CHAPTER 9
INFERRING POPULATION MEANS
SYMBOLS... SAMPLE STATISTICS & POPULATION
PARAMETERS
SAMPLING DISTRIBUTIONS... UNBIASED
ESTIMATORS...
AS N INCREASES, SD (OR SE) DECREASES...
More specifically, SE = σ/ n
 ...or in a few minutes... SE = s/ n


Example: considera distribution with s = 100;
what is our SE if n = 1? If n = 4? If n = 25?
CENTRAL LIMIT THEOREM...

http://blog.minitab.com/blog/michelleparet/explaining-the-central-limit-theorem-withbunnies-and-dragons-v2

http://onlinestatbook.com/stat_sim/sampling_
dist/
≈ NORMAL DISTRIBUTIONS & CENTRAL LIMIT
THEOREM
POPULATION DISTRIBUTION & SAMPLING
DISTRIBUTIONS...
GUINNESS BEER...
What would cause the head brewer of the famous Guinness brewery in
Dublin, Ireland, not only to use statistics but also to invent new
statistical methods? The search for better beer, of course.
William S. Gosset joined Guinness as a brewer in 1899. He soon
became involved in experiments and in statistics to understand the
data from those experiments. What are the best varieties of barley
and hops for brewing? How should they be grown, dried, and stored?
The results of the field experiments, as you can guess, varied. Gosset
faced the problem in using the z test to introduce the reasoning of
statistical tests: he didn't know the population standard deviation σ.
He observed that just replacing σ by s in the z statistic
and calling the result roughly Normal wasn't accurate enough.
GUINNESS BEER...
After much work, Gosset developed what we
now call the t distributions. His new t test
identified the best barley variety, and Guinness
promptly bought up all the available seed.
 Guinness allowed Gosset to publish his
discoveries, but not under his own name. He
used the name “Student,” so Gosset's t test is
sometimes called “Student's t” in his honor.

ONE-SAMPLE T-DISTRIBUTION
t-distribution, with SRS, size n, ≈ Normal, then
t=
𝑥−µ
𝑠
𝑛
with degrees of freedom = n – 1
remember, unknown σ is the reason for t-distribution
t-distribution has fatter tails (due to unknown σ); more
“wiggle room”
‘T’ IS LIKE ‘Z’... WELL, KIND OF... FOR MEANS

t-distribution  for means  unknown σ

t  # of standard deviations away from center

farther away from center, the more unlikely the
event (example, a female’s height is 5 standard
deviations above mean)  less area in that tail
of density curve
INFERENCE FOR MEANS (T PROCEDURES)...
DIFFERENT FLAVORS

Confidence Intervals




Primarily 2-sided; middle/center portion of density curve
1-sample or 2-sample
If 2-sample, then dependent (paired data) or independent (2
populations)
Hypothesis Testing



1-sided, 2-sided
1-sample or 2-sample
If 2-sample, then dependent (paired data) or independent (2
populations)
T DISTRIBUTIONS... STILL NEED CONDITIONS
MET
Random sample or random assignment
 Normal (or ≈ Normal) distribution.

If population is Normal, then sample is automatically
Normal.✔
 If n ≥ 30, then Central Limit Theorem says we have an ≈
Normal distribution. ✔
 If n ≤ 30, then need to show that sample is ≈ symmetric
with no outliers; then we can proceed in using the tdistribution.✔


Independence

Population must be at least 10 times sample size.
LET’S TRY A ONE-SAMPLE T-TEST ...

The mean weight of all 20-year-old women is 128
pounds; and the weights of all 20-year-old women
are Normally distributed. A random sample of 40
vegetarian women who are 20 years old showed a
sample mean of 122 pounds with a standard
deviation of 15 pounds. The women’s
measurements were independent of each other.

Determine if the weight for all 20-year-old
vegetarian women is different than 128 pounds,
using a significance level of 5%.
ONE-SAMPLE T-TEST ...

