Transcript GG 313 Lecture 7

GG 313 Lecture 7
Chapter 2: Hypothesis Testing
Sept. 13, 2005
CRUISE
Non- required planning meeting for all
involved
This Friday 1:30 PM,
POST 703
Let’s go over the homework and finish up the last of
Chapter 2 first.
Homework being returned:
• Do your calculations in Matlab or Excel and
DOCUMENT them.
• You should be able to determine what a reasonable
answer is. If your answer isn’t reasonable, its probable
that it’s wrong! I take off points for unreasonable
answers.
Today’s homework:
1) Hawaiian data: What did you get for a correlation
coefficient?
Are age and distance correlated?
(We’ll come back to these data set later.)
2) chromium: You can use the functions in Matlab, but
to be sure you understand them, I strongly suggest that
you try your own functions and compare the two.
Any bad points? Is nickel correlated?
3) Generate the plots
Inferences about Population Mean
We estimate the mean of a population by calculating the
mean of our sample. Knowing from the central limit
theorem that the sample mean estimate should be
normally distributed about the population mean, we can
get some idea of the precision of our estimate of the
population mean.
Here’s one place where I found Paul’s notes near
impossible to understand. Read his notes, and try the
following explanation.
If we took many trials, our sample mean, x, would have a
normal distribution about the population mean, µ. From
earlier, the standard deviation of the sample mean is equal
to the population standard deviation divided by the √n:
sx 

(see Eqn. 1.101)
n
Thus, the deviation of our sample mean from the
population mean has a normal distribution defined by a
of 0=x-µ, and a variance of 2/n. Changing to
mean
normalized coordinates,
zi 
xi  

xi
xi n


s
s/ n
We have used s as an approximation to . This
ASSUMES that n is LARGE (>30).

xi n
zi 

s
We want a zi that will give us appropriate confidence limits for
our estimate of the mean. Zi=2, for example is 2 standard
deviationsfrom the mean, with a probability that the estimate
will be within zi=±2 of 95%, thus, this is the 95% confidence
interval. We want to solve for the xi appropriate for that zi.
xi n
2
,
s

2 s
xi 
n
If we wanted a 97% confidence interval, we would use zi=3,
and so forth.
Paul calls this special value of zi z/2, and the special value
of xi he calls E.
The probability that x is greater than the confidence
interval z/2 is .
s
E  z / 2 
n
When n~30 or more, and the population is infinite,
we can substitute the sample standard deviation, sx,
for . What sample size do
we need to be confident
that our error will be no larger than E?
z / 2  s 
n  
 and
 E 
x 
z
2

