the boundaries are
-0.84, -.26, .26, and 0.84.
5) Place the data into the proper bins: (see figure
above)
6) Get the 2 value for the data. Using eqn. 2.21, we
get:
2
k
 
2
i1
O
j
 E j
Ej
= 3.04
7) Get the theoretical 2 value .We have 5 bins and
used one degree of freedom to compute s, thus we
have5-2=3 degrees of freedom. For a 95%
probability, and 3 degrees of freedom, we get 2
=5.99.
8) Compare: Since the H0 value of 2 is less than the
theoretical value, we cannot reject the null hypothesis
that the data have a normal distribution with 95%
confidence.
As part of homework:
1) Stand 8’ from a vertical wall and toss coins such
that they hit the wall. Measure how far from the wall
the coin lands. If the coin doesn’t hit the wall before
the floor, ignore the result and toss again.
2) Do this again at 12’ from the wall.
3) Plot your data. Easy cumulative: y axis = number of
events where the distance measured is <x.
Testing Correlation Coefficients
In nearly all situations we find that two data sets are
correlated to some extent; it is rare to find two samples
that have a correlation of zero. But what correlation is
SIGNIFICANT?
Rceall the correlation coefficient between two samples, x
and y (eqn.1.107)



 x y
2
xy
 x  xy  y 
 x  x   y  y 
n
i1
n
i1
i
i
2
i
n
i1
2
Eqn. 2.28
i
Where  is the correlation of the population that we estimate
from samples yielding r, the sample correlation coefficient.

We ant to test the significance of r.
For a null hypothesis, we choose that the population
correlation coefficient is zero, =0. If this were the case,
then the distribution of r would have a mean of zero, µ=0,
and a standard deviation of

2
1
r
 
n  2
We use the t-statistic:


t
r 


r
(1 r 2 ) /(n  2)

r n 2
1 r 2
Eqn. 2.29
Example:
Toss two dice one red, one green and get the following
numbers in 6 trials:
Red (x)
Green (y)
4
5
2
2
4
6
2
1
6
4
Using 2.28, we get r=0.66.
That seems high for a
measurement where we
expect no correlation. Is it
significant?
We choose =0.05 and find t/2=3.182. Applying 2.29
gives t=1.52, hence the observed 0.66 correlation is most
likely random. The value of r would have to be much
higher to get 95% confidence that the results were
2
correlated:
r n 2
3r
2
t / 2 
1 r
2
 3.182 
1 r
2
 r  0.88
Consider the figure below from a proposal. It shows the
correlation coefficient between small volcanic earthquakes
correlated against a reference quake late on day 8.
It’s obvious that the high correlations are significant, but are
the low ones?
For a plot like this, it would be good to include horizontal
lines to show significance. I don’t know the value of n (the
length of each earthquake record), so I can’t add it, but it
might look like this:
Implying that nearly all the correlations shown are
significant.