Hypothesis Tests
One Sample Means
I tell if they
really
ExampleHow
2: Acan
government
agency
are
underweight?
has received numerous
complaints that a particular
A hypothesis
restaurant
has
been
selling
test will help me
underweight
decide!
Takehamburgers.
a sample & find x.The
restaurant advertises that it’s
pattiesButare
how“adoquarter
I know ifpound”
this x is(4
one
that I expect to happen or is it one
ounces).
that is unlikely to happen?
What are hypothesis tests?
Calculations that tell us if a value, x,
Is it one of the
occurs by random
chance
or
not
–
if
sample means that
it is statistically
aresignificant
likely to occur?
Is it . . .
Statistically significant means that it
Is it one occurrence
that
–isaNOT
random
due
to
a random chance occurrence!
isn’t likely
to
natural
variation?
occur?
– a biased occurrence due to some
other reason?
Nature of hypothesis tests How does a murder trial work?
• First begin by supposing the
“effect”
NOTthat
present
First - is
assume
the
innocent
• Next,person
see ifisdata
provides
Then – must
have sufficient
evidence
against
the
evidence to prove guilty
supposition
Hmmmmm …
Example:
murder
Hypothesis tests
use
the same process!
trial
Steps:
Notice the steps are the
same except we add
hypothesis statements –
which you will learn today
1) Assumptions
2) Hypothesis statements &
define parameters
3) Calculations
4) Conclusion, in context
Assumptions for z-test
(t-test):
YEA –
•
•
These are the same
Have an SRS
of context
assumptions
as confidence
intervals!!
Distribution is (approximately)
normal
– Given
– Large sample size
– Graph data
• s is known (unknown)
Example 1: Bottles of a popular cola
are
supposed
to bottles
contain 300 mL of
•Have
an SRS of
cola.
There
is some variation
from
•Sampling
distribution
is approximately
normalto
because
boxplot
is
bottle
bottle.theAn
inspector,
who
symmetrical
suspects
that the bottler is under• s is unknown
filling, measures the contents of six
randomly selected bottles. Are the
assumptions met?
299.4 297.7 298.9 300.2 297 301
Writing Hypothesis statements:
• Null hypothesis – is the statement
being tested; this is a statement of
“no effect” or “no difference”
H0:
• Alternative hypothesis – is the
statement that we suspect is true
Ha:
The form:
Null hypothesis
H0: parameter = hypothesized value
Alternative hypothesis
Ha: parameter > hypothesized value
Ha: parameter < hypothesized value
Ha: parameter = hypothesized value
Example 2: A government agency has
received numerous complaints that a
particular restaurant has been selling
underweight hamburgers. The
restaurant advertises that it’s patties
are “a quarter pound” (4 ounces).
State the hypotheses :
H0: m = 4
Ha: m < 4
Where m is the true
mean weight of
hamburger patties
Example 3: A car dealer advertises
that is new subcompact models get
47 mpg. You suspect the mileage
might be overrated.
State the hypotheses :
H0: m = 47
Ha: m < 47
Where m is the
true mean mpg
Example 4: Many older homes have electrical
systems that use fuses rather than circuit
breakers. A manufacturer of 40-A fuses
wants to make sure that the mean amperage at
which its fuses burn out is in fact 40. If the
mean amperage is lower than 40, customers
will complain because the fuses require
replacement too often. If the amperage is
higher than 40, the manufacturer might be
liable for damage to an electrical system due
to fuse malfunction. State the hypotheses :
H0: m = 40
Ha: m = 40
Where m is the true
mean amperage of
the fuses
Facts to remember about hypotheses:
• ALWAYS refer to populations
(parameters)
• The null hypothesis for the
“difference” between populations is
usually equal to zero
H0: mx-y= 0
• The null hypothesis for the correlation
(rho) of two events is usually equal to
zero.
H0: r= 0
Activity: For each pair of
hypotheses,
indicate
are not
Must use parameter
(population)
x
Must bewhich
NOT equal!
is &
a statistics
(sample)
legitimate
explain why
a) H 0 : m  15 ; H a : m  15

is
the
population
b) H 0 : x  123 ; H a : x  123
proportion!
Must use same
c)r H
:   .1as; for
H
:   .1
is 0parameter
number
H0a! population
correlation coefficient – but H0
d) H 0 : mMUST
 .4;beH“=“
a : !m  .6
e) H 0 : r  0 ; H a : r  0
P-values • Assuming H0 is true, the
probability that the test
statistic would have a value as
extreme or more than what
is actually observed
In other words . . . is it far
out in the tails of the
distribution?
Level of Significance
Activity
Level of significance • Is the amount of evidence
necessary before we begin to doubt
that the null hypothesis is true
• Is the probability that we will
reject the null hypothesis, assuming
that it is true
• Denoted by a
– Can be any value
– Usual values: 0.1, 0.05, 0.01
– Most common is 0.05
Statistically significant –
• The p-value is as small or smaller
than the level of significance (a)
• If p > a, “fail to reject” the null
hypothesis at the a level.
• If p < a, “reject” the null
hypothesis at the a level.
Facts about p-values:
• ALWAYS make decision about the null
hypothesis!
• Large p-values show support for the
null hypothesis, but never that it is
true!
• Small p-values show support that the
null is not true.
• Double the p-value for two-tail (=)
tests
• Never accept the null hypothesis!
Never “accept” the null hypothesis!
Never “accept” the null
hypothesis!
Never “accept” the
null hypothesis!
