AP Statistics
Tuesday, 09 February 2016
• OBJECTIVE TSW explore Hypothesis Testing.
• Student to Ms. Havens: “Is either yesterday’s test or
the previous test graded yet?”
• Ms. Havens to student: “No.”
• Student to Ms. Havens: “Do you know when they will
be graded?”
• Ms. Havens to student: “No.”
• Ms. Havens is sorry to not be able to give student a
good answer, but she will try to have them graded by
the end of the week.
Hypothesis Tests
One-Sample Means
can I tell if they
really
Example:How
A government
agency
are
underweight?
has received numerous
complaints that a particular
Hypothesis
restaurant
has
been
selling
testing will help
underweight
me 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
occurs by random chance or not.
If it is statistically significant, is it
...
– a random occurrence due to
variation?
– 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
or Ha: parameter < hypothesized value
or 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
You MUST indicate
what μ represents!
Where m is the true
mean weight of
hamburger patties
Example 3: A car dealer advertises
that his 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
AP Statistics
Wednesday, 10 February 2016
• OBJECTIVE TSW explore Hypothesis Testing.
• TEST: Sampling Distributions is graded.
– You will get it back after lunch.
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,
are not
Must use indicate
parameter
(population)
Must bewhich
NOT
equal!
x is&aexplain
statisticwhy:
(sample)
legitimate
a) H0 : m  15 ; Ha : m  15

is
the
population
b) H0 : x  123; Ha : x  123
proportion!
Must use same
c) rHis0 parameter
:number
  0.1;
H0a! population
:   0.1
as for
H
correlation coefficient – but H0
d) H0 : mMUST
 0.4;beH“=“
a :!m  0.6
e) H0 : r  0 ; Ha : r  0
P-values • 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 • 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 a 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 0.05, would you
reject or fail to reject H0 for
the given p-values?
a)
b)
c)
d)
0.03
0.15
0.45
0.023
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 = 0.06305
2) left-tail test
z = -2.4; n = 15
p = 0.008198
3) two-tail test
p = 0.03045
t = 2.3; n = 25
Assignment
• WS Hypothesis Testing #1
– Due on Tuesday, 16 February 2016.
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
H0: m = 15
concentration
of lead is 15 ppb (parts
Ha: m
> 15 or greater. Suppose a
per
billion)
t=2.1
Where
m
is
the
true
mean
concentration
community randomly
selects
of 25
of leadsamples
in drinking
water
water
and
computes
Since the p-value < a,
I reject Ha0. t-test
There is
statistic
2.1. Assume
that
lead
sufficient
evidence
to suggest
that
the
P-value =of
tcdf(2.1,10^99,24)
concentrations
are of
normally
mean
concentration
lead in drinking
=0.0232
water is greater
thanthe
15 ppb.
distributed.
Write
hypotheses,
calculate the p-value & write the
appropriate conclusion for a = 0.05.
Example 6: A certain type of frozen
dinners states that the dinner
H0: m240
= 240calories.
calories A random
contains
Ha: of
m > 12
240of
calories
sample
these frozen dinners
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.
=0.0420
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.
ASSUMPTIONS
•
•
•
•
SRS (given)
If a is not given,
Normal distribution (given)
include it!
s unknown
Ho: m = 240 calories
Ha: m > 240 calories, where m is the true mean
caloric content of frozen
dinners
p-value = tcdf(1.9, ∞, 11) = 0.04197 < a = 0.05
Since p ≤ a, we reject H0. There is evidence to
suggest that the true mean caloric content of frozen
dinners is greater than 240 calories.
Formulas:
s known:
statistic - parameter
test statistic 
standard deviation of statistic
z=
x m
σ
n
Formulas:
s unknown:
statistic - parameter
test statistic 
standard deviation of statistic
t=
x m
s
n
AP Statistics
Friday, 12 February 2016
• OBJECTIVE TSW explore the aspects of hypothesis
testing.
• ASSIGNMENTS DUE TUESDAY
–
–
–
–
WS
WS
WS
WS
Hypothesis Testing #1
Hypothesis Testing #2
Hypothesis Testing #3
Matched Pairs  due 02/19/16
• LOOKING AHEAD
– Tuesday, 02/16/2016: QUIZ: Hypothesis Testing
– Wednesday, 02/17/2016: ASSESSMENT: Hypothesis
Testing
REVIEW: Hypothesis Testing
– Friday, 02/19/2016: TEST: Hypothesis Testing
Hypothesis Testing WS #1
1a) No, H0 must be =
c)
b) No, must use parameter μ
2a)
b) H0: μ = 30 ppm Where μ is the true mean nitrate concentration in the
Ha: μ > 30 ppm water
c)
d) H0: μ = $42,500 Where μ is the true mean household income of mall
Ha: μ > $42,500 shoppers
e)
3a) Yes, the normal probability (quantile) plot is approximately
linear so the distribution is approximately normal.
b)
Hypothesis Testing WS #2
1)
3) fail to reject Ho
5)
2) reject Ho
4)
6) p-value = .0256
7) P-value = .092896
8)
9) p – value = tcdf(2.056, 1E99, 14)(2) = .02946(2) =
.05892
Since the p-value > α, I fail to reject the null hypothesis.
