Transcript degrees of freedom

Statistics and Quantitative
Analysis U4320
Segment 6
Prof. Sharyn O’Halloran
I. Introduction

A. Review of Population and Sample Estimates
Population
Sample
N
Mean

X

2 
i
X 
i 1
N
N
Variance
n
 (X
i
i 1
N
X
n
n
 )2
s2 
i
i 1
 (X
i
 X )2
i 1
n 1
B. Sampling

1.
Samples


The sample mean is a random variable with a normal distribution.
If you select many samples of a certain size then on average you
will probably get close to the true population mean.
I. Introduction

(cont.)
2. Central Limit Theorem

Definition

If a Random Sample is taken from any population with mean_
and standard deviation_, then the sampling distribution of the
sample means will be normally distributed with



1) Sample Mean E() = _, and
2) Standard Error SE() = _/_n
As n increases, the sampling distribution of tends toward the true
population mean_.
I. Introduction

(cont.)
C. Making Inferences

To make inferences about the population from a given sample, we have to
make one correction, instead of dividing by the standard deviation, we
divided by the standard error of the sampling process:
Z

X
.
SE
Today, we want to develop a tool to determine how confident we are that
our estimates lie within a certain range.
II. Confidence Interval

A. Definition of 95% Confidence Interval
known)

(_
1. Motivation




We know that, on average, is equal to_.
We want some way to express how confident we are that a given is
near the actual_ of the population.
We do this by constructing a confidence interval, which is some
range around that most probably contains_.
The standard error is a measure of how much error there is in the
sampling process. So we can say that is equal to__ the standard
error

 = X_Standard Error
II. Confidence Interval

(cont.)
2. Constructing a 95% Confidence Interval

a. Graph


First, we know that the sample mean is distributed
SE=normally,
.
n
with mean_ and standard error
SE

II. Confidence Interval

(cont.)
b. Second, we determine how confident we want to be in
our estimate of_.



Defining how confident you want to be is called the -level.
So a 95% confidence interval has an associate - level of .05.
Because we are concerned with both higher and lower values, the
relevant range is _/2 probability in each tail.
= (1-.95) = .05 5% level
SE



II. Confidence Interval

(cont.)
C. Define a 95% Confidence Range


Now let's take an interval around  that contains 95% of the area
under the curve.
So if we take a random sample of size n from the population, 95%
of the time the population mean _ will be within the range:
[-z.025 * SE , z.025 * SE]



What is a z value associated with a .025 probability?
From the z-table, we find the z-value associated with a .025
probability is 1.96.
So our range will be bounded by:
[-1.96 * SE , 1.96 * SE].
II. Confidence Interval

(cont.)
Now, let's take this interval of size [-1.96 * SE , 1.96 * SE] and use it as a
measuring rod.
SE
..
..

-1.96*SE

1.96*SE
d. Interpreting Confidence Intervals
SE
..
..
X3
-1.96*SE
X2

X1
1.96*SE
II. Confidence Interval


(cont.)
What's the probability that the population mean  will fall within
the interval  1.96 * SE?
e. In General


We get the actual interval as 1.96 * SE on either side of the
sample mean .
We then know that 95% of the time, this interval will contain .
This interval is defined by:
X - (Z.025 * /n) <  < X + (Z .025 * /n)
 = X  Z /2 * SE
II. Confidence Interval

(cont.)
For a 95% confidence interval:
 = X  Z.025 * SE
 = X  1.96 * SE

Examples


1. Example 1: Calculate a 95% confidence Interval
Say we sample n=180 people and see how many times they ate at a fastfood restaurant in a given week. The sample had a mean of 0.82 and the
population standard deviation  is 0.48. Calculate the 95% confidence
interval for these data.
II. Confidence Interval

(cont.)
Answer:
SE = .48 / 180 = 0.036 .
z.025 * SE = 1.96 * .036 = .07
C.I. = .82  .07
[.75 <  < .89]
OR
SE = .036
..
..
-1.96*.036= .75

Why 95%?
.82
1.96*.036=.89
II. Confidence Interval

2. Example
interval

2:
Calculating
a
90%
(cont.)
confidence
A random sample of 16 observations was drawn from a normal
population with  = 6 and = 25. Find a 90% ( = .10) confidence
interval for the population mean, _.

