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7
Hypothesis Testing
Elementary Statistics
Larson
Farber
Section 7.1
Introduction to
Hypothesis Testing
A statistical hypothesis is a claim about a population.
Null hypothesis H0
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,
you are true
If I am false,
you are true
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.
claim
A consumer magazine claims the proportion of cell
phone calls made during evenings and weekends is at
most 60%.
claim
Hypothesis Test Strategy
Begin by assuming the equality condition in the null hypothesis is true.
This is regardless of whether the claim is represented by the null
hypothesis or by the alternative hypothesis.
Collect data from a random sample taken from the population
and calculate the necessary sample statistics.
If the sample statistic has a low probability of being drawn
from a population in which the null hypothesis is true, you will
reject H0. (As a consequence, you will support the alternative
hypothesis.)
If the probability is not low enough, fail to reject H0.
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,
Maximum probability of committing a type I error.
Types of Hypothesis Tests
Ha is more probable
Ha is more probable
Ha is more probable
Right-tail test
Left-tail test
Two-tail test
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.
P-value = indicated area
Area in
left tail
Area in
right tail
z
z
For a right tail test
For a left tail test
If z is negative,
twice the area
in the left tail
If z is positive,
twice the area
in the right tail
z
z
For a two-tail test
Finding P-values: 1-tail Test
The test statistic for a right-tail test is z = 1.56. Find the
P-value.
Area in right tail
z = 1.56
The area to the right of z = 1.56 is 1 – .9406 = 0.0594.
The P-value is 0.0594.
Finding P-values: 2-tail Test
The test statistic for a two-tail test is z = –2.63. Find the
corresponding P-value.
z = –2.63
The area to the left of z = –2.63 is 0.0043.
The P-value is 2(0.0043) = 0.0086.
Test Decisions with P-values
The decision about whether there is enough
evidence to reject the null hypothesis can be
made by comparing the P-value to the value of
the level of significance of the test.
If
If
,
reject the null hypothesis.
fail to reject the null hypothesis.
Using P-values
The P-value of a hypothesis test is 0.0749. Make your
decision at the 0.05 level of significance.
Compare the P-value to
. Since 0.0749 > 0.05, fail to reject H0.
If P = 0.0246, what is your decision if
1) Since
, reject H0.
2) Since 0.0246 > 0.01, fail to reject H0.
Interpreting the Decision
Claim
Decision
Claim is H0
Reject H0
Fail to
reject H0
Claim is Ha
There is enough
evidence to
reject the claim.
There is enough
evidence to
support the
claim.
There is not
enough
evidence to
reject the claim.
There is not
enough
evidence to
support the
claim.
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 the equality condition in H0 is true
and that the experiment is repeated an infinite number of
times.
4. Find the test statistic and standardize it.
Perform the calculations to standardize your sample statistic.
5. Calculate the P-value for the test statistic.
This is the probability of obtaining your test statistic or
one that is more extreme from the sampling distribution.
6. Make your decision.
If the P-value is less than
(the level of significance)
reject H0.
If the P value is greater , fail to reject H0.
7. Interpret your decision.
If the claim is the null hypothesis, you will either reject
the claim or determine there is not enough evidence to
reject the claim.
If the claim is the alternative hypothesis, you will either
support the claim or determine there is not enough
evidence to support the claim.
Section 7.2
Hypothesis Testing for
the Mean
(n  30)
The z-Test for a Mean
The z-test is a statistical test for a population mean. The z-test
can be used:
(1) if the population is normal and s is known or
(2) when the sample size, n, is at least 30.
The test statistic is the sample mean and the standardized
test statistic is z.
When n  30, use s in place of
.
The z-Test for a Mean (P-value)
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 mg and a standard deviation of 10 mg. At
= 0.05, do you have enough evidence to reject the company’s
claim?
1. Write the null and alternative hypothesis.
2. State the level of significance.
= 0.05
3. Determine the sampling distribution.
Since the sample size is at least 30, the sampling distribution is normal.
4. Find the test statistic and standardize it.
n = 52
s = 10
Test statistic
5. Calculate the P-value for the test statistic.
Since this is a right-tail test, the P-value
is the area found to the right
of z = 1.44 in the normal distribution.
From the table P = 1 – 0.9251
P = 0.0749.
Area in right tail
z = 1.44
6. Make your decision.
Compare the P-value to .
Since 0.0749 > 0.05, fail to reject H0.
7. Interpret your decision.
There is not enough evidence to reject the claim
that the mean sodium content of one serving of its cereal is
no more than 230 mg.
Rejection Regions
Sampling distribution for
Rejection Region
z
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 critical value separates the rejection region from the
non-rejection region.
Critical Values
The critical value z0 separates the rejection region from
the non-rejection region. The area of the rejection region
is .
Rejection
region
Rejection
region
z0
z0
Find z0 for a right-tail
test with = .05.