n
Since this statistic is normally distributed, and thus
symmetric, the interval -z/2< z < z/2 is where z will have
its value with probabiliity 1- .
Plugging in for z:
x 
z / 2 
< z / 2 or
s
n
x  z / 2  s
  < x  z / 2  s
n
n
This shows the confidence interval on µ at the 1- 
confidence level. This is a commonly used statistic for
estimation of the population mean. Often the level
used is 95%, or 2.
Example:
A 30-grain sample of a sediment is obtained, and the
mean computed. The mean, x, is 10.5 microns, and the
standard deviation is s=1.2 microns. At the 95%
confidence interval, what is the uncertainty in the mean
grain size of the sediment?
From the above equation, the 95% confidence interval is
z=2, thus the uncertainty is:
z / 2  s
n
 2  1.2
30
 0.44
So the mean grain size, based on the above analysis is
10.5±0.4 microns with 95% confidence.
What if our sample size is smaller? Above, we insisted
that we had a fairly large sample size, effectively
guaranteeing that the sample mean is a normally
distributed statistic. If we use smaller sample sizes we
need to assume that the population we are sampling is
close to normally distributed. Then we can use the
distribution below to estimate the uncertainty in the
mean:
x 
t
s/ n
This is known as the Student t-distribution. The
transformation looks identical to the one for z, but  has
been replaced by s. The t distribution shape depends on
the number
 of degrees of freedom =n-1. If n is large, then
the t-statistics are the same as the z-statistics (normal
distribution). There are tables of t values for different
combinations of degrees of freedom and confidence level.
Student's distribution arises when (as in nearly all practical
statistical work) the population standard deviation is
unknown and has to be estimated from the data. We use it
at this point to estimate the uncertainty in the population
mean.
For a detailed discussion, see:
http://en.wikipedia.org/wiki/Student's_t-distribution
The derivation of the t-distribution was first published in
1908 by William Sealey Gosset, while he worked at a
Guinness brewery in Dublin. He was not allowed to publish
under his own name, so the paper was written under the
pseudonym Student.To get values of the student’s tdistribution, use Matlab functions tcdf (cumulative) and ppdf
(pdf).
These are student’s T distributions for =1 (lowest in
middle) to =6 (highest). It does not depend on µ or .
What we really want is the values of t that bound the
confidence interval we want. So we want to use the
cumulative function, tcdf. tcdf(3.078,1) = 0.9 .
Compare this with the table on the next page. It gives the
probability that the value will be less than t=3.078 is 0.9,
given 1 degree of freedom. This allows us to use the
Matlab function to obtain the table values for t. But this is
actually backwards from what we really want, so lets try
the function tinv(probability,degrees of freedom).
tinv(0.9,1)=3.0777. This gives us exactly what we want the t statistic for a given probability and given degrees of
freedom.
This is the table Paul uses in his examples:
Probability of exceeding the critical value
0.10
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
0.05 0.025
3.078
1.886
1.638
1.533
1.476
1.440
1.415
1.397
1.383
1.372
1.363
1.356
1.350
1.345
1.341
1.337
1.333
1.330
1.328
1.325
1.323
1.321
1.319
1.318
1.316
1.315
1.314
1.313
1.311
1.310
0.01 0.005 0.001
6.314 12.706
2.920 4.303
2.353 3.182
2.132 2.776
2.015 2.571
1.943 2.447
1.895 2.365
1.860 2.306
1.833 2.262
1.812 2.228
1.796 2.201
1.782 2.179
1.771 2.160
1.761 2.145
1.753 2.131
1.746 2.120
1.740 2.110
1.734 2.101
1.729 2.093
1.725 2.086
1.721 2.080
1.717 2.074
1.714 2.069
1.711 2.064
1.708 2.060
1.706 2.056
1.703 2.052
1.701 2.048
1.699 2.045
1.697 2.042
31.821 63.657 318.313
6.965 9.925 22.327
4.541 5.841 10.215
3.747 4.604 7.173
3.365 4.032 5.893
3.143 3.707 5.208
2.998 3.499 4.782
2.896 3.355 4.499
2.821 3.250 4.296
2.764 3.169 4.143
2.718 3.106 4.024
2.681 3.055 3.929
2.650 3.012 3.852
2.624 2.977 3.787
2.602 2.947 3.733
2.583 2.921 3.686
2.567 2.898 3.646
2.552 2.878 3.610
2.539 2.861 3.579
2.528 2.845 3.552
2.518 2.831 3.527
2.508 2.819 3.505
2.500 2.807 3.485
2.492 2.797 3.467
2.485 2.787 3.450
2.479 2.779 3.435
2.473 2.771 3.421
2.467 2.763 3.408
2.462 2.756 3.396
2.457 2.750 3.385
http://www.itl.nist.gov/div898/handbook/eda/section3/eda3672.htm
This table contains the upper critical values of the
Student's t-distribution. The upper critical values are
computed using the percent point function. Due to
the symmetry of the t-distribution, this table can be
used for both 1-sided (lower and upper) and 2-sided
tests using the appropriate value of , the significance
level, demonstrated with the graph below which
plots a t distribution with 10 degrees of freedom..

