Hypothesis Testing

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Hypothesis Testing
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Elementary Statistics Larson Farber
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Section 7.1
Introduction to Hypothesis Testing
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A statistical hypothesis is a claim about a
population.
Alternative hypothesis Ha contains a statement
of inequality such as lt , ¹ or gt
Null hypothesis H0 contains a statement of
equality such as ³ , or .
If I am false, you are true
If I am false, you are true
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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
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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.
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Errors and Level of Significance
Actual Truth of H0
H0 True
H0 False
Do not reject H0
Decision
Type II Error
Type I Error
Reject H0
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.
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Types of Hypothesis Tests
Right-tail test
Ha is more probable
Left-tail test
Ha is more probable
Ha is more probable
Two-tail test
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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 positive, twice the area in the right tail
If z is negative, twice the area in the left tail
z
z
For a two-tail test
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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.
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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.
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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 reject the null hypothesis.
If fail to reject the null
hypothesis.
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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 gt 0.05,
fail to reject H0.
If P 0.0246, what is your decision if
1) Since , reject H0. 2)
Since 0.0246 gt 0.01, fail to reject H0.
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Interpreting the Decision
Claim
Claim is H0
Claim is Ha
There is enough evidence to support the claim.
There is enough evidence to reject the claim.
Reject H0
Decision
There is not enough evidence to support the claim.
There is not enough evidence to reject the claim.
Fail to reject H0
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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.
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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.
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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.
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Section 7.2
Hypothesis Testing for the Mean (n ? 30)
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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 .
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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 companys 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.
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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
Area in right tail
P 0.0749.
z 1.44
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6. Make your decision.
Compare the P-value to . Since 0.0749 gt
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.
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Rejection Regions
Sampling distribution for
Rejection Region
Critical Value z0
z
z0
A critical value separates the rejection region
from the non-rejection region.
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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
z0 1.645
Rejection region
Rejection region
z0 2.33
z0 2.575 and z0 2.575
z0
z0
Find z0 and z0 for a two-tail test with
.01.
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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.
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5. Find the rejection region.
4. Find the critical value.
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.
Rejection Region
z0
6. Find the test statistic.
Perform the calculations to standardize your
sample statistic.
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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.
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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 companys 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.
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Since Ha contains the gt symbol, this is a
right-tail test.
4. Find the critical value.
5. Find the rejection region.
z0
1.645
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 H0
8. Interpret your decision.
There is not enough evidence to reject the
companys claim that there is at most 230 mg of
sodium in one serving of its cereal.
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Using the P-value of a Test to Compare Areas
0.05
Area to the left of z 0.1093
z0 1.645
Rejection area 0.05
z 1.23
P 0.1093
z0
z
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.
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Section 7.3
Hypothesis Testing for the Mean (n lt 30)
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The t Sampling Distribution
Find the critical value t0 for a left-tailed test
given 0.01 and n 18.
d.f. 18 1 17
Area in left tail
t0 2.567
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
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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
universitys claim? 11.8 8.6 12.6 7.9
6.4 10.4 13.6 9.1
1. Write the null and alternative hypothesis
0.01
2. State the level of significance
3. Determine the sampling distribution
Since the sample size is 8, the sampling
distribution is a t-distribution with 8 1 7
d.f.
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Since Ha contains the ? symbol, this is a
two-tail test.
4. Find the critical values.
5. Find the rejection region.
t0
t0
3.499
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 0.01
8. Interpret your decision.
There is not enough evidence to reject the
universitys claim that faculty spend a mean of
11 classroom hours.
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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
not 11.000 Variable N Mean StDev
SE Mean T P Hours 8
0.050 2.485 0.879 1.08 0.32
Minitab reports the t-statistic and the P-value.
Since the P-value is greater than the level of
significance (0.32 gt 0.01), fail to reject the
null hypothesis at the 0.01 level of significance.
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Section 7.4
Hypothesis Testing for Proportions
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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 is normal.
The standardized test statistic is
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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
spokespersons claim at 0.05. What can you
conclude?
1. Write the null and alternative hypothesis.
2. State the level of significance.
0.05
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3. Determine the sampling distribution.
1036(.40) gt 5 and 1036(.60) gt 5. The sampling
distribution is normal.
4. Find the critical value.
Rejection region
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.
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Section 7.5
Hypothesis Testing for Variance and Standard
Deviation
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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
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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 administrators
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.
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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
administrators claim that the standard deviation
is less than 30.