Bring a bag of M&M's for day 1 chapter 11.
Chapter 10
Comparing Two Means
Target Goal:
I can use two-sample t procedures to
compare two means.
10.2a
h.w: pg. 626: 29 – 32, pg. 652: 35, 37, 57
Two-Sample Problems

Comparative studies are more convincing
than single single-sample investigations, so
one-sample inference (for matched pairs) is
not as common as comparative (two-sample)
inference.
In a comparative study:
the units or sample sizes can be different.
1) We may want to compare two treatments, or we may
want to compare two populations.
2) The samples must be chosen randomly and
independently in order to perform statistical
inference.
 Because match pairs are NOT chosen independently,
we will NOT use two-sample inference for a
matched pairs design.
 Recall: for a matched pairs design, apply the onesample t procedures to the observed differences.

Otherwise, we may use two-sample inference
to compare two treatments or two populations.
Significance Tests for µ1 – µ2

The null hypothesis is that there is no
difference between the two parameters.

The alternative hypothesis could be that
Before you begin, check your assumptions!
Conditions for Comparing Two Means
Both samples must be a
 Random sample or randomized experiment.
 must be chosen independently
 both populations must be normally
distributed.
(Check the data for outliers or skewness.)

Two-Sample t Interval for a Difference Between Means
When the Random, Normal, and Independent conditions are met, an
approximate level C confidence interval for ( x1  x2 ) is
s12 s2 2
( x1  x2 )  t *

n1 n2
where t * is the critical value for confidence level C for the t distribution with
degrees of freedom from either technology or the smaller of n1  1 and n2  1.
Comparing Two Means
Confidence Intervals for µ1 – µ2. On page 634

Two-Sample t Interval for a Difference Between Means
Random The data are produced by a random sample of size n1 from
Population 1 and a random sample of size n2 from Population 2 or by
two groups of size n1 and n2 in a randomized experiment.
Normal Both population distributions are Normal OR both sample
group sizes are large ( n1  30 and n2  30).
Independent Both the samples or groups themselves and the individual
observations in each sample or group are independent. When sampling
without replacement, check that the two populations are at least 10 times
as large as the corresponding samples (the 10% condition).
Comparing Two Means
Confidence Intervals for µ1 – µ2. On page 634
The Sampling Distribution of x1  x2 :

If these assumptions hold, then the difference
in sample means is an unbiased estimator of
the difference in population means,
so x1  x2 is equal to 1  2 .
The variance of x1  x2 is the sum of the
variances of x1 and x2 ,
which is 
2
x1  x 2


2
1
n1

2
2
n2
Note: variances add, standard deviations don’t.
Furthermore, if both populations are normally
distributed, then x1  x2 is also normally
distributed.
Standardizing x1  x2

Subtract the mean and divide by the standard
deviation:
x1  x2    1  2 

z
 12  22

n1 n2


If we do not know  1 and  2 , we will
substitute with the standard error SE using
and s2 for the standard deviation.
n
This gives the standardized t value:
x1  x2    1  2 

t
s12 s22

n1 n2
s1
n
Ex: Calcium and blood pressure


Does increasing the amount of calcium in our diet
reduce blood pressure? The subjects were 21 healthy
black men. A randomly chosen group of 10 received
calcium supplement for 12 weeks. A control group
of 11 men received a placebo (double blinded
experiment).
The response variable is the decrease in systolic
(heart contracted) blood pressure. An increase
appears as a negative response (drop in blood
pressure).
Group 1: 7, -4, 18, 17, -3, -5,1,10,11,-2
Group 2: -1, 12, -1,-3, 3, -5, 5, 2, -11, -1, -3
 Calculate summary statistics: (Enter data into
lists.)
x
Group
Treatment
n
s
1
Calcium
10
5.000 8.743
2
Placebo
11
-0.273 5.901
Is this outcome good evidence that calcium decreases
blood pressure?
Perform a significance test.
Step 1. State - Identify the population of interest and the
parameter you want to draw a conclusion about. State the null
and alternative hypothesis in words and symbols.
H 0 : 1  2 : The mean decrease in blood
pressure for those taking calcium is the same as
the mean decrease in blood pressure for those
taking a placebo.
H a : 1  2: The mean decrease in blood
pressure for those taking calcium is greater than
the mean decrease in blood pressure for those
taking a placebo.
Step 2. Plan –If the conditions are met, we will
construct a two-sample t test for 1  2 .
Verify the conditions (random, indep, normal).
 Random: Because of randomization, we must
assume the calcium and placebo groups as
SRS.
 Normal: Examine the stem plot in notes. Are
there outliers?
Step 2. Plan –If the conditions are met, we will
construct a two-sample t test for 1  2 .

Normal: Examine the stem plot in notes. Are
there outliers? no

Are there any departures from normality that
would prevent us from using a t-test?
Calcium is a little irregular but this is not
unusual for a small sample.

Independent: Yes, due to random assignment, the
populations can be viewed as independent.
Individual observations in each group should also be
independent.
Proceed with t procedures.

Step 3. Carry out the inference procedure. Compute the
test statistic and P-value

To test
H 0 : 1  2
compute: H 0 : 1  2  0
x1  x2   (1  2 )  (5.00  (0.273))  0 


t
s12 s22

n1 n2
8.7432 5.9012

10
11
= 1.604
Note: x1-x2 as a statistic, measures the advantage of calcium over a placebo.

For df, use the smaller of n1 – 1, n2 – 1.
There are 9 df,

Find P
tcdf(
P = .072

Step 4: Interpret your results:


The experiment found evidence that calcium
reduces blood pressure but not at the
traditional 5% and 1% levels as the p value is
.071.
We fail to reject Ho at these levels.
We will use graphing calculators to perform twosample T tests!





2-SampTTest for a hypothesis test
2-SampTInt for a confidence interval
Confirm the P-value and construct a 90%
confidence interval
STAT:TESTS:
Pooled: no
Use data if you have it or Stats if you are
given x1 , x2 , s1 , s2 .




CI = (-.4767, 11.022).
The 90 % C.I. covers 0 (both neg. and pos.
changes).
We can not reject Ho against the two sided
alternative at the α = 0.10 level of
significance.
Read pg. 625 - 637