Now let’s do the example in Paul’s notes:
We have density samples: (2.2 2.25 2.25 2.3 2.3 2.3 2.35).
What is the 95% confidence interval on the sample mean?
We can get the mean: 2.28, s=0.05, n=7, and degrees of
freedom=6 easily. Now we just plug in:
The t-value for =6 and /2=0.025 is given by:
tinv(0.975,6)=2.447.
So the limits are 2.447*0.05/√7, so µ=2.28±0.046
In words, the mean density of our samples is 2.28±0.046
gm/cm^3 with a confidence of 95%.
Try another example now:
You have 4 rock sample ages of 14.2,
13.6. 11.7, and 12.1 million years. What
is the mean age, and what is the 90%
confidence interval for that age?
Chapter 2 Testing Hypotheses
In many situations you can give quantitative answers
to the question: “Does your data satisfy your
hypothesis?”
That’s what this chapter is about.
You cannot PROVE your hypothesis, but you might be
able to reject the opposite of your hypothesis.
For example you have two rock samples and
hypothesize that they have different densities. You can
likely show that they do not have the same densities
within some bounds.
This is the NULL HYPOTHESIS.
Example:
Claim: a particular sandstone has a density of 2.35
gm/cm^3.
We are given 50 samples from another outcrop. We
hypothesize that if the mean density of these samples
are not between 2.25 and 4.45, then the samples are
from a different lithological unit.
What is the probability of making a wrong decision?
There are two possibilities - we could have the mean
of our sample be outside of the range we have
selected, thus incorrectly rejecting the idea that the
samples are from the same sandstone, or we could
have the mean inside the range, and accept the idea,
but be incorrect; the sample mean being a poor
representative of another unit.
Other examples:
Hypothesis
Alcohol has a harmful effect
On reaction time.
null hypothesis
Alcohol has no effect
The date of isochron A is
older than isochron B.
Isochron B is older than A.
Strain increases before
earthquakes
Strain does not increase.
Let’s do the first case: The null hypothesis is that the
sample is from the unit, but the sample mean is outside the
bounds:
x  2.25 or x  2.45
We can reject the null hypothesis if it occurs in less than a
small fraction of samples, like only 5%.
N=50, s==0.42, so:

s
0.42
sx 

 0.06
n
50
This is the standard deviation of the mean.
We now normalize to get the normal scores:
zi 
x i 

2.25  2.35
z0 
 1.67
0.06
2.45  2.35
z0 
1.67
0.06
We get the area under the tails (eqn. 1.128):
1 
a 
1 
1.67 
P(z  a)  1 erf ( ) P(z  1.67)  1 erf (
) .0475
2 
2 
2 
2 
Multiplying by to to get both tails: p=0.095=9.5%.
Thus the probability that we will erroneously say that the
sample is not the same as the known formation is 9.5%.
Calling a result false when it is actually true is called a
type I error.
Let’s look at the other possibility. Suppose that the true
density of our sandstone sample is µ=2.53, and that it is
not the same formation as the 2.35 reference sandstone.
What’s the probability of having our sample mean density
fall between our limits and thus erroneously accepting the
hypothesis?
In this case, we want to know the probability of making
a type II error.
Computing the normal scores as above:
z0 
2.25  2.53
 4.67
0.06
z1 
2.45  2.53
 1.33
0.06
We then use eqn 1.130 to calculate the probability of the
sample mean falling between the two limits:
 a 

1   b 
1.33
4.67 
Pc (a  z  b)  erf   erf   0.5erf (
 erf (

2   2 
 2 

2
2 
 0.092  9.2%
So we have a 9.2% probability of calling the rocks the
same when they really are not.
There is always the possibility of making an erroneous
conclusion, and these methods allow us to quantify the
probabilities. There are for possible outcomes in our
hypothesis testing, we can accept or reject the null
hypothesis, and the null hypothesis can be true or false:
Type II errors are hard to detect and most desirable to avoid..