μ = 128; n = 40;
x = 122; s = 15
Ho:
μ = 128
Ha:
μ ≠ 128
where μ is the mean weight of all vegetarian women

Conditions: SRS, Normal, Independent

Input into Minitab; your results should be (≈):
t = - 2.529
P-value = 0.0156

Reject null hypothesis. With an alpha level of 5% and a P-value of
about 1%, there is sufficient evidence to suggest that the weights for
all 20-year-old vegetarian women are significantly different than 128
pounds.

HOW ABOUT A T-INTERVAL ...



μ = 128; n = 40;
x
= 122; s = 15
Ho: μ = 128
Ha: μ ≠ 128
Reject null hypothesis. With an alpha level of 5% and a
P-value of
 about 1%, there is sufficient evidence to
suggest that the weights for all 20-year-old vegetarian
women are significantly different than 128 pounds.
Let’s calculate a t-interval and see what we get... Does
our confidence interval confirm our findings?
HOW ABOUT A T-INTERVAL ...

Ho: μ = 128
Ha: μ ≠ 128
Reject null hypothesis. With an alpha level of 5% and a Pvalue of about 1%, there is sufficient evidence to suggest
that the weights for all 20-year-old vegetarian women are
significantly different than 128 pounds.

We are 95% confident that the interval from 117.2 to 126.8
captures the true mean weight for all 20-year-old vegetarian
women.

Why can we create a confidence interval to confirm our
findings in our hypothesis test? Do we need to do this every
time? Can we do this with every hypothesis test?
SWEETNESS OF DIET COLAS ...
Diet colas use artificial sweeteners to avoid sugar. These
sweeteners gradually lose their sweetness over time.
Manufacturers therefore test new colas for loss of sweetness
before marketing them. Trained tasters are randomly
assigned to sip the cola along with drinks of standard
sweetness and score the cola on a “sweetness scale” of 1 to
10.
The cola is then stored for a month at high temperature to
imitate the effect of four months' storage at room
temperature. Each taster scores the cola again after storage.
Our data are the differences (score before storage minus
score after storage) in the tasters' scores. The bigger these
differences, the bigger the loss of sweetness.
Most are positive. That is, most tasters found a loss
of sweetness. But the losses are small, and two
tasters (the negative scores) thought the cola
gained sweetness.
Assume that the population is Normally distributed.
Are these data good evidence that cola loses
sweetness in storage? Or is this variation just due
to sampling variability/just by coincidence?
Are these data good evidence that the cola loses
sweetness over time?
Hypotheses:
Ho: μ before - after = 0
Ha: μ before - after > 0
(mean sweetness loss for tasters is +; cola lost sweetness)
t-test (because we don’t know σ; we can calculate/we have s,
but not σ; and we are dealing with means)
SRS: Random assignment
Normal/Big Sample: Since population is Normally distributed,
our sample is automatically Normal
Independence:
We must assume the population is at least (10) (10); safe
assumption; and must assume one taste tester doesn’t
influence any other
Let’s use Minitab to run the t-test
Input data into Minitab
Results (≈): t = 2.6967
P-value = 0.0123
Reject Ho. Assuming an α = 0.05, and based on a Pvalue of ≈ 0.01, there is evidence that supports
the alternative hypothesis, that cola does seem to
lose sweetness over time.
NOTES...

If there is no α referenced, safe to assume an
α = 0.05 (this is the default if none is
referenced in problem)

Standard α levels are usually 0.05 or 0.01
(sometimes even 0.10)
LET’S GO BACK TO ALPHA = 0.05 FOR COLA
SWEETNESS TEST... P-VALUE = 0.0123
At α = 0.05, reject Ho; don’t believe μ = 0; but rather do believe μ > 0
(cola lost sweetness)
Remember, if we do everything absolutely correct, we can still,
sometimes, get an extreme sample (just by chance) which may lead
us to an incorrect decision (type I; type II errors)
No guarantee we have made the ‘correct’ decision; for the majority of
the time, we will never know if we made an error or not.
Also... What would happen if our α level was 1%? Would our decision
change? Why?
Can we create a confidence interval to confirm our findings in our
hypothesis test?
ABOUT CONDITIONS...
HEAT THROUGH GLASS...
How well materials conduct heat matters when
designing houses. Conductivity is measured in
terms of watts of heat power transmitted per
square meter of surface per degree Celsius of
temperature difference on the two sides of the
material. In these units, glass has conductivity
about 1.
Here is a SRS of 11 measurements of the heat
conductivity of a particular type of glass:

Is there evidence that the conductivity of this type
of glass is greater than 1? Carry out an
appropriate test. We may assume that the
population is Normally distributed.

