Chapter 11.1
Inference for the Mean of
a Population.
Example 1: One concern employers have
about the use of technology is the amount of
time that employees spend each day making
personal use of company technology, such as
phone, e-mail, internet, and games. The
Associated Press reports that, on average,
workers spend 72 minutes a day on such
personal technology uses. A CEO of a large
company wants to know if the employees of
her company are comparable to this survey. In
a random sample of 10 employees, with the
guarantee of anonymity, each reported their
daily personal computer use. The times are
recorded at right.
Employee
Time
1
2
3
4
5
6
7
8
9
10
66
70
75
88
69
71
71
63
89
86
When the
standard
deviation the
of astatistic
statistic is
we use
this estimator,
estimated
from
the
data,
the
result
is
called
that
results
does
not
have
a
normal
Doesisthe
data provide
evidence
that the mean for
What
different
about
this
problem?
the
standard
error
of
the
statistic,
and
is
distribution,
instead
it
has
a
new
this company is greater than 72 minutes?
given
by s/√n.called the t-distribution.
distribution,
Time for some Nspiration!
One-Sample z-statistic
s known:
statistic - parameter
test statistic 
standard deviation of statistic
z=
x  m
s
n
One-sample t-statistic:
s unknown:
statistic - parameter
test statistic 
standard deviation of statistic
t=
x m
s
n
The variability of the t-statistic is controlled by the
Sample Size.
The number of degrees of freeom is equal to
n-1
.
ASSUMING NORMALITY?
Use a Box
and
1. SRS is extremely important.
Whisker to
2. Check for skewness.
check.
3. Check for outliers.
4. If necessary, make a cautionary statement.
5. In Real-Life, statisticians and researchers try
very hard to avoid small samples.
Example 2: The Degree of Reading Power (DRP) is a
test of the reading ability of children. Here are
DRP scores for a random sample of 44 third-grade
students in a suburban district:
40 26 39 14
42 18
25
43 46 27 19
47 19
26
35 34 15
44 40 38 31
46 52 25 35 35 33 29
34 41
49 28 52 47 35
48 22 33 41
51
27 14
54 45
At the a = .1, is there sufficient evidence to
suggest that this district’s third graders reading
ability is different than the national mean of 34?
• I have an SRS of third-graders
SRS?
Normal?
•Since the sample size is large, the sampling distribution
is
How do you
approximately
normally
distributed
Name the Test!!
One Sample t-test for mean
OR
know?
Do you
•Since the histogram is unimodal
withs?no outliers, the
know
What are your
sampling distribution is approximately normally
hypothesis
distributed
• s is unknown
statements? Is
H0: m = 34
a key word?
where m is the true mean there
reading
Ha: m = 34
ability of the district’s third-graders
35.091  34
Plug values
t
 .6467
into formula.
11.189
44
p-value = tcdf(.6467,1E99,43)=.2606(2)=.5212
Use tcdf to
calculate p-value.
a = .1
Compare your p-value to
a & make decision
Since p-value > a, I fail to reject the null
hypothesis.
Conclusion:
There is not sufficient evidence to suggest that the
true mean reading ability of the district’s third-graders
is different than the national mean of 34.
Write conclusion in
context in terms of Ha.
Back to Example 1. The times are recorded below.
Employee 1
2 3 4 5 6 7 8 9 10
Time
66 70 75 88 69 71 71 63 89 86
Does this data provide evidence that the mean for this
company is greater than 72 minutes?
• I have an SRS of employees
SRS?
•Since the histogram has no outliers and is roughly
Normal?
symmetric, the sampling distribution is approximately
How do you
normally distributed
Do you
know s?
know?
What are your
hypothesis
• s is unknown, therefore we are using a 1 sample t-test
statements? Is
H0: m = 72 where m is the truethere
# ofamin
keyspent
word?on PT
Ha: m = 72 time spent by this company’s employees
Use tcdf to
calculate p-value.
74.8  72
Plug values
t
 .937
9.45
into formula.
