Some Examples of Statistical Inference
Example – 1: Do pregnant women who smoke have
babies with lower birth weight than those who do not
smoke?
A researcher thinks so. To test her
conjecture she has recorded the birth weight of babies
born at Shands in a given period, together with the
smoking status of the mother and summarized the
data as shown below:
Smoking
Status
1. Smokers
2. Nonsmokers
Sample Statistics
Sample
Sample
Sample
Standard
sizes (ni)
Means ( X i ) Deviations
(Si)
134
2733 grams 599 grams
5974 3118 grams 672 grams
Do the above data support the conjecture of the
researcher?
ALWAYS: Before you dive-in to answer any
problem, there are some questions you should ask
yourself and find the answers in the statement of the
problem.
Here are the questions for this problem:
Chapter 9 Examples, Fall 2007
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1. What type of a problem is this?
a. How many populations, parameters, samples
are there in this problem? Define the
parameter(s).
There are 2 populations (set of all pregnant
women who smoke is population 1 and the
set of all pregnant women who do not smoke
is population 2) and we have 2 independent
samples, one from each population.
There are two parameters of interest:
µ1 = µS = mean birth weight of babies born
to ALL pregnant women who smoke
and
µ2 = µN = mean birth weight of babies born
to ALL pregnant women who do not
smoke
We are interested in the difference between
these means, i.e., µ1 – µ2.
b. What type of data (random variable) do we
have: quantitative or categorical? Why?
c. Are the samples dependent or independent?
Why?
d. Hence, what type of problem do we have?
Chapter 9 Examples, Fall 2007
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Comparing two population means using two
independent samples.
2.
What are hypotheses to be tested?
Ho: µ1 – µ2 = 0 vs. Ha: µ1 – µ2 < 0
How can you tell?
3. What is the test statistic for this problem?
Since we have quantitative data, (and the
population variance is unknown) the test
statistic to use is
Estimator  Number in Ho
T
~ t( df )
Est. SE ( Estimator )
4. In this problem, what are “Estimator”, “Number
in Ho”, “Est. SE(Estimator) and “df” ( = the
degrees of freedom)?
Since we are interested in µ1 – µ2 (unknown),
this difference is estimated by X 1  X 2 . [That is,
the “estimator” of µ1 – µ2 is X 1  X 2 ].
Looking back to Ho, we see that we have
Ho: µ1 – µ2= 0. Hence “Number in Ho” is 0.
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We also know that the estimated SE (Estimator)
S12 S22
is Est. SE ( X 1  X 2 ) 
. [Look at the

n1 n2
table of formulas at the end of summary to
Chapters 8 and 9.]
Finally, since the two sample sizes are very
different we will use df = smaller of (n1 – 1) and
(n2 – 1) = 133. [Note that the df used by
computer is different. Why?]
So,
(X  X2)  0
T 1
~ t(133) .
2
2
S1 S2

n1 n2
Since the df is too large, we will use the normal
distribution. (Why?)
5. Are all the assumption needed for this procedure
satisfied?
a) First we need two independent random
samples. Although the samples may be
accepted as independent (with well defined
populations), it is not clear whether these
samples are random or not. The method of
selection as well as the period of time the
Chapter 9 Examples, Fall 2007
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samples were selected may have
implications
for
some
special
characteristics for the pregnant women.
We will assume that the samples are
random and representative of the
populations. Our results will be valid if
this assumption is justifiable.
b) The type of random variable (birth weight)
is quantitative. So this assumption is
satisfied.
c)
We do not know if the populations have
normal distributions. [If we had sample
data we could see if there are any extreme
values indicating skewed distributions.]
Since we have no way of knowing this, we
will assume that the populations are
normally distributed and our results will be
valid if this assumption is approximately
true.
Chapter 9 Examples, Fall 2007
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6. What is the calculated value of the test statistic?
Tcal 
( X1  X 2 )  0
2
1
2
2
S
S

