Transcript P. STATISTICS LESSON 11 – 1 (DAY 1)

CHAPTER 11
Section 11.2 – Comparing Two Means
COMPARING TWO MEANS



Comparing two populations or two treatments is one of
the most common situations encountered in statistical
practice. We call such situations two-sample problems.
A two sample problem can arise from a randomized
comparative experiment that randomly divides subjects
into two groups and exposes each group to a different
treatment.
Comparing random samples separately selected from
two populations is also a two sample problem. Unlike
the matched pairs designs studied earlier, there is no
matching of the units in the two samples and the two
samples can be of different sizes.
TWO – SAMPLE PROBLEMS


The goal of inference is to compare the responses
to two treatments or to compare the
characteristics of two populations.
We have a separate sample from each treatment
or each population.
EXAMPLE 11.9 - TWO-SAMPLE PROBLEMS
1.
A medical researcher is interested in the effect on blood
pressure of added calcium in our diet. She conducts a
randomized comparative experiment in which one group
of subjects receives a calcium supplement and a control
group receives a placebo.
2.
A psychologist develops a test that measures social
insight. He compares the social insight of male college
students with that of female college students by giving the
test to a sample of students of each gender.
3.
A bank wants to know which of two incentive plans will
most increase the use of its credit cards. It offers each
incentive to a random sample of credit card customers and
compares the amount charged during the following six
months.
CONDITIONS FOR COMPARING TWO MEANS



We have two SRSs, from two distinct populations.
Both populations are normally distributed. The
means and standard deviations of the populations
are unknown.
The samples are independent (That is, one
sample has no influence on the other.)
Matching violates independence, for example.
 We measure the same variable for both samples.

ORGANIZING THE DATA
Call the variable we measure x1 in the first population
and x2 in the second . We know parameters in this
situation.
Populatio
n
Variable
Mean
Standard
deviation
1
x1
μ1
σ1
2
x2
μ2
σ2
ORGANIZING DATA (PART 2)
There are four unknown parameters, the two means
and the two standard deviations.
Population
Sample
Size
Sample
Mean
1
𝒏𝟏
𝒙𝟏
Sample
Standard
deviation
𝒔𝟏
2
𝒏𝟐
𝒙𝟐
𝒔𝟐
THE SAMPLING DISTRIBUTION OF 𝑥1 − 𝑥2


The mean of 𝑥1 − 𝑥2 is 𝜇1 − 𝜇2 . That is, the difference
of sample means is an unbiased estimator of the
difference of population means.
The variance of the difference is the sum of the
variance of 𝑥1 − 𝑥2 which is
𝜎2
1
𝑛1


+
𝜎2
2
𝑛2
Note that the variances add. The standard deviations do
not.
If the two populations are both normal, then the
distribution of 𝑥1 − 𝑥2 is also normal.
THE SAMPLING DISTRIBUTION OF 𝑥1 − 𝑥2
(CONTINUED…)


Because the statistic 𝑥1 − 𝑥2 has a normal
distribution, we can standardize it to obtain a
normal z statistic.
The two-sample z statistic is standardized by:
𝑧=
(𝑥1 −𝑥2 ) − (𝜇1 − 𝜇2 )
𝜎2 1 𝜎2 2
+
𝑛1
𝑛2
STANDARD ERROR



Because we don’t know the population standard deviations,
we estimate them by the sample standard deviations from our
two samples.
The result is the standard error, or estimated standard
deviation:
𝑠2 1 𝑠2 2
SE =
+
𝑛1
𝑛2
The two-sample t statistic:
𝑡=
(𝑥1 −𝑥2 ) − (𝜇1 − 𝜇2 )
𝑠2 1 𝑠2 2
+
𝑛1
𝑛2
DEGREES OF FREEDOM FOR TWO-SAMPLE PROBLEMS


The two-sample t statistic does not have a t distribution
since we replaced two standard deviations by the
corresponding standard errors.
Therefore, we use two methods for calculating degrees
of freedom:

Option 1: Use procedures based on the statistic t with critical
values from a t distribution (used by calculator).