Note: Just looking at the data, all values are
greater than 1... But is this just because we
happened to get a sample that tested high?
Would this happen if we took another sample?
That’s why we conduct hypothesis tests.
H o: μ = 1
Ha: μ > 1
where μ = mean heat conductivity transmitted per
square meter of surface per degree Celsius
difference on the two sides of the glass
Appropriate procedure: t-test (why?)
Conditions: Random, Normal/Big Sample,
Independence
H o:
μ=1
t = 8.95 and df = 10
Ha:

μ>1
p-value = 0.00000216
≈0
Interpretation:
Reject Ho. Because p-value is less than any
reasonable α (say α = 0.05), we can conclude that
the mean heat conductivity for this type of glass is
greater than 1.
POSSIBLE ERRORS... AND HOW ABOUT IF ALPHA
LEVEL CHANGED?
Ho:
μ=1
Ha: μ > 1
p-value = 0.00000216 ≈ 0
 Reject Ho
 we conclude that the mean heat
conductivity for this type of glass is > 1
How about if α was changed to 10%? 1%?
FLORAL SCENTS & LEARNING...
Some claim that listening to Mozart improves students'
performance on tests. Perhaps pleasant scents have a
similar effect. To test this idea, 21 subjects worked a
paper-and-pencil maze while wearing a mask. The mask
was either unscented or carried a floral scent.
The response variable is their average time on three trials.
Each subject worked the maze with both masks, in a
random order. The randomization is important because
subjects tend to improve their times as they work a
maze repeatedly. The following gives the subjects'
average times with both masks.
The 21 differences form a single sample. They appear in the
“Difference” columns. The first subject, for example, was 7.37
seconds slower wearing the scented mask, so the difference is
negative. Positive differences show that the subject did better/faster
when wearing the scented mask.
TASK: DETERMINE IF FLORAL SCENT IMPROVES
TEST PERFORMANCE
H0: μ( unscented – scented)= 0
Ha: μ(unscented – scented) > 0
Where μ is the mean difference in the population
from which the subjects were drawn. The null
hypothesis says that no improvement occurs,
and Ha says that unscented times are longer
than scented times on average.
Assume a Normal distribution.
TASK: DETERMINE IF FLORAL SCENT IMPROVES
TEST PERFORMANCE
H0: μ( unscented – scented)= 0
Ha: μ(unscented – scented) > 0
We could perform a paired t test or a t test (why?)
SRS, Normal, Independence
Input values from the ‘difference’ column into
Minitab & perform the t-test
TASK: DETERMINE IF FLORAL SCENT IMPROVES
TEST PERFORMANCE
H0: μ( unscented – scented)= 0
Ha: μ(unscented – scented) > 0
Interpretation:
Fail to reject Ho (don’t believe Ha). Since P-value
of 0.3652 is larger than any reasonable α (say
α = 0.05), data doesn’t support that floral
scent improves test performance.
(outside rejection zone)
TASK: DETERMINE IF FLORAL SCENT IMPROVES
TEST PERFORMANCE
H0: μ( unscented – scented)= 0
Ha: μ(unscented – scented) > 0
Possible errors in context:
Reject Ho when Ho is really true. We would determine that
floral scent improves test performance when it really
doesn’t. Consequences?
Fail to reject Ho, when Ha is really true. We would determine
that floral scent does not improve test performance when
it really does. Consequences?
TASK: DETERMINE IF FLORAL SCENT IMPROVES
TEST PERFORMANCE
H0: μ( unscented – scented)= 0
Ha: μ(unscented – scented) > 0
What if alpha level was something other than
5%? Would our decision change? Why or why
not?
RIGHT’IES VS. LEFT’IES
The design of controls and instruments affects how easily
people can use them. A student project investigated this
effect by asking a SRS of 25 right-handed students to turn a
knob (with their right hands) that moved an indicator. There
were two identical instruments, one with a right-hand thread
(the knob turns clockwise) and the other with a left-hand
thread (the knob turns counterclockwise).
The following table gives the times in seconds each subject
took to move the indicator a fixed distance. The project
designers hoped to show that right-handed people find righthand threads easier to use. Carry out a significance test at
the 5% significance level to investigate this claim.
CARRY OUT PAIRED T-TEST AT ALPHA = 0.05
μ(right – left)
PAIRED T-TEST; ALPHA 5%; SAMPLE STATISTICS
IN SECONDS...
Right threat sample mean = 104.12
 Right threat sample standard deviation = 15.80