There is not sufficient evidence to suggest that the true
mean diameter of the catheters is not equal to 2.00 mm.
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? (Full write-up.)
Assumptions:
SRS?
Normal?
•I have an SRS of milk from one producer How do you
•The freezing temperature of milk is a normal know?
distribution. (given)
• s is known
Do you
What are your
H0: μ = -0.545
know s?
hypothesis
Ha: μ > -0.545
statements? Is
there a keyofword?
where μ is the true mean freezing temperature
milk
 .538   .545 
z
 1.9566
.008
5
Plug values
into formula.
p-value = normalcdf(1.9566, 1E99) = 0.0252
Use normalcdf to
calculate p-value.
α = .05
Compare your p-value
to α & make decision
Since p-value < α, 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 = 0.1, is there sufficient evidence to
suggest that this district’s third graders
reading ability is different than the national
mean of 34? (Full write-up.)
• I have an SRS of third-graders
SRS?
Normal?
•Since the sample size is large, the sampling How do you
know?
distribution is approximately normally distributed
Do you
• s is unknown
know s?
What are your
H0: μ = 34
Ha: μ ≠ 34
hypothesis
where μ is the true mean reading
statements? Is
there a key word?
ability of the district’s third-graders
35.091  34
t
 .6467
11.189
44
Plug values
into formula.
p-value = 2*tcdf(0.6467, 1E99, 43) = 2*0.2606 = 0.5212
Use tcdf to
calculate p-value.
α = 0.1
Compare your p-value
to α & make decision
Since p-value > α, I fail to reject the null hypothesis.
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.
Conclusion:
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? (Include assumptions.)
Assumptions
•Have an SRS of weeks
•Distribution of sales is approximately normal due to
large sample size
• s unknown
H0: μ = 1323
where μ is the true mean cookie
sales per week
Ha: μ ≠ 1323
1208  1323
t
 2.29 p-value  0.0295
275
30
Since p-value < α of 0.05, I reject the null
hypothesis. There is sufficient evidence to
suggest that the sales of cookies are different
from the earlier figure.
Example 9 (Continued): 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.
(Just compute the interval.)
CI = ($1105.30, $1310.70)
Based on this interval, is the mean weekly sales rate
statistically different from the reported $1323?
The sales rate is statistically different, since the
reported mean of $1323 is not in the interval.
Assignment
• WS Hypothesis Testing #3
– Due on Tuesday, 16 February 2016.
• WS Matched Pairs
– Due on Friday, 19 February 2016.
Matched Pairs Test
A special type of tinference
Matched Pairs – Two Forms
•
•
•
•
Form #1
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
Form #2
• Individual persons or
items receive both
treatments
• Order of treatments
is 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
collected
the following data on 15
way – just
be careful when
Ha
randomly writing
selected
days over the past month.
(Note: days were not consecutive.)
Day
1 2
3 4
Morning
8 9
7 9 10 13 10 8
Afternoon
8 10 9 8 9
5
6
7
11 8
Since you have two values for each
day, they are dependent on the day
– making this data matched pairs
8
9
10
11 12 13 14 15
2 5
7 7 6 8 7
10 4 7
8 9 6 6 9
First, you must find the
differences for each
day.
Day
1 2
3
Morning
8 9
7 9 10 13 10 8
Afternoon
8 10 9 8 9
Differe
nces
0 -1
4 5
1 1
2
6
7
8
9
10
11 12 13 14 15
2 5
7 7 6 8 7
11 8 10 4 7 8 9 6 6 9
I subtracted:
- Morning
–
afternoon
2 2 -2
-2
0 2
2
1 2
2
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 differences is approximately
Notice that, even with the
normal.
granularity in this plot, it still
displays a nice linear
relationship!
Differenc
es
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: μD = 0
Ha: μD < 0
Be careful writing your Ha!
Think about
how you–
If you subtract
afternoon
subtracted: M-A
Noticemorning;
we used μDthen
for differences
Ha: μD>0
If
afternoon
is more should
& it equals 0 since the null should be
the is
differences
be + or -?
that there
NO difference.
Don’t look at numbers!!!!
Where μD is the true mean
difference in whale sightings
from morning minus afternoon
-2
Differenc
es
0 -1
-2
1
1
2
2
-2
-2 -2
Finishing the hypothesis test:
x m
0.4  0
t

 0.945
s
1.639
n
15
p  0.1803
df  14
a  0.05
-1
-2 0
2
In your calculator,
perform a t-test
Notice
thatthe
if you
using
subtracted
differencesA-M,
(L3)
then your test
statistic
t = + 0.945, 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
Assignment
• WS Matched Pairs
– Due on Friday, 19 February 2016 (TEST day).