Transcript Document

7
Chapter
Hypothesis Testing
Elementary Statistics
Larson
Farber
A Statistical Hypothesis
A claim about a population.
Null hypothesis H0
Statistical hypothesis that
contains a statement of
equality, such as , = or .
Alternative hypothesis Ha
contains a statement of
inequality, such as <,  or >.
Complementary Statements
If I am false,
If I am false,
you are true
you are true
H0
Ha
Writing Hypotheses
Write the claim about the population. Then, write its
complement. Either hypothesis, the null or the alternative, can
represent the claim.
A hospital claims its ambulance response time is less
than 10 minutes.
H0 :   10 min
Ha :   10 min
claim
 A consumer magazine claims the proportion of cell phone
calls made during evenings and weekends is at most 60%.
H0 : p  0.60 claim
Ha :
p  0.60
Decision
Errors and Level of Significance
Actual Truth of H0
Do not
reject H0
Reject H0
H0 True
Correct
Decision
Type I
Error
H0 False
Type II
Error
Correct
Decision
A type I error: Null hypothesis is actually true but the
decision is to reject it.
Level of significance, a
Maximum probability of committing a type I error.
Rejection Regions
Sampling distribution for x
Rejection Region
z
0
z0
Critical Value z0
The rejection region is the range of values for which the null
hypothesis is not probable. It is always in the direction of the
alternative hypothesis. Its area is equal to a.
A critical value separates the rejection region from the
non-rejection region
Types of Hypothesis Tests
Right-tail test Ha: >value
Rejection
region
0
z0
Reject H0 if z > z0
otherwise fail to reject H0.
Left-tail test Ha: <value
Rejection
region
z0
0
Rejection
region
Rejection
region
-z0
Reject H0 if z < z0
otherwise fail to reject H0.
0
z0
Two-tail test Ha: value
Reject H0 if z >z0 or z <-z0
otherwise fail to reject H0.
Interpreting the Decision
Claim is H0
There is enough
evidence to reject
the claim
There is not enough
evidence to reject
the claim
Interpreting the Decision
Claim is Ha
There is enough
evidence to support
the claim
There is not enough
evidence to support
the claim
8 Steps in a Hypothesis Test
1. Write the null and alternative hypothesis
Write H0 and Ha as mathematical statements. Remember H0
always contains the = symbol.
2. State the level of significance
This is the maximum probability of rejecting the null hypothesis
when it is actually true. (Making a type I error.)
3. Identify the sampling distribution
The sampling distribution is the distribution for the test statistic
assuming that H0 is true and that the experiment is repeated an
infinite number of times.
4. Find the critical
5. Find the
value
rejection region
The critical value separates the rejection region of the sampling
distribution from the non-rejection region.
6. Find the test statistic
Perform the calculations to standardize your sample statistic.
7. Make your decision
If the test statistic falls in the critical region, reject H0.
Otherwise, fail to reject H0.
8. Interpret your decision
Critical Values
The critical value z0 separates the rejection region from the
non-rejection region. The area of the rejection region is a.
Rejection
region
Rejection
region
z0
0
Find z0 for a left-tail
test with a =.01
z0=-2.33
0 z0
Find z0 for a right-tail
test with a =.05
z0=1.645
Rejection
region
Rejection
region
-z0 0
z0
Find - z0 and z0 for a two-tail test with a =.01
-z0=-2.575
and z0 =2.575
The z-test for a Mean
 A cereal company claims the mean sodium content in
one serving of its cereal is no more than 230 mg. You work
for a national health service and are asked to test this claim.
You find that a random sample of 52 servings has a mean
sodium content of 232 milligrams and a standard deviation
of 10 mg. At a= 0.05, do you have enough evidence to reject
the company’s claim?
1. Write the null and alternative hypothesis
H0:   230 mg.(claim)
Ha:  > 230 mg.
2. State the level of significance
a= 0.05
3. Determine the sampling distribution
Since the sample size is at least 30, the sampling distribution is normal.
Since Ha contains the > symbol, this is a right tail test.
4. Find the critical value
Rejection
region
0 z0
1.645
5. Find the rejection region
6. Find the test statistic
n=52 x = 232
s=10
232  230
2
z