Left threat sample mean = 117.44
 Left threat sample standard deviation = 27.26

ALPHA = 0.05; TIME IN SECONDS
Ho:
Ha:
μ(right – left) = 0
μ(right – left) < 0
We want to assess if right-handed people find
right-hand threads easier (faster) to use.
CONDITIONS...

1-sample t-procedure

SRS – Stated in problem.

Normal/Big Sample – n = 25, so CLT; so
distribution/sample is ≈ Normal

Independence – We must assume the population of
students is at least (10) (25). This is reasonable to
assume. Also one subject does not influence the other.
CALCULATIONS...

Use Minitab to run the test
Right threat sample mean = 104.12
 Right threat sample standard deviation = 15.80

Left threat sample mean = 117.44
 Left threat sample standard deviation = 27.26

INTERPRETATION...
Ho:
μ(right – left) = 0
Ha:
μ(right – left) < 0
Reject Ho. With a p-value of 0.0039 (well below
any reasonable α of, say, 0.05), data does
supports Ha, right-ies find right-handed threads
easier & faster
Possible errors? What if we changed our alpha
level?
NOW WE ARE ON TO TWO-SAMPLE MEAN
PROCEDURES...
CALCIUM & 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 to
determine if calcium in the diet reduces blood pressure.
The subjects in part of the experiment were 21 healthy black men. A
randomly assigned group of 10 of the men received a calcium
supplement for 12 weeks. 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 (because the researchers set it up as
‘before’ minus ‘after’.
Assume the population from which these samples were randomly taken is
Normal.

Here is the data from group 1, n = 10
(calcium group; before - after):

Here is data from group 2, n = 11 (placebo
group; before - after):

Remember, an increase in blood pressure
appears as a negative response.
DOES INCREASING CALCIUM IN THE DIET REDUCE
BLOOD PRESSURE?
An increase in blood pressure appears as a negative
response so ...
+  decrease in blood pressure
-  increase in blood pressure.
Group 1 received calcium; group 2 received placebo.
H o:
H a:
μ1 = μ 2
μ1 > μ 2
OR
OR
μ1 – μ2 = 0
μ1 – μ2 > 0
CONDITIONS...

2-sample t-procedure

SRS – Subjects were randomly assigned (good)

Normality/Big Sample – Generally we must look at n1 & n2
separately; stated both were from a Normal population

Independence – Blood pressure for one subject/group
should not influence the blood pressure for other
subjects/group; also assume each population at least 10
times each sample size
CALCULATIONS...

Use Minitab; 2-sample t-procedure

Don’t ever pool (don’t ever assume equal variances)

t = 1.60
P-Value = 0.0644
df = 15
INTERPRETATION...

Fail to reject Ho. With a P-value of 0.0644, df = 15,
and at any reasonable α level (5% or 1%), there is little
evidence to show that calcium reduces blood
pressure.

WHAT IF.... Our α = 0.10. Would this change our
decision?? Why?

