IDEA: Compare two groups/populations based on samples from each of them.
Examples.



Compare average height of men and women. Draw sample of men heights: x1, x2, …, xm and a sample of women heights: y1, y2, …, yn.
Test Ho:
Ho:
μx = μy vs Ha:
μx ≠ μy or Ha:
μx > μy
Compare proportions of Democrats in two cities,
Compare weights of people before and after a diet, etc.
General considerations for the samples : Dependent or independent samples.
Example. Comparing weights of people before and after a diet we have dependent (same people ) samples of weights. Comparing weights of people in two cities we have independent samples. Analysis methods will differ for dependent and independent samples.

Observations come as matched pairs (X,Y).
X and Y are NOT independent, X and Y are dependent .
Examples.

X is score on a test before studying studying hard for the same student; hard; Y is score on the test after

X is score on a test or in sports before training after training program; program, Y score

X is weight before program; weight loss program, Y is weight after the

X and Y are heights of twins or siblings.

Hypotheses of interest : does training make a difference?
μx = score before training; μy = score after training.
Ho:
μx = μy vs Ha:
μx < μy
(no difference) (score after training is higher)
Data are pairs of observations
: (x1, y1), (x2, y2), …, (xn, yn).
Typically, we work with differences: d=X-Y , and phrase hypotheses in terms of differences:
.
Hypotheses e.g. Ho:
μd = 0
Ha:
μd < 0
Data: d1, d2, …, dn.
obs before after difference
1
2
.
n
.
x1 x2 xn
.
y1 y2 yn
.
d1=x1-y1 d2=x2=y2 dn=xn-yn

To test Ho, we do one sample t-test . Need sample mean and standard deviation of d’s:
n d

1
n i n


1
d i
and
s
2
d

i


1
(
d n i


1
d
)
2
.
Compute the test statistic :
t

s d
/
d n
.
Under Ho the test statistic has t(n-1) distribution .
Make decision in exactly the same way as for the one sample t-test.
A (1-
α)100% CI for d:
d

t

/ 2
(
n

1)
s d n
.

The amount of lactic acid in the blood was examined for 10 men, before and after a strenuous exercise, with the results in the following table.
(a) Test if exercise changes the level of lactic acid in blood. Use significance level
α=0.01.
(b) Find a 95% CI for the mean change in the blood lactose level.
Before
After
15
33
16
20
13
30
13
35
17
40
20
37
13
18
16
26
14
21
18
19

Solution
. Take d=“After level” – “before level” of lactic acid.
Data for d: 18, 4, 17, 22, 23, 17, 5, 10, 7, 1. Sample stats:
d

12.4 and
s d
2

63.156.
STEP1.
Ho:
μd = 0 vs Ha: μd ≠ 0
STEP 2.
Test statistic:
t

s d
/
d n

12.4
7.95 / 10

4.93.
STEP 3 . Critical value? df=n-1=9, t
α/2
=t
0.005
=3.69.
STEP 4 . DECISION: t = 4.93 > 3.69 = t
0.005
, so reject Ho.
STEP 5.
Exercise changes lactic acid level.

(b) Find a 95% CI for the mean change in the blood lactose level.
d

t

/ 2
s d n
It is the familiar formula for the 95% CI for the mean, this time mean difference
μd
. Need percentile from the t distribution with n-1 degrees of freedom.
n=10, n-1=9,
α=0.05, so t
α/2
=t
0.025
=2.262, so the 95% CI for
μd is:
7.947

10

(6.716,18.086).

set lactic-acid.MPJ

set
Paired T-Test and Confidence Interval
Paired T for before - after
N Mean StDev SE Mean before 10 15.50 2.37 0.75
after 10 27.90 8.17 2.58
Difference 10 -12.40 7.95 2.51
95% CI for mean difference: (-18.08, -6.72)
T-Test of mean difference = 0 (vs not = 0): T-Value = -4.93 P-Value =
0.001
Ho Ha
Histogram of Differences
(with Ho and 95% t-confidence interval for the mean)
3
Conclusion : Reject Ho, lactic acid level changes after exercise.
Note
: CI for “Before –after”
2
1
0
-25 -20
[
_
X
-15 -10
Differences
]
-5
Ho
0