Two-Sample
Inference Procedures
with Means
Of the following situations, decide which should
be analyzed using one-sample matched pair
procedure and which should be analyzed using
two-sample procedures?
A pharmaceutical company wants to test its new
weight-loss drug. Before giving the drug to a
random sample, company researchers take a weight
measurement on each person. After a month of
using the drug, each person’s weight is measured
again.
Matched pair
Of the following situations, decide which should
be analyzed using one-sample matched pair
procedure and which should be analyzed using
two-sample procedures?
A researcher wants to know if a population of
brown rats on one city has a greater mean
length than a population in another city. She
randomly selects rats from each city and
measures the lengths of their tails.
Two independent samples
Of the following situations, decide which should
be analyzed using one-sample matched pair
procedure and which should be analyzed using
two-sample procedures?
A researcher wants to know if a new vitamin supplement
will make the tails of brown rats grow longer. She takes
50 rats and divides them into 25 pairs matched by
gender and age. Within each pair, she randomly selects
one rat to receive the new vitamin. After six months, she
measures the length of the rat’s tail.
Matched pair
Of the following situations, decide which should
be analyzed using one-sample matched pair
procedure and which should be analyzed using
two-sample procedures?
A college wants to see if there’s a difference in time
it took last year’s class to find a job after graduation
and the time it took the class from five years ago to
find work after graduation. Researchers take a
random sample from both classes and measure the
number of days between graduation and first day of
employment
Two independent samples
Matched Pairs
(Special type of onesample means)
Differences of Paired Means (Matched Pairs)
CONDITIONS:
1) The samples are paired. The sample differences can be
viewed as a random sample from a population of
differences.
2) The sample distribution of differences is approximately normal
- the populations of differences is known to be normal, or
- the number of sample difference is large (n  30), or
- graph data to show approximately normal
3) 10% rule – The sample of differences is not more than
10% of the population of differences.
Differences of Paired Means (Matched Pairs)
CI  statistic
xd 
df  n  1
 critical value
t*
SD
of statistic
 sd 


 n 

Differences of Paired Means (Matched Pairs)
Parameter:
md
= true mean difference in …
Hypothesis Statements:
H0: md = hypothesized value
Ha: md < hypothesized value
Ha: md> hypothesized value
Ha: md ≠ hypothesized value
Differences of Paired Means (Matched Pairs)
Hypothesis Test:
Test statistic

statistic
- parameter
SD of statistic
t 
df  n  1
xd  m d
sd
n
#18 Summer School.
Having done poorly on their Math final exams in
June, six students repeat the course in summer
school and take another exam in August. If we
consider these students to be representative of
all students who might attend this summer
school in other years, do these results provide
evidence that the program is worthwhile?
June
54
49
68
66
62
62
Aug.
50
65
74
64
68
72
Page 590: 18
Parameters and Hypotheses
μd = the true mean difference in scores between June and August for students who repeated
the course in summer school
Ho: μd = 0
Ha: μd > 0
Assumptions (Conditions)
1) The samples must be paired and random. The samples are from the same student
so they are paired and we will assume the sample differences are a random sample of
the population of differences..
2) The sample distribution should be approximately normal.
The normal probability plot is fairly linear and the boxplot shows no outliers, so we will
assume that the sample distribution of differences is approximately normal.
3) The sample should be less than 10% of the population. The population should
be at least 60 students, which we will assume.
4)  is unknown
Since the conditions are met, a t-test for the matched pairs is appropriate.
Calculations
mx  md  0
x d  5.333
n6
s d  7.4475
 = 0.05
df  5
t
xd  m x
sd
n

 (0 )
5.333
7.4475
 1.7541
6
p  value  P ( t  1.7541)  .06988
.0 6 9 8 8  .0 5
Decision:
Since p-value > , I fail to reject the null hypothesis at the .05 level.
Conclusion:
There is not sufficient evidence to suggest that the true mean difference in scores is
different from June to August.
This suggests that program may not be worthwhile.