Possible errors? Could we calculate a confidence
interval to confirm our findings?
A SIDE NOTE...ROBUSTNESS OF TDISTRIBUTIONS...



t-distributions/procedures (1-sample, matched
pair, 2-sample have the quality of being considered
“robust”
“Robust”... think of the energizer bunny... it can
take a lot of abuse (i.e., not all conditions have to
be met exactly) and still be accurate, precise,
produce good numbers; just need be careful about
outliers and extreme skewness
2-sample t-distributions are especially robust when
the 2 samples have similar shapes and n for both
samples are equal (n1 = n2)
RED WINE VS. WHITE WINE


Observational studies suggest that moderate use of
alcohol reduces heart attacks, and that red wine may
have special benefits. One reason may be that red
wine contains polyphenols, substances that do good
things to cholesterol in the blood and so may reduce
the risk of heart attacks.
In an experiment, healthy men were assigned at
random to drink half a bottle of either red or white
wine each day for two weeks. The level of polyphenols
in their blood was measured before and after the twoweek period. Here are the percent changes in level for
the subjects in both groups:

Is there good evidence that red wine drinkers’ mean
polyphenol levels were different from white wine drinkers’
mean polyphenol levels? Assume both populations are
approximately Normal.

We want to test:
Ho:
μR = μW
Ha :
μR ≠ μW
or
or
μR – μW = 0
μR – μW ≠ 0
where μR & μW are the mean percent change in polyphenols for
men who drink red and white wine, respectfully.
Ho:
μR = μW
or
μR – μW = 0
Ha:
μR ≠ μW
or
μR – μW ≠ 0
Conditions:
SRS – random assignment stated in the problem
Normality – Both distributions assumed Normal (what if they
hadn’t told us that?)
Independence - Independence – We must assume that EACH
population (red and white wine subjects) are at least (10)
times the sample size; and that one group does not
influence the other.
Ho:
Ha:
μR = μW
μR ≠ μW
or
or
μR – μW = 0
μR – μW ≠ 0
Calculations:
2-sample t-test (do not assume variances =)
t = 3.81
p = 0.0019
df = 14
HO:
HA:
ΜR = ΜW
ΜR ≠ ΜW
OR
OR
ΜR – ΜW = 0
ΜR – ΜW ≠ 0
Interpretation:
Reject the null hypothesis. With t = 3.81, p ≈
0, & df = 14, we have strong evidence at
any reasonable α level that red wine
drinkers’ polyphenol levels are different
from white wine drinkers’ polyphenol levels.
HO:ΜR = ΜW
HA: ΜR ≠ ΜW
OR
OR
ΜR – ΜW = 0
ΜR – ΜW ≠ 0
Let’s look at our 95% (why 95%?) confidence interval
and confirm our hypothesis decision from our
confidence interval. Why can we do this?
We are 95% confident that the interval from 2.32% to
8.21% capture the true mean percentage difference
between all red-wine drinkers and all white wine
drinkers’ levels of polyphenol.
Or, in other words, on average, all red-wine drinkers
have from 2.32% to 8.21% higher polyphenol levels
than all white wine drinkers.
HO:ΜR = ΜW
HA: ΜR ≠ ΜW
OR
OR
ΜR – ΜW = 0
ΜR – ΜW ≠ 0
... and think about ‘what if’ we changed our
alpha level?
... and what types of errors could we make...
even if we did everything perfectly correct?
THE EFFECT OF LOGGING...
How badly does logging damage tropical rain
forests? One study compared 2 forest plots in
Borneo that had never been logged with similar
plots nearby that had been logged 8 years
earlier. Plots can be considered randomly
assigned. Here are the data on the number of
tree species in 12 unlogged plots and 9 logged
plots:
Does logging significantly change the mean
number of species in a plot after 8 years? Give
appropriate statistical evidence to support
your conclusion. Assume both populations are
Normally distributed.
We want to test Ho: μU = μL OR
Ha: μU ≠ μL OR
μU – μL = 0
μU – μL ≠ 0
where μU & μL are the mean number of species in