10
p-value = tcdf(.937,1E99,9)=.1866(2)=.3732
Compare your p-value to
a & make decision
Since p-value > 15%, I fail to reject the null
hypothesis that this company’s employees spend
72 minutes on average on Personal Technology
uses.
Conclusion:
There is not sufficient evidence to suggest that the
true amount of time spent on personal technology use by
employees of this company is more than the national
mean of 72 min.
Write conclusion in
context in terms of Ha.
Now for the fun calculator stuff!
Example 3: The Wall Street Journal
(January 27, 1994) reported that based
on sales in a chain of Midwestern grocery
stores, President’s Choice Chocolate Chip
Cookies were selling at a mean rate of
$1323 per week. Suppose a random sample
of 30 weeks in 1995 in the same stores
showed that the cookies were selling at
the average rate of $1208 with standard
deviation of $275. Does this indicate that
the sales of the cookies is different from
the earlier figure?
Assume:
•Have an SRS of weeks
Name the Test!!
•Distribution of sales
is approximately
due to
One
Sample t-testnormal
for mean
large sample size
• s unknown
H0: m = 1323
Ha: m ≠ 1323
where m is the true mean cookie sales
per week
1208  1323
t 
 2.29
275
30
p  value  .0295
Since p-value < a of 0.05, I reject the null hypothesis.
There is sufficient to suggest that the sales of cookies
are different from the earlier figure.
Example 3: President’s Choice Chocolate Chip
Cookies were selling at a mean rate of $1323
per week. Suppose a random sample of 30
weeks in 1995 in the same stores showed
that the cookies were selling at the average
rate of $1208 with standard deviation of
$275. Compute a 95% confidence interval for
the mean weekly sales rate.
CI = ($1105.30, $1310.70)
Based on this interval, is the mean weekly
sales rate statistically different from the
reported $1323?
What do you notice about the decision from the
Remember
p-value
= .01475
confidence
intervalyour,
& the
hypothesis
test?
a = .02,
we would
reject H03. if a =
What decision At
would
you make
on Example
.01?
CI H
= ($1100,
$1316).
You would fail A
to96%
reject
0 since the p-value > a.
What confidence level would be correct to use?
Since $1323 is not in the interval, we would
reject H0.
Does that confidence interval provide
the same
In
a one-sided
test,use
all of
a (2%)
goes into level
that tail
(lower
decision?
You
should
a
99%
confidence
for
aThe
98%
CI
=
($1084.40,
$1331.60)
tail).
two-sided
test at a = we
.01. would fail
Since
$1323 hypothesis
is in the interval,
InIfa H
CI,
tails have
equal
area –would the
: the
m <probabilities
1323,
what
decision
aTail
toatbetween
reject
H 0.
test
give
ain=the
.02?
sohypothesis
there should
also
be 2%
significant
level (a) and
CIthe
= ($1068.6
, $1346.40)
- Since a$1323
in this.02
upper
tail
= .02 is.96
the confidence
leveltoMUST
Why
are
we
getting
different
answers?
interval
we
would
fail
reject
H0.
Now,
what96%
confidence
level
isthat
appropriate
for this
That
leaves
in
the
middle
&
match!)
alternative
hypothesis?
should be your confidence level
Ex4: The times of first sprinkler activation (seconds) for a
series of fire-prevention sprinklers were as follows:
27
41
22
27
23
35
30
33
24
27
28
22
24
Construct a 95% confidence interval for the mean activation
time for the sprinklers.
Matched Pairs
Test
A special type of
t-inference
Matched Pairs – two forms
• Pair individuals by
certain
characteristics
• Randomly select
treatment for
individual A
• Individual B is
assigned to other
treatment
• Assignment of B is
dependent on
assignment of A
• Individual persons
or items receive
both treatments
• Order of
treatments are
randomly assigned
or before & after
measurements are
taken
• The two measures
are dependent on
the individual
Is this an example of matched pairs?
1)A college wants to see if there’s a
difference in time it took last year’s
class to find a job after graduation and
the time it took the class from five years ago
to find work after graduation. Researchers
take a random sample from both classes and
measure the number of days between
graduation and first day of employment
No, there is no pairing of individuals, you
have two independent samples
Is this an example of matched pairs?