n1 n2

(2733  3118)  0
2
2
 7.34
599 672

134 5974
7. What is the p-value of the test?
Before you write the formula for the p-value
ALWAYS look at Ha. That determines the tail
to look at.
Ha: µ1 – µ2 < 0 so the p-value is
P(T < Tcal) = P(T < –7.34).
Since the df is large we may use the normal
approximation to the t-distribution and write
P-value = P(T < –7.34) = P(Z < – 7.34) = 0
(almost). [Sketch a graph and see the answer
without looking at any table.]
8. What is the decision?
Since p-value < any reasonable level of
significance (α), we will reject Ho.
Remember that the decision rule is always:
“Reject Ho when p-value  α”
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9. What is the conclusion of the test?
The observed data strongly support the
researcher’s conjecture that on the average
pregnant women who smoke have babies with
lower birth weight than those who do not smoke
[assuming that the samples are representative of
the two populations and that the distributions of
the populations are not too far from the normal
distribution).
10. What is the CI? Interpret what you have found.
Suppose we have observed the following output
from Minitab:
Two-Sample T-Test and CI
Sample
1
2
N
134
5972
Mean
2733
3118
StDev
599
672
SE Mean
51.7
8.7
Difference = mu(1) – mu(2)
Estimate for difference = –385.000
95% CI for difference (– 488.738, – 281.262)
T-Test of difference = 0 (vs <):
T-Value = – 7.34 P-Value = 0.000 DF = 140
Why is df = 140 and not 134?
What does p-value = 0.000 mean?
Why is SE Mean = 51.7 or 8.7?
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Now we can see that the 95% CI for µ1 – µ2,
calculated using the general formula for CI, is
CI   Estimator  ME 

S12 S 22 
*
  ( X1  X 2 )  t 



n1 n2 


5992 6722 
  (2733  3118)  1.96 



134 5972 

 ( 488.738,  281.262).
What does this tell us?
We are 95% confident that the mean birth weight
of babies born to mothers who smoke is between
281.262 grams and 488.738 grams less than the
mean birth weight of babies born to mothers who
do not smoke.
Note that this is a (– , –) type of confidence
interval meaning that there is a significant
difference between the two population means
(since zero is not in the CI); actually, we can say
more: we can state that the second population
mean is larger than the first one (with 95%
confidence).
Why is t(140) = 1.96?
Chapter 9 Examples, Fall 2007
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11. Do the results of the CI and significance test
agree? Must they agree? Why or why not?
Although the results do agree here (because we
rejected Ho) they do not NEED TO AGREE ALL
THE TIME.
In general, the results of significance test and CI
MUST AGREE WHEN Ha is 2-SIDED
[What is a 2-sided hypothesis?]
Example – 2: Many children are diagnosed each year
with asthma. In an effort to educate these children
about their condition, an educational video was
developed. To test the effectiveness of this video, ten
randomly selected children, of elementary school
age, who had been recently diagnosed, were chosen
to participate in a study. A nurse asked the children a
series of questions about asthma, then showed them
the video and asked the same questions again. The
children’s scores were as follows:
Child 1 2 3 4 5 6 7 8 9 10
Before 61 60 52 74 64 75 42 63 53 56
After 67 62 54 83 60 89 44 67 62 57
Chapter 9 Examples, Fall 2007
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1) What type of a problem is this?
a) What is the random variable? Is it
continuous or discrete?
b) How many populations, samples,
parameters are there?
c) Are samples dependent or independent?
d) Hence?
2) Assumptions?
a) Random sample?
b) Normal population?
c) What else?
3) Parameter(s) of interest?
µd = µbefore – µafter
4) Hypotheses?
Ho: µd = 0 vs. Ha: µd < 0 [Why?]
5) Test Statistic T 
X d  0
~ t( n1)
Sd / n
6) What is n in this problem? Why?
7) The p-value?
First look at Ha: µd < 0.
Thus, P-Value = P(T  Tcal)
X  0
4.5  0