Find z0 for a left-tail
test with = .01.
z0 = –2.33
Rejection
region
z0
Rejection
region
z0
Find –z0 and z0 for a two-tail test with
z0 = 1.645
–z0 = –2.575
and z0 = 2.575
= .01.
Using the Critical Value to Make Test Decisions
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 the equality condition in H0 is true
and that the experiment is repeated an infinite number of
times.
4. Find the critical value.
Rejection Region
z0
6. Find the test statistic.
5. Find the
rejection region.
The critical value
separates the rejection
region of the sampling
distribution from the
non-rejection region.
The area of the critical
region is equal to the
level of significance of
the test.
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.
If the claim is the null hypothesis, you will either
reject the claim or determine there is not enough
evidence to reject the claim.
If the claim is the alternative hypothesis, you will
either support the claim or determine there is not
enough evidence to support the claim.
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 mg and a standard deviation of 10
mg. At
= 0.05, do you have enough evidence to reject
the company’s claim?
1. Write the null and alternative hypothesis.
2. State the level of significance.
= 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.
Rejection
region
z0
1.645
4. Find the critical value.
5. Find the rejection region.
6. Find the test statistic and standardize it.
n = 52
= 232 s = 10
7. Make your decision.
z = 1.44 does not fall in the rejection region, so fail to reject H 0
8. Interpret your decision.
There is not enough evidence to reject the company’s claim that
there is at most 230 mg of sodium in one serving of its cereal.
Using the P-value of a Test to Compare
Areas
= 0.05
z0 = –1.645
Rejection area
0.05
z0
z = –1.23
P = 0.1093
z
For a P-value decision, compare areas.
If
reject H0.
If
fail to reject H0.
For a critical value decision, decide if z is in the rejection region
If z is in the rejection region, reject H0. If z is not in the rejection
region, fail to reject H0.
Section 7.3
Hypothesis Testing for
the Mean
(n < 30)
The t Sampling Distribution
Find the critical value t0 for a left-tailed test given
and n = 18.
= 0.01
d.f. = 18 – 1 = 17
t0 = –2.567
Area in
left tail
t0
Find the critical values –t0 and t0 for a two-tailed test given
= 0.05 and n = 11.
–t0 = –2.228 and t0 = 2.228
d.f. = 11 – 1 = 10
t0
t0
Testing
–Small Sample
A university says the mean number of classroom hours per
week for full-time faculty is 11.0. A random sample of the
number of classroom hours for full-time faculty for one week is
listed below. You work for a student organization and are asked
to test this claim. At
= 0.01, do you have enough evidence
to reject the university’s claim?
11.8 8.6 12.6 7.9 6.4 10.4 13.6 9.1
1. Write the null and alternative hypothesis
2. State the level of significance
= 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.
4. Find the critical values.
5. Find the rejection region.
–t0
–3.499
t0
3.499
6. Find the test statistic and standardize it
n=8
= 10.050 s = 2.485
7. Make your decision.
t = –1.08 does not fall in the rejection region, so fail to reject H0 at
8. Interpret your decision.
There is not enough evidence to reject the university’s claim that
faculty spend a mean of 11 classroom hours.
= 0.01
Minitab Solution
Enter the data in C1, ‘Hours’.
Choose t-test in the STAT menu.
T-Test of the Mean
Test of
= 11.000 vs
Variable N
Hours
8
Mean
0.050
not = 11.000
StDev
2.485
SE Mean
T
0.879
–1.08
Minitab reports the t-statistic and the P-value.
Since the P-value is greater than the level of
significance (0.32 > 0.01), fail to reject the null
hypothesis at the 0.01 level of significance.
P
0.32
Section 7.4
Hypothesis Testing for
Proportions
Test for Proportions
p is the population proportion of successes. The
test statistic is
.
(the proportion of sample successes)
If
and
the sampling distribution for
The standardized test statistic is:
is normal.
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
= 0.05. What can you
conclude?
1. Write the null and alternative hypothesis.
2. State the level of significance.
= 0.05
3. Determine the sampling distribution.
1036(.40) > 5 and 1036(.60) > 5. The sampling distribution is
normal.
Rejection
region
4. Find the critical value.
5. Find the rejection region.
1.645
6. Find the test statistic and standardize it.
n = 1036
x = 456
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.
Section 7.5
Hypothesis Testing for
Variance and
Standard Deviation
Critical Values for
s2 is the test statistic
for the population
variance. Its sampling
distribution is a c2
distribution with n – 1
d.f.
Find a c20 critical value for a left-tail test when n = 17 and
= 0.05.
c20 = 7.962
Find critical values c20 for a two-tailed test when n = 12,
= 0.01.
c2L = 2.603 and c2R = 26.757
The standardized test statistic is
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
= 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.
2. State the level of significance.
= 0.01
3. Determine the sampling distribution.
The sampling distribution is c2 with 10 – 1 = 9 d.f.
4. Find the critical value.
5. Find the rejection region.
2.088
6. Find the test statistic.
n = 10
s = 28.8
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.