unlogged and logged plots, respectfully
Ho: μU = μL
H a : μU ≠ μ L
OR
OR
μU – μL = 0
μU – μL ≠ 0
Conditions:
SRS; and both unlogged and logged plots were randomly assigned
Normal/Big Sample: Stated in problem
Independence – we must assume each population (unlogged and
logged) is at least (10) times each sample size; and that one
area does not effect the other.
Ho: μU = μL
Ha: μU ≠ μL
OR
OR
μU – μL = 0
μU – μL ≠ 0
Use Minitab to run your 2-sample t procedure
t = 2.114
P-value = 0.0519
HO: ΜU = ΜL
HA: ΜU ≠ ΜL
OR
OR
ΜU – ΜL = 0
ΜU – ΜL ≠ 0
Interpretation:
Fail to reject null hypothesis. Assuming an α
= 0.05, and t = 2.11, and p = 0.0519 (df =
14), there is evidence to determine that
logging does not significantly changed the
number of tree species in a plot of land.
What do you think about this decision?
HO: ΜU = ΜL
HA: ΜU ≠ ΜL
OR
OR
ΜU – ΜL = 0
ΜU – ΜL ≠ 0
Let’s look at our 95% confidence interval from
this data. What additional information does
the confidence interval give us?
We are 95% confident that the interval from 0.036 to 7.703 captures the true population
difference of all tree species between all
unlogged and logged plots of land in Borneo
after 8 years.
EACH DAY I AM GETTING BETTER IN
MATH....
A “subliminal” message is below our threshold of awareness
but may nonetheless influence us. Can subliminal
messages help students learn math? A group of students
who had failed the mathematics part of the City University
of New York Skills Assessment Test agreed to participate in
a study to find out.
All received a daily subliminal message, flashed on a screen
too rapidly to be consciously read. The treatment group of
10 students (assigned at random) was exposed to “Each
day I am getting better in math.” The control group of 8
students was exposed to a neutral message, “People are
walking on the street.” All students participated in a
summer program designed to raise their math skills, and
all took the assessment test again at the end of the
program.
EACH DAY I’M GETTING BETTER IN
MATH...
Is there good evidence at the 5% level that the
treatment brought about a greater
improvement in math scores than the neutral
message? Assume all conditions have been
checked and met.
EACH DAY I’M GETTING BETTER IN
MATH... ENTER STATISTICS INTO MINITAB
sample mean improvement/change for treatment
group = 11.4
sample standard deviation treatment group = 3.17
sample size n = 10 for treatment group
sample mean improvement/change for control
group =8.25
sample standard deviation for control group = 3.69
sample size n = 8 for control group
HO: ΜPOST - PRE FOR TREATMENT GROUP = Μ POST – PRE FOR CONTROL GROUP
HA:ΜPOST - PRE FOR TREATMENT GROUP > Μ POST – PRE FOR CONTROL GROUP
t = 1.91
p-value = 0.038
df = 14
Reject null hypothesis. At the 5% level, and a Pvalue of 0.038, there is good evidence that the
positive subliminal message brought about
greater improvement in math scores than the
control.
Errors? Different alpha? Confidence interval?
IS OUR MEAN RESTING HEART RATE DIFFERENT
FROM THE NATIONAL MEAN RESTING HEART
RATE?

Class activity: Let’s gather our data and
calculate a 95% confidence interval based on
our class data. You may assume that the
population is Normally distributed; and that
conditions have been met and checked.
IS OUR BODY TEMPERATURE DIFFERENT FROM
THE MEAN HEALTHY BODY TEMPERATURE FOR
ADULTS IN US (WHICH IS 98.6 DEGREES F)?

Do you think our temperatures are higher or
lower than the mean healthy body temperature
of adults in US?

Class activity: Let’s gather our data and
perform a hypothesis test.
NEXT UP? EXAM #3...

Homework quiz ....
Will cover chapters 7, 8, & 9
 List of review topics, randomly assigned to
teams of 2 students


Not sure if we will have a 4th exam... Need to
see how are timing is...