 1.44
10
1.387
52
7. Make your decision
z = 1.44 does not fall in the rejection region, so fail to reject H0
8. Interpret your decision
There is not enough evidence to reject the company’s claim that there is
at most 230mg of sodium in one serving of its cereal.
P-Values
The P-value is the probability of obtaining a sample
statistic with a value as extreme or more extreme than the
one determined by the sample data. Reject H0 if P < a.
P-value = indicated area
Area in
left tail
z
0
For a left tail test
If z is
negative,
twice the
area in the
left tail
Area in right tail
0 z
For a right tail test
If z is positive
twice the area
in the right tail
z
0
z
For a two-tail test
Using P-Values
The standardized z-score for the hypothesis
The
standardized z-score for the hypothesis test on the sodium
test on the sodium content of cereal was z =
content
cerealthe
was P-value
z = 1.44. Find
P-value
and make
your
1.44.ofFind
andthe
make
your
decision
at theat0.05
level
decision
the 0.05
levelof
ofsignificance.
significance.
Since this was a right-tail test, the P-value is the area to the right of the z
score. The area to the right of z = 1.44 is 0.0749 so the P-value is .0749.
Since 0.0749 > 0.05, fail to reject H0.
If
0.0246,
what
is your
decision
if
If
P P= =
0.0246,
what
is your
decision
if
1) a=0.05
1) a=0.05
2) a=0.01
2) a=0.01
1) Since 0.0246 < 0.05, reject H0.
2) Since 0.0246 > 0.01, fail to reject H0 .
Using the P-value of a test compare areas
If
  0.05
z = -1.23
Rejection area
0.05
z0 z 0
For a P-value decision, compare areas.
P  a fail to reject H0.
If P  a reject H0. If
For
  0.05
z0 = -1.645. If z = -1.23
The P-value is 0.1093. The actual area to the left of the z-score is 0.1093.
With a standardized z-score of -1.23, fail to reject the null hypothesis at
the 0.05 level of significance.
The t Sampling Distribution
Find
the critical
value t0 for
a left-tailed
given a = 0.01
Find
the critical
value
t0 fortest
a left-tailed
and
n =given
18.
test
a = 0.01 and n = 18.
d.f.= 18-1 = 17
t0 = - 2.567
Area in
left tail
to 0
Find
the critical
values - tvalues
test a
0 and t0 for
Find
the critical
- ta0two-tailed
and t0 for
given
a = 0.05 and
11. a = 0.05 and n = 11.
two-tailed
testn =
given
-t0 =-2.228 and t0=2.228
d.f.= 11-1 = 10
0
-t0
t0
Testing µ-small Sample
A university
mean
number
of per
classroom

A university
says thesays
mean the
number
of classroom
hours
week for
hours per week for full-time faculty is 11.0. A
full-time
faculty is 11.0. A random sample of the number of classroom
random sample of the number of classroom hours
hours
for full-time faculty
faculty forfor
one week
listed is
below.
You below.
work for a
for full-time
one is
week
listed
You organization
work for and
a student
and
student
are askedorganization
to test this claim At
a =are
0.01,asked
do you
to test this claim At a = 0.01, do you have enough
have
enough evidence
to reject
the university’s claim?
evidence
to reject
the university’s
claim?
11.88.6 8.6
13.6 9.1
11.8
12.6 12.6
7.9 6.4 7.9
10.4 6.4
13.6 10.4
9.1
1. Write the null and alternative hypothesis
H0:  = 11.0 (claim)
Ha:   11.0
2. State the level of significance
a= 0.01
3. Determine the sampling distribution
Since the sample size is 8, the sampling distribution is a t-distribution
with 8 - 1 = 7 d.f.
Since Ha contains the  symbol, this is a two- tail test.
Rejection
region
Rejection
region
-t0 0
t0
-3.499
3.499
4. Find the critical values
5. Find the rejection region
6. Find the test statistic
n=8 x = 10.050 s =2.485
10.050  11.0  0.95
t