2) In a taste test, a researcher asks people
in a random sample to taste a certain brand
of spring water and rate it. Another
random sample of people is asked to
taste a different brand of water and rate it.
The researcher wants to compare these
samples
No, there is no pairing of individuals, you
have two independent samples – If you would
have the same people taste both brands in
random order, then it would be an example
of matched pairs.
Is this an example of matched pairs?
3) A pharmaceutical company wants to test
its new weight-loss drug. Before giving the
drug to a random sample, company
researchers take a weight measurement
on each person. After a month of using
the drug, each person’s weight is
measured again.
Yes, you have two measurements that are
dependent on each individual.
A whale-watching company noticed that many
customers wanted to know whether it was
better to book an excursion in the morning or
the afternoon.
To test
this question, the
You may subtract
either
company
thewhen
following data on 15
way – collected
just be careful
writing Hadays over the past
randomly selected
month. (Note: days were not
consecutive.)
Day
1
2
Morning
8 9
3
4
5
6
7
8
9
10
11 12 13 14 15
7 9 10 13 10
8
2
5
7 7 6 8 7
After8 10 9 8 9 11 8
noon
Since you have two values for
10
4 7 8 9 6 6 9
First, you must find
the differences for
each day.
each day, they are dependent
on the day – making this data
matched pairs
Day
1
2
3
Morning
8
9
7 9 10 13 10
Afternoon
8 10
4
5
9 8 9
6
7
8
9
10
11 12 13 14 15
8
2
5
7 7 6 8 7
11
8 10 4 7 8 9 6 6 9
I subtracted:
Differenc
0 -1 -2 1 1 Morning
2 2 – -2
-2 -2 -1 -2 0 2 -2
afternoon
es
You could subtract the other
way!
• Have an SRS of days for whale-watching
You need to state assumptions using the
• s unknown
differences!
Assumptions:
•Since the boxplot doesn’t show any outliers, we can
assume the distribution is approximately normal.
Notice the skewness of the
boxplot, however, with no outliers,
we can still assume normality!
Differences
0
-1
-2
1
1
2
2
-2
-2
-2
-1 -2
0
2
Is there sufficient evidence that more whales are
sighted in the afternoon?
H0: mD = 0
Ha: mD < 0
Be careful writing your Ha!
Think about
how you–
If you subtract
afternoon
subtracted: M-A
Hdifferences
mD>0should
Notice morning;
we
mthen
a:more
D foris
Ifused
afternoon
& it equals
since the nullbeshould
the0 differences
+ or -?
be that there
NOat
difference.
Don’t islook
numbers!!!!
Where mD is the true mean
difference in whale sightings
from morning minus afternoon
-2
Differences
0
-1
-2
1
1
2
2
-2
finishing the hypothesis test:
x m
.4  0
t 

 .945
s
1.639
n
15
p  .1803
df  14
a  .05
-2
-2
-1 -2
0
2
In your calculator,
perform
t-test
Notice athat
if
the
youusing
subtracted
differences
(L3)
A-M, then your
test statistic
t = + .945, but pvalue would be
the same
Since p-value > a, I fail to reject H0. There is
insufficient evidence to suggest that more whales are
sighted in the afternoon than in the morning.
-2
Ex: The effect of exercise on the amount of lactic acid
in the blood was examined in journal Research
Quarterly for Exercise and Sport. Eight males were
selected at random from those attending a week-long
training camp. Blood lactate levels were measured
before and after playing 3 games of racquetball, as
shown in the table.
What is the parameter of interest in
this problem?
Construct a 95% confidence
interval for the mean change in
blood lactate level.
Player
1
2
3
4
5
6
7
8
Before
13
20
17
13
13
16
15
16
After
18
37
40
35
30
20
33
19
Based on the data, would you conclude that there is a
significant difference, at the 5% level, that the mean
difference in blood lactate level was over 10 points?
Player
1
2
3
4
5
6
7
8
Before
13
20
17
13
13
16
15
16
After
18
37
40
35
30
20
33
19