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Independent Samples:
Comparing Means
Lecture 40
Section 11.4
Mon, Apr 10, 2006
Independent Samples




In a paired study, two observations are made on
each subject, producing one sample of bivariate
data.
Or we could think of it as two samples of
paired data.
Often these are “before” and “after”
observations.
By comparing the “before” mean to the “after”
mean, we can determine whether the intervening
treatment had an effect.
Independent Samples




On the other hand, with independent samples,
there is no logical way to “pair” the data.
One sample might be from a population of
males and the other from a population of
females.
Or one might be the treatment group and the
other the control group.
The samples could be of different sizes.
Independent Samples




We wish to compare population means 1 and
2.
We do so by comparing sample meansx1
andx2.
More specifically, we will usex1 –x2 as an
estimator of 1 – 2.
If we want to know whether 1 = 2, we test to
see whether 1 – 2 = 0 by computingx1 –x2.
The Distributions ofx1 andx2



Let n1 and n2 be the sample sizes.
If the samples are large, thenx1 andx2 have
(approx.) normal distributions.
However, if either sample is small, then we will
need an additional assumption.

The populations are normal.
Further Assumption

We will also assume that the two populations
have the same standard deviation.



Call it .
If this assumption is not supported by the
evidence, then it should not be made.
If this assumption is not made, then the
formulas become much more complicated. See p.
658.
The Distribution ofx1 –x2


Suppose thatx1 andx2 have normal
distributions with means 1 and 2 and standard
deviations 1/n1 and 2/n2 (according to the
Central Limit Theorem for Means).
Thenx1 –x2 is a normal random variable with
the following properties:
The mean is 1 – 2.
2
2
 The variance is


 x x 2   x 2   x 2  1  2

1
2
1
2
n1
n2
The Distribution ofx1 –x2

If we assume that 1 = 2, (call it ), then
the standard deviation may be simplified to
2
2
1 1



n1 n2
n1 n2

That is,

1 1 

x1  x2 is N  1   2 , 
 .
n1 n2 

The Distribution ofx1




x1 is N  1 ,

n1 

0
1
The Distribution ofx2

 
x2 is N   2 ,

n2 

0
2
1
The Distribution ofx1 –x2

1 1 

x1  x2 is N  1   2 , 
 
n1 n2 

0
1 – 2
2
1
The Distribution ofx1 –x2

If


1
1
x1  x2 is N  1   2 , 
 ,
n1 n2 

then it follows that

x1  x2   1   2 
Z
1 1


n1 n2
The t Distribution


Let s1 and s2 be the sample standard deviations.
Whenever we use s1 and s2 instead of , then we
will have to use the t distribution instead of the
standard normal distribution, unless the sample
sizes are large.
Estimating 



Individually, s1 and s2 estimate .
However, we can get a better estimate than
either one if we “pool” them together.
The pooled estimate is
n1  1s1  n2  1s2
2
sp 
n1  n2  2
2
.
x1 –x2 and the t Distribution



If we use sp instead of , and the sample sizes
are small, then we should use t instead of Z.
The number of degrees of freedom is
df = df1 + df2 = n1 + n2 – 2.
That is
x  x   1   2 
t (n  n  2)  1 2
1
2
sp
1 1

n1 n2
Hypothesis Testing


See Example 11.4, p. 699 – Comparing Two
Headache Treatments.
State the hypotheses.
H0: 1 = 2
 H1: 1 > 2


State the level of significance.

 = 0.05.
The t Statistic

Compute the value of the test statistic.

The test statistic is
x1  x2
t
1 1
sp

n1 n2
with df = n1 + n2 – 2.
Computations
9s1  9s2
sp 
 5.052.
18
22.6  19.4
t
 1.416.
1 1
5.052

10 10
2
2
Hypothesis Testing

Calculate the p-value.
The number of degrees of freedom is
df = df1 + df2 = 18.
 p-value = P(t > 1.416)
= tcdf(1.416, E99, 18)
= 0.0869.

Hypothesis Testing

State the decision.


Accept H0.
State the conclusion.

At the 5% level of significance, the data do not
support the claim that Treatment 1 is more effective
than Treatment 2.
The TI-83 and Means of
Independent Samples






Enter the data from the first sample into L1.
Enter the data from the second sample into L2.
Press STAT > TESTS.
Choose either 2-SampZTest or 2-SampTTest.
Choose Data or Stats.
Provide the information that is called for.

2-SampTTest will ask whether to use a pooled
estimate of . Answer “yes.”
The TI-83 and Means of
Independent Samples


Select Calculate and press ENTER.
The display shows, among other things, the
value of the test statistic and the p-value.
Paired vs. Independent Samples

The following data represent students’ calculus test
scores before and after taking an algebra refresher
course.
Student 1 2 3 4 5 6 7 8
Before 85 63 94 78 75 82 45 58
After
92 68 98 83 80 88 53 62
Paired vs. Independent Samples

Perform a test of the hypotheses
H0: 2 – 1 = 0
 H1: 2 – 1 > 0

treating the samples as independent.
Paired vs. Independent Samples

Had we performd a test of the “same”
hypotheses
H0: D = 0
 H1: D > 0


treating the samples as paired, then the p-value
would have been 0.000005688.
Why so small?
Paired vs. Independent Samples

Why is there a difference?
100
90
80
70
60
50
40
1
2
3
4
5
6
7
8
Confidence Intervals



Confidence intervals for 1 – 2 use the same
theory.
The point estimate isx1 –x2.
The standard deviation ofx1 –x2 is
approximately
sp
1 1
  
 n1 n2 
Confidence Intervals

The confidence interval is
or
1 1
x1  x2  z     
 n1 n2 
( known, large samples)
1 1
x1  x2  z  s p   
 n1 n2 
( unknown, large samples)
or
x1  x2  t  s p
1 1

n1 n2
( unknown, normal pops.,
small samples)
Confidence Intervals

The choice depends on
Whether  is known.
 Whether the populations are normal.
 Whether the sample sizes are large.

Example





Find a 95% confidence interval for 1 – 2 in
Example 11.4, p. 699.
x1 –x2 = 3.2.
sp = 5.052.
Use t = 2.101.
The confidence interval is
3.2  (2.101)(2.259) = 3.2  4.75.
The TI-83 and Means of
Independent Samples

To find a confidence interval for the difference
between means on the TI-83,
Press STAT > TESTS.
 Choose either 2-SampZInt or 2-SampTInt.
 Choose Data or Stats.
 Provide the information that is called for.


2-SampTTest will ask whether to use a pooled estimate of
. Answer “yes.”
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