 2.78
Where Tcal  d
Sd / n 5.13/ 10
Chapter 9 Examples, Fall 2007
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So, p-value= P(T  – 2.78)=P(T  +2.78) [Why?]
Looking at the row with df = 9 we see that
P(T  2.262) = 0.025
P(T  2.78) = p-value
and
P(T  2.821)= 0.010.
Hence 0.010 < p-value < 0.025
8) Decision?
Since 0.01 < p-value < 0.05 < 0.10, we will reject
Ho at α = 0.05 and α = 0.10, but not at α = 0.01.
9) Conclusion?
The observed data indicate that the video is
effective in increasing the knowledge of elementary
school aged children’s about asthma at 5% level of
significance.
The following output is obtained from Minitab.
Interpret the results:
Chapter 9 Examples, Fall 2007
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Paired T-Test and CI: before, after
Paired T for before – after
before
after
Difference
N
10
10
10
Mean
StDev SE Mean
60.0000 10.0000 3.1623
64.5000 13.2267 4.1826
– 4.50000 5.12619 1.62104
95% CI for mean difference: (– 8.16705, – 0.83295)
T-Test of mean difference = 0 (vs < 0):
T-Value = – 2.78 P-Value = 0.011
First note that the p-value here is consistent with
what we found “by hand,” it is in fact between 1%
and 2.5%. Hence we will reject Ho at 5% and 10%
level of significance but not at 1% level of
significance.
Next, note that the confidence interval has both ends
negative, which indicates that the mean of the
population of scores before watching the video is
larger than the mean of the population of scores after
watching the video, i.e., we are 95% confident that
watching the video is effective in increasing the
children’s knowledge.
Chapter 9 Examples, Fall 2007
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Actually, we could use
1 – p-value = 1 – 0.011 = 0.989
And state that we are 98.9% confident that the video
increases the average level of children’s knowledge
on asthma. But we cannot make it at 99% level of
confidence. [So what? What is so special about
99%?]
Let’s have a look at some other cases. Suppose in
another study (of the same problem) we found some
different p-values, as shown below. What can we
conclude in each case?
a) When p-value = 0.03 and
α = 0.10,
α = 0.05,
α = 0.01,
p-value < α, Reject Ho. Results significant.
p-value < α, Reject Ho. Results significant.
p-value > α, Do NOT Reject Ho.
Results NOT significant
 We can reject Ho at α = 0.10 and 0.05 but NOT at 0.01
 Results are significant at 10% and 5% level of
significance but NOT at 1% level of significance.
 We can be 90% and 95% confident that there is a
significant difference but NOT 99% confident
 Taking 1 – p-value = 1 – 0.03 = 0.97, we can be
97% confident that there is a significant difference
between the population means.
Chapter 9 Examples, Fall 2007
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b) When p-value = 0.09 and
α = 0.10,
α = 0.05,
α = 0.01,
p-value < α, Reject Ho. Results significant.
p-value > α, Do NOT Reject Ho.
Results NOT significant
p-value > α, Do NOT Reject Ho.
Results NOT significant
 We can reject Ho at α = 0.10 but NOT at 0.01 or 0.05
 Results are significant at 10% level of significance but
NOT at 1% or 5% level of significance.
 We can be 90% confident that there is a significant
difference but NOT 95% or 99% confident
 Taking 1 – p-value = 1 – 0.09 = 0.91, we can be
97% confident that there is a significant difference
between the population means.
c) When p-value = 0.12 and
α = 0.10, 0.05 05 0.01, p-value > α, Do NOT Reject Ho.
Results NOT significant
 Results are not significant at any reasonable level of
significance.
 We can NOT reject Ho at α = 0.10 or 0.05 or 0.01(or at
any reasonable level of significance).
 Results are NOT significant at 10% and 5% and 1%
level of significance.
 There is no significant difference between the population
means at any reasonable level of significance.
 There is not enough evidence to indicate any difference
between the population means.
Chapter 9 Examples, Fall 2007
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Example – 3:
[Agresti and Franklin, problem 9.6, modified]
A Swedish study selected a random sample of 684
patients who had suffered a stroke and asked them to
take a low does of aspirin daily. Another independent
sample of 676 stroke patients was given placebo to be
taken daily. The Minitab output of the analysis of the
data is shown below, where X is the number of
deaths due to heart attack during a follow up study of
about 3 years. Was aspirin effective to reduce heart
attacks among patients who suffered a stroke?
Test and CI for Two Proportions
Sample
X
N
Sample p
1
28
684
0.040936
2
18
676
0.026627
Difference p(1) – p(2)
Estimated difference = 0.0143085
95% CI for difference:(–0.00486898, 0.0334859)
Test for difference = 0 (vs not = 0):
Z = 1.46 p-value = 0.144
Interpret this output.
Chapter 9 Examples, Fall 2007
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1. What is/are the population(s) and samples?
Population 1: The set of heart stroke patients who
take a low does of aspirin every day.
Sample 1: The 684 patients selected from the first
population
Population 2: The set of all heart stroke patients who
take placebo (i.e. nothing) [Why?]
Sample 2: The 676 patients selected from the second
population
A question: Do such populations physically exist?
1. What type of study do we have here? Why?
What type of variable do we have?
How many populations?
How many samples?
Hence?
Define the parameters of interest?
p1 = Proportion who die in the population of all
patients who take a low does of aspirin
daily.
p2 = Proportion who die in the population of all
patients who take nothing (placebo)
Chapter 9 Examples, Fall 2007
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2. What are the hypotheses?
Ho: p1 – p2 = 0 vs. Ha: p1 – p2 > 0
How can we tell?
3. What is the test statistic?
pˆ1  pˆ 2 