First, find Z.10/2 in the standard normal tables
Z.05 = 1.64
II. Confidence Interval

(cont.)
Second, calculate the 90% confidence interval
 = X  Z.05 * /n
 = 25  1.64 * 6/16
SE= 1.5
 = 25  1.64 * 1.5 = 25  2.46
22.53 <  < 27.46

90% of the time, the mean will lie with in this range.
II. Confidence Interval

(cont.)
What if we wanted to be 99% of the time sure that the mean falls with
in the interval?
Z.005.= 2.58
25  2.58 * 1.5
25  3.87
21.13 <  < 28.87

What happens when we move from a 90% to a 99% confidence
interval?
.
.
.
.
21.13 22.53
27.46
28.87
II. Confidence Interval

(cont.)
C. Confidence Intervals (_ unknown)

1.Characteristics of a Student-t distribution

a.Shape the student t-distribution
Normal Distribution
Student-t
t.025

Z.025

Z.025
t.025
The t-distribution changes shape as the sample size gets larger,
and in the limit it becomes identical to the normal.
II. Confidence Interval


(cont.)
b. When to use t-distribution
 i.  is unknown
 ii. Sample size n is small
2. Constructing Confidence Intervals Using t-Distribution

A. Confidence Interval


95% confidence interval is:
X  t.025
s
.
n
B. Using t-tables

Say our sample size is n and we want to know what's the cutoff
value to get 95% of the area under the curve.
II. Confidence Interval

i) Find Degrees of Freedom


Degree of freedom is the amount of information used to calculate
the standard deviation, s. We denote it as d.f. _ d.f. = n-1
ii) Look up in the t-table



(cont.)
Now we go down the side of the table to the degrees of freedom
and across to the appropriate t-value.
That's the cutoff value that gives you area of .025 in each tail,
leaving 95% under the middle of the curve.
iii) Example:

Suppose we have sample size n=15 and t.025 What is the critical
value? 2.13
II. Confidence Interval

Answer:
(cont.)
X = (64 + 66 + 89 + 77) / 4 = 74
s2 = (64-74)2 + (66-74)2 + (89-74 )2 + (77-74 )2 / 3 = 132.7
SE =
s
132.7

 5.76
n
4
d.f. = 3
t.025 = 3.18
t.025 * SE = 3.18 * 5.76 = 18
 = X  18
56 <  < 92 (not very precise with a sample of only size 4)
..
..
56
74
92
II. Confidence Interval

III. Differences of Means

A. Population Variance Known



(cont.)
Now we are interested in estimating the value (1 - 2) by the sample
means, using 1 - 2.
Say we take samples of the size n1 and n2 from the two populations.
And we want to estimate the differences in two population means.
To tell how accurate these estimates are, we can construct the familiar
confidence interval around their difference:
( 1 -  2) = ( X 1 - X 2)  z.025
 12
n1

 22
n2
.
II. Confidence Interval



This would be the formula if the sample size were large and we
knew both 1 and 2.
B. Population Variance Unknown
1. If, as usual, we do not know 1 and 2, then we use the sample
standard deviations instead. When the variances of populations are
not equal (s1  s2):
( 1 -  2) = ( X 1 - X 2)  t.025

s12 s22
 .
n1 n2
Example: Test scores of two classes where one is from an inner city school and the other is from
an affluent suburb.
II. Confidence Interval

2. Pooled Sample Variances, s1 = s2

(cont.)

