AP Statistics
Section 11.2
“Comparing Two Means”
Last section, we looked at
inference based on one sample or on the differences in a matched pairs design
(still one set of data). We will now
look at having TWO samples/sets of data.
This is a much more common situation than just having one set!
Look at Example
11.9 on p. 648 in your text for examples of two-sample problems.
In order for a
comparison of two means to be valid (i.e. in order for us to use two-sample
inference reliably), we need to address the following two assumptions:
(1) There
should be 2 INDEPENDENT SRS’s!
**If
the SRS’s are not independent, then DO NOT USE the t-test! The t-test is NOT robust in this sense!
(2) The
two populations should be approximately normally distributed.
**You
can check this with graphs (dotplot, stemplot, histogram, etc.).
If
this assumption is violated, you can still proceed with the t-test. It IS robust to this assumption!
In examining 2 different
populations or samples, we will use subscripts to distinguish between the
two. (Because it often does not matter
which you label as population 1 and population 2, it is IMPERATIVE that you
clearly state which is which in the beginning of the problem!)
Population Variable Mean St. Dev. Sample
Size Sample Mean Sample St. Dev.
1 x1 
n1 
s1
2 x2 
n2 
s2
To compare two
means, we can perform inference in two ways …
(1) Establish
a CI for their difference 
(2) Test
a hypothesis of no difference
H0: 
OR H0: 
(Notice
these two statements are equivalent.)
Just as we used 
as an unbiased
estimator of 
, we will use 
as an unbiased
estimator of 
.
Using the rules of
means & variances for random variables that we examined in Chapter 7, we
can examine characteristics of the sampling distribution of .
(1)
is the mean of the distribution.
(2) The
variance of is the SUM of the
individual variances (and the standard deviation is the square root of this):
*This
is coming from each individual standard error being = to 
, and therefore, each
individual variance being = to 
.
**Also
remember that variances add – standard deviations (standard errors) to
not! To combine two populations/samples,
you must find the
variances first, then add, and then finally take the square root
for the difference’s standard deviation/error.
(3) The
distribution of 
is normal if both
populations are normal OR if the sample size is large enough for the Central
Limit Theorem to “kick in” (to become evident).
Because of the
normal distribution, if and are known, we can standardize using a
z test statistic:
One sample z:
Two-sample z:
Since it is not
realistic to say that we would know and , we will need to use s1
and s2 as estimates. As a
result, we will need to find a t test statistic instead:
The Two-Sample
t:
NOTE: When
we introduced s in place of , we introduced more
uncertainty and ended up with a t distribution with n-1 degrees of freedom
instead of a z distribution (Standard Normal).
When we introduce two ’s, we introduce even more uncertainty …
For this reason,
the t distribution for two samples is not exactly the same as the t
distribution for one sample. The
two-sample distribution is a t distribution with degrees of freedom given by:
instead of df = n – 1.
*If you use your
calculator to run a two-sample t test, it will produce results based on this
accurate number of degrees of freedom.
**If you choose to
use the one-sample t-table & do it by hand, you will want to choose the
smaller degrees of freedom between n1 – 1 and n2 –
1. This will provide a conservative
estimate of the p-value.
(Better to overestimate
your error and p-values, than underestimate it!
Remember, lower degrees of freedom imply smaller sample sizes, which
imply more variability – more area in the tails.)
The College Board
will accept BOTH methods as correct on the AP Exam. J
The t procedures for
two-samples will remain basically the same as those for the one-sample t:
CI:
For
hypothesis testing, the two-sample t test statistic:
Notice that the “
” is missing from the
two-sample t test statistic above. The
reason for this is that the hypothesis H0 suggests that , which means that , which means that you would be
subtracting 0 when you are subtracting .
(EX) The heights (measured in inches) of 20
randomly selected women & 30 randomly selected men were independently
obtained from a student body of a certain college in order to estimate the
difference in their mean heights.
Assume the heights
are approximately normally distributed for both populations. Here is the sample data:
POPULATION n x-bar s
1: Male 30 69.8 1.92
2: Female 20 63.8 2.18
Find a 95%
Confidence Interval for the difference between mean heights,
.
**Do by hand
first.
**With calc: STAT, TESTS, 2SampleTInt
Notice that with
the calculator, you are asked whether or not you want the variances “pooled.”
What this means …
If
population variances are equal (if ), then the
2 sample variances can be averaged or “pooled” to estimate the common
population variance. The result is
called the “pooled 2-sample t.”
Its advantage …
If
population variances are the same, then the distribution is exactly a t
distribution with df = n1
+ n2 – 2.
On the computer
output on p. 666 of your text, the “EQUAL” choice is pooled; the “UNEQUAL”
choice is not (it’s a regular 2-sample t).
(EX) Suppose we are interested in comparing the academic success of
college students who belong to fraternities with the academic success of those
who do not belong. The reason for the
comparison centers on the recent concern that fraternity members, on average,
are achieving academically at a lower level than non-fraternity students. (Cumulative GPA is used to measure academic
success.)
Random samples of
size 40 are taken from each population.
Sample results …
POPULATION n x-bar s
1: Non-fraternity 40 2.21 0.59
2: Fraternity 40 2.03 0.68
Complete a
hypothesis test using 
. Assume GPA’s for
both groups are approximately normally distributed.
On calc: STAT, TESTS, 2SampleTTest
ASSIGNMENT:
p. 649 (11.37,
11.38)
p. 657-658 (11.40
– 11.43)
p. 665-666 (11.48)