Z
~ N (0,1)
1 1
pˆ (1  pˆ )   
 n1 n2 
4. What are the assumptions? Are they satisfied?
a) Independent Random Samples
b) Categorical variable
c) Observed number of “Success”s  10 in both
samples
d) Observed number of “Failure”s  10 in both
samples
5. What is the p-value of the test?
First look at Ha: p1 – p2 > 0.
So p-value = P(Z  1.46)
But, P(Z  1.46) = P(Z  – 1.46) by symmetry of
the normal distribution.
So, p-value =0.0749 [from z-tables].
But the computer output gives p-value = 0.144. Is
something wrong?
Chapter 9 Examples, Fall 2007
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How do we find the calculated value of the test
statistic?
pˆ1  pˆ 2 

Z cal 
1 1
pˆ (1  pˆ )   
 n1 n2 

 0.040936  0.026627 
1 
 1
0.0338(1  0.0338) 


 684 676 
 1.46
What is p̂ and how is it found?
If Ho is true, then p1 = p2 and we may just call it p.
Then, p is estimated by “pooling” the two samples,
so p̂ is called the “pooled sample estimate.” It is
calculated as
X  X2
28  18
pˆ  1

 0.0338
n1  n2
684  676
6. What is the decision?
Since the p-value = 0.0749 > 0.05 > 0.01, we fail to
reject Ho at 1% and 5% levels of significance.
However, at 10% level of significance we reject Ho.
7. What is the conclusion?
Chapter 9 Examples, Fall 2007
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The observed data do not give sufficient support to
the claim that aspirin is effective in reducing heart
attacks (p-value = 0.0749).
8. What does the confidence interval tell us?
A 95% CI for the difference of population
proportions is found as (– 0.00486898, 0.0334859)
Since the CI contains zero, we fail to reject Ho.
Remember that a definition of a confidence interval
is “the set of all acceptable hypotheses.” Since
zero is in the CI, Ho: p1 – p2 = 0 is an acceptable
hypothesis and hence we do not reject it.
9. How was the confidence interval found?
The general formula for a CI is
( Estimator ± (t* or z*)  SE(Estimator) ).
In this problem parameter of interest is p1 – p2 and
it is estimated by pˆ1  pˆ 2 and hence

pˆ1 (1  pˆ1 ) pˆ 2 (1  pˆ 2 ) 
*
ˆ
ˆ
CI   ( p1  p2 )  z 


n
n
1
2



0.040936(0.959064) 0.026627(0.973373) 
  (0.040936  0.026627)  1.96 


684
676


  0.014309  0.0191778
 ( 0.00486898 , 0.0334859)
Chapter 9 Examples, Fall 2007
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10. Do the results of the CI and significance test
agree? Must they agree? When MUST the two
agree?
Some more work for you: Look at other examples and
problems in text. Make sure you can identify the
problems and solve them by hand. Also, identify the
problem type for the following examples. Make sure
to ask the relevant questions before you decide on the
type of problem.
Example – 4: There is no known cure for
fibromyalgia – a mysterious ailment with symptoms
that include stiffness, fatigue, and pain.
But
acupuncture may help, according to research from the
Mayo Clinic. Twenty-five patients underwent six
sessions of acupuncture, while 25 received a
simulated version. A month later, those who go the
real thing had less pain and felt significantly less tired
and anxious.
Example – 5: Nearly 20% of pregnancies end in
miscarriage, often for no apparent reason.
Example – 6: It’s well known that older moms have
a higher chance of miscarrying, but a recently
published study of 14,000 women shows that the
father’s age matters too. If he’s 40 or older, the
mother’s risk of miscarrying is three times higher
than if the dad is under 25.
Chapter 9 Examples, Fall 2007
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Example – 7: If you suffer from high blood pressure
and high cholesterol, make sure you’re being treated
for both.
Having the two problem together
dramatically increases the risk of hear attack and
stroke. In a study of nearly 3,000 men and women
conducted by the University of California, Irvine it
was found that less than one third of high-risk
patients are actually prescribed medication for both
problems.
Example- 8: In case you need encouragement: It
takes only 14 days to start seeing good results from
flossing, says a New York University study of 51 sets
of twins. All the participants brushed regularly, but
one twin in each pair also flossed. After two weeks,
the flossing twins had significantly less gum
bleeding.
Chapter 9 Examples, Fall 2007
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