( is unknown)
If both samples come from the same population (e.g., test scores
for two classes in the same school), we can assume that they
have the same population variance . Then the formula becomes:
( 1 -  2) = ( X 1 - X 2)  t.025
s p2 s 2p
 ,
n1 n2
or just
( 1 -  2) = ( X 1 - X 2)  t.025 s p

1 1
 .
n1 n2
In this case, we say that the sample variances are pooled. The
formula for s 2 is:
p
s
2
p
(X

 X1 )2   ( X2  X2 )2
(n1  1)  (n2  1)
1
II. Confidence Interval


(cont.)
The degrees of freedom are (n1-1) + (n2-1), or (n1+n2-2).
3. Example:

Two classes from the same school take a test. Calculate the 95%
confidence interval for the difference between the two class means.
X1
64
66
89
77
X1/n= 296/4 = 74
X2
56
71
53
X2/n= 180/3= 60
II. Confidence Interval

(cont.)
Answer:
X 1 = 74
X 2 = 60
( X 1 - X 2) = 14
n1 = 4; n2 = 3
s =  ( X
2
p
1
 X1 )2   ( X2  X2 )2
=
(n1  1)  (n2  1)
(64 - 74)2 + (66 - 74)2 + (89 - 74)2 + (77 - 74)2  (56  60)2  (71  60)2  (53  60)2
4 -1  3 -1
II. Confidence Interval
s 2p = (398 + 186) / (3 + 2) = 117.
sp = 10.8
1
1
+
 10.8 
n1
n2
SE = s p 
1 1
  8.26
4 3
d.f. = 5
t.025 = 2.57
t.025 * SE = 2.57 * 8.26 = 21
( 1 -  2) = ( X 1 - X 2)  21 = 14  21
= -7  (1 - 2  35.
..
..
-7
 1 - 2
35
(cont.)
II. Confidence Interval

(cont.)
C. Matched Samples

1. Definition


Matched samples are ones where you take a single individual and
measure him or her at two different points and then calculate the
difference.
2. Advantage

One advantage of matched samples is that it reduces the variance
because it allows the experimenter to control for many other
variables which may influence the outcome.
II. Confidence Interval

(cont.)
3. Calculating a Confidence Interval


Now for each individual we can calculate their difference D from one
time to the next.
We then use these D's as the data set to estimate , the population
difference.

The sample mean of the differences will be denoted .

The standard error will just be:
SE = sD / n.

Use the t-distribution to construct 95% confidence interval:
 = D  t.025 * sD/n.
II. Confidence Interval

(cont.)
4. Example:
Student
Trimbl e
Wild e
Giannos
Ames
X1 (Fall)
X2 (Spring)
D = X1-X2
64
66
89
77
57
57
73
65
7
9
16
12
D = (7 + 9 + 16 + 12) / 4 = 11
d.f. = n-1 = 3
s2D = (7-11) 2 + (9-11) 2 + (16-11) 2 + (12-11) 2 /3= 46/3
sD = 3.91
SE = sD / 4 = 3.91 / 2 = 1.96
t.025 = 3.18
t.025 * SE = 3.18 * 1.96 = 6
II. Confidence Interval
(cont.)
So  = D  t.025 * sD/n.
= 11  6 = 5 to 17.
5    17

Notice that the standard error here is much smaller than in most of our
unmatched pair examples.
IV. Confidence Intervals for Proportions

Just before the 1996 presidential election, a Gallup poll of about
1500 voters showed 840 for Clinton and 660 for Dole. Calculate
the 95% confidence interval for the population proportion  of
Clinton supporters.

Answer: n= 1500

Sample proportion P:
P=
840
.56
1500
IV. Confidence Intervals for Proportions

(cont.)
Create a 95% confidence interval:
 = P  sampling allowance
 = P  1.96

P(1 P)
,
n
where  and P are the population and sample proportions, and n is the
sample size.
 = .56  1.96
.56(1.56)
,
1500
 = .56  .03.

That is, with 95% confidence, the proportion for Clinton in the whole
population of voters was between 53% and 59%.