At an a level of .05, would you
reject or fail to reject H0 for
the given p-values?
a) .03
b) .15
c) .45
d) .023
Reject
Fail to reject
Fail to reject
Reject
Calculating p-values
• For z-test statistic –
– Use normalcdf(lb,ub)
– [using standard normal curve]
• For t-test statistic –
– Use tcdf(lb, ub, df)
Draw & shade a curve &
calculate the p-value:
1) right-tail test
t = 1.6; n = 20
P-value = .0630
2) left-tail test
z = -2.4; n = 15
P-value = .0082
3) two-tail test
t = 2.3; n = 25
P-value = (.0152)2 = .0304
Writing Conclusions:
1) A statement of the decision
being made (reject or fail to
reject H0) & why (linkage)
AND
2) A statement of the results in
context. (state in terms of Ha)
“Since the p-value < (>) a,
I reject (fail to reject)
the H0. There is (is not)
sufficient evidence to
suggest that Ha.”
Be sure to write Ha in
context (words)!
Example 5: Drinking water is considered
unsafe if the mean concentration of lead is
H0: m = 15
greater
than 15 ppb (parts per billion).
Ha: m > 15a community randomly selects of
Suppose
t=2.1
Where
m
is
the
true
concentration
25 water samples andmean
computes
a t-test
of lead inofdrinking
water that lead
statistic
2.1.
Assume
Since the p-value < a, I reject H0. There is
concentrations
are normally
distributed.
sufficient
evidence
to suggest
that the
P-value = tcdf(2.1,10^99,24)
Write
the
hypotheses,
the pmean
concentration
ofcalculate
lead in drinking
=.0232
value
& write
the appropriate
water
is greater
than 15 ppb.conclusion
for a = 0.05.
Example 6: A certain type of frozen
dinners states that the dinner
H0: m240
= 240calories. A random
contains
Ha: of
m > 12
240of these frozen dinners
sample
t=1.9
Where
m
is
the
mean caloric
was selected fromtrue
production
to see
Since
the p-value
<frozen
a, I reject
H0. There is
content
of
the
dinners
if the caloric content was greater
sufficient evidence to suggest that the
than
stated
on
the
box.
The
t-test
P-value
=
tcdf(1.9,10^99,11)
true mean caloric content of these frozen
statistic
was
calculated
to
be
1.9.
=.0420
dinners is greater than 240 calories.
Assume calories vary normally. Write
the hypotheses, calculate the p-value
& write the appropriate conclusion
for a = 0.05.
Formulas:
s known:
statistic - parameter
test statistic 
standard deviation of statistic
z=
x  m
s
n
Formulas:
s unknown:
statistic - parameter
test statistic 
standard deviation of statistic
t=
x m
s
n
Example 7: The Fritzi Cheese Company buys milk from
several suppliers as the essential raw material for its
cheese. Fritzi suspects that some producers are
adding water to their milk to increase their profits.
Excess water can be detected by determining the
freezing point of milk. The freezing temperature of
natural milk varies normally, with a mean of -0.545
degrees and a standard deviation of 0.008. Added
water raises the freezing temperature toward 0
degrees, the freezing point of water (in Celsius). The
laboratory manager measures the freezing
temperature of five randomly selected lots of milk
from one producer with a mean of -0.538 degrees. Is
there sufficient evidence to suggest that this
producer is adding water to his milk?
Assumptions:
SRS?
•I have an SRS of milk from one producer
•The freezing temperature of milk is a normal
distribution. (given)
• s is known
Do you
Normal?
How do you
know?
What are your
hypothesis
Ha: m > -0.545
statements? Is
thereofamilk
key
where m is the true mean freezing temperature
word?
H0: m = -0.545
know s?
 .538   .545 
z 
 1.9566
.008
5
Plug values
into
formula.
p-value = normalcdf(1.9566,1E99)=.0252
Use normalcdf to
calculate p-value.
a = .05
Compare your p-value
to a & make decision
Since p-value < a, I reject the null hypothesis.
There is sufficient evidence to suggest that the true
mean freezing temperature is greater than -0.545. This
suggests that the producer is adding water to the milk.
Conclusion:
Write conclusion in
context in terms of Ha.
Example 8: 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:
(data on note page)
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
know?
OR
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.
Example 9: 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
•Distribution of sales is approximately normal due to
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 9: 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,
would
H0.10 if a =
What decisionAtwould
youwe
make
on reject
Example
.01?
CI H
= ($1100,
$1316).
You would fail A
to96%
reject
0 since the p-value > a.
What Since
confidence
would
beinterval,
correctwe
towould
use?
$1323level
is not
in the
reject H0.
that confidence
interval
provide
thetail
same
InDoes
a one-sided
test,
all
of
a
2%
goes
into
that
(lower
You
should
use
a
99%
confidence
level
for
adecision?
The
98%
CI
=
($1084.40,
$1331.60)
tail).
two-sided
test at a = we
.01. would fail
Since
$1323 hypothesis
is in the interval,
In If
a CI,
the
tails
have
equal
area – would the
HTail
:
m
<
1323,
what
decision
a
probabilities
to between
reject
H 0.
so hypothesis
there shouldtest
alsogive
be 2%
in
the
at(a)
a =and
.02?
the
significant
level
CItail
= ($1068.6 , $1346.40) - Since a$1323
in this.02
upper
= .02 is.96
the confidence
leveltoMUST
Why
are
we
getting
different
answers?
interval
we
would
fail
reject
H0.
That
leaves
96%
in
the
middle
&
that
Now,
what
confidence
level
is
appropriate
for this
match!)
alternative
should
be your hypothesis?
confidence level
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 bean 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 normal probability plot is approximately
linear, the distribution of difference is approximately
Notice the granularity in this
normal.
plot, it is still displays a nice
linear relationship!
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