 1.08
2.485
0.878
8
7. Make your decision
t = -1.08 does not fall in the rejection region, so fail to reject H0 at α= 0.01
8. Interpret your decision
There is not enough evidence to reject the university’s claim that faculty
spend a mean of 11 classroom hours.
Minitab Solution
Enter the data in C1, ‘Hours’.
Choose t-test in the STAT menu.
T-Test of the Mean
Test of mu = 11.000 vs mu not = 11.000
Variable N Mean StDev SE Mean T
Hours
8 0.050 2.485 0.879 -1.08
P
0.32
Minitab reports the t-statistic and the P-value.
Since the P-value is greater than the level of significance (0.32 > 0.01),
reject the null hypothesis at the 0.01 level of significance.
Test for Proportions
p is the population proportion of successes. The test
x
ˆ
p

statistic is
n
If np  5 and nq  5 the sampling distribution for p̂ is normal.
ˆ  p
p
z 
The standardized test statistic is:
pq
n
 A communications industry spokesperson claims that over
40% of Americans either own a cellular phone or have a
family member who does. In a random survey of 1036
Americans, 456 said they or a family member owned a cellular
phone. Test the spokesperson’s claim at a = 0.05. What can
you conclude?
Test for Proportions
 A communications industry spokesperson claims that over
40% of Americans either own a cellular phone or have a
family member who does. In a random survey of 1036
Americans, 456 said they or a family member owned a cellular
phone. Test the spokesperson’s claim at a = 0.05. What can
you conclude?
1. Write the null and alternative hypothesis
H0: p  0.40
Ha: p > 0.40 (claim)
2. State the level of significance
a= 0.05
3. Determine the sampling distribution
1036(.40) > 5 and 1036(.60)>5 . The sampling distribution
is normal.
4. Find the critical value
Rejection
region
5. Find the rejection region
0 1.645
6. Find the test statistic
n =1036 x = 456
z 
. 44  . 40
0 . 04

0 . 01522
(. 40 )(. 60 )
1036
 2 . 63
7. Make your decision
z = 2.63 falls in the rejection region, so reject H0
8. Interpret your decision
There is enough evidence to support the claim that over 40% of
Americans own a cell phone or have a family member who does.
Critical Values for c
s2 is the test statistic for
the population variance.
Its sampling distribution
is a c2 distribution with
n-1 d.f.
0
1 0
2 0
3 0
2
4 0
Find a c20 critical value for a left-tail test when n =17 and a = 0.05.
c20 =7.962
Find critical values c20 for a two-tailed test when n = 12, a = 0.01.
c2L=2.603 and c2R=26.757
The standardized test statistic is
c2 
( n  1) s 2
2
Test for

 A state school administrator says that the standard deviation of test
scores for 8th grade students who took a life-science assessment test is
less than 30. You work for the administrator and are asked to test this
claim. You find that a random sample of 10 tests has a standard
deviation of 28.8. At a = 0.01, do you have enough evidence to
support the administrator’s claim? Assume test scores are normally
distributed.
1. Write the null and alternative hypothesis
H0:   30
Ha:  < 30 (claim)
2. State the level of significance
3. Determine the sampling distribution
The sampling distribution is c2 with 10 - 1 = 9 d.f.
a= 0.01
4. Find the critical value
0
1 0
2 0
3 0
4 0
2.088
5. Find the rejection
region
6. Find the test statistic
n=10
s = 28.8
c 
2
(n  1) s 2
2
(10  1)28.8 2

 8.2944
2
30
7. Make your decision
c2 = 8.2944 does not fall in the rejection region, so fail to reject
H0
8. Interpret your decision
There is not enough evidence to support the administrator’s claim
that the standard deviation is less than 30.