Option 2: Use procedures based on the statistic t with
critical from the smaller n – 1.
TWO-SAMPLE T DISTRIBUTIONS


The statistic t has the same interpretation as any z
or t statistic: it says how far 𝑥1 − 𝑥2 is from its
mean in standard deviation units.
When we replace just one standard deviation in a z
statistic by a standard error we must replace the z
distribution with the t distribution.
CONFIDENCE INTERVAL FOR A TWO-SAMPLE T

The confidence interval for 𝜇1 − 𝜇2 is given by:
(𝑥1 − 𝑥2 ) ± 𝑡 ∗


𝑠2 1
𝑛1
+
𝑠2 2
𝑛2
𝑡 ∗ is the upper (1-C)/2 critical value for the t(k)
distribution with k the smaller of 𝑛1 − 1 and 𝑛2 −
1
To test the hypothesis 𝐻0 : 𝜇1 =𝜇2 , compute the
two-sample t statistic:
(𝑥 −𝑥 )
𝑡= 1 2
𝑠2
𝑠2
1+ 2
𝑛1
𝑛2
Example: Calcium and Blood Pressure

Does increasing the amount of calcium in our diet reduce blood pressure? Examination of a
large sample of people revealed a relationship between calcium intake and blood pressure.
The relationship was strongest for black men. Such observational studies do not establish
causation. Researchers therefore designed a randomized comparative experiment. The
subjects were 21 healthy black men who volunteered to take part in the experiment. They
were randomly assigned to two groups: 10 of the men received a calcium supplement for 12
weeks, while the control group of 11 men received a placebo pill that looked identical. The
experiment was double-blind. The response variable is the decrease in systolic (top number)
blood pressure for a subject after 12 weeks, in millimeters of mercury. An increase appears
as a negative response Here are the data:
We want to perform a test of
H0: µ1 - µ2 = 0
Ha: µ1 - µ2 > 0
where µ1 = the true mean decrease in systolic
blood pressure for healthy black men like the
ones in this study who take a calcium
supplement, and µ2 = the true mean decrease in
systolic blood pressure for healthy black men like
the ones in this study who take a placebo.
We will use α = 0.05.
Example: Calcium and Blood Pressure
If conditions are met, we will carry out a two-sample t test for µ1- µ2.
• Random The 21 subjects were randomly assigned to the two treatments.
• Normal With such small sample sizes, we need to examine the data to see if it’s
reasonable to believe that the actual distributions of differences in blood pressure when
taking calcium or placebo are Normal. Hand sketches of calculator boxplots and
Normal probability plots for these data are below:
The boxplots show no clear evidence of skewness and no outliers. The Normal
probability plot of the placebo group’s responses looks very linear, while the Normal
probability plot of the calcium group’s responses shows some slight curvature. With no
outliers or clear skewness, the t procedures should be pretty accurate.
• Independent Due to the random assignment, these two groups of men can be
viewed as independent. Individual observations in each group should also be
independent: knowing one subject’s change in blood pressure gives no information
about another subject’s response.
Example: Calcium and Blood Pressure
Since the conditions are satisfied, we can perform a two-sample t test for the
difference µ1 – µ2.
Tes t statistic
:
(x  x 2 )  ( 1  2 ) [5.000 (0.273)] 0
t 1

 1.604
2
2
2
2
8.743
5.901
s1
s

 2
10
11
n1 n 2
P-value Using the conservative df = 10 – 1 = 9,
we can use Table C to show that the P-value is
between 0.05 and 0.10.
Because the P-value is greater than α = 0.05, we fail to reject H0. The
experiment provides some evidence that calcium reduces blood pressure,
but the evidence is not convincing enough to conclude that calcium
reduces blood pressure more than a placebo.
Example: Calcium and Blood Pressure
We can estimate the difference in the true mean decrease in blood pressure for
the calcium and placebo treatments using a two-sample t interval for µ1 - µ2. To
get results that are consistent with the one-tailed test at α = 0.05 from the
example, we’ll use a 90% confidence level. The conditions for constructing a
confidence interval are the same as the ones that we checked in the example
before performing the two-sample t test.
With df = 9, the critical value for a 90% confidence interval is t* = 1.833.
The interval is:
s12 s2 2
8.7432 5.9012
(x1  x 2 )  t *

 [5.000 (0.273)] 1.833

n1 n 2
10
11
 5.273 6.027
 (0.754,11.300)
We are 90% confident that the interval from -0.754 to 11.300 captures the difference in true
mean blood pressure reduction on calcium over a placebo. Because the 90% confidence
interval includes 0 as a plausible value for the difference, we cannot reject H0: µ1 - µ2 = 0
against the two-sided alternative at the α = 0.10 significance level or against the one-sided
alternative at the α = 0.05 significance level.
THE POOLED TWO-SAMPLE PROCEDURES



Procedures that average use the statistical term
“pooled.”
Pooled two-sample t procedures is a situation
where the variances of both the samples are
assumed to be the same and the sample sizes are
the same. This rarely happens and the same
results will occur with regular t procedures.
On the print out from a computer (p.666), use the
unequal line for variances, degrees of freedom,
and probability.

Homework:
Technology toolbox on p.660-661
 P.649 #’s 37, 38, 40, 43, & 50
 Chapter 11 review: p.675 #’s 64, 68, & 72
 Ch.10/11 take home test due Thursday 3/27


DO NOT WORK ON THE TEST TOGETHER!!!