Download Statistical Hypotheses Writing Hypotheses Hypothesis Test Strategy

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Statistical Hypotheses
Section 7.1
A statistical hypothesis is a claim about a population.
Introduction to
Hypothesis Testing
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
Schrodinger’s cat quantum mechanics
thought experiment (1935)
Writing Hypotheses
Write the claim about the population. Write its complement.
Either hypothesis, can represent the claim.
Example: A hospital claims its ambulance response
time is less than 10 minutes.
claim
Example: A consumer magazine claims the proportion
of cell phone calls made during evenings and weekends
is at most 60%.
claim
If I am false,
you are true
Accept or reject null, never prove null is true
Hypothesis Test Strategy
1) Assume the equality condition in the null hypothesis is true,
regardless of whether the claim is represented by the null or
alternative hypothesis.
2) Collect data from a random sample taken from the population and
calculate the necessary sample statistics.
3a) 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.
(i.e. you will support the alternative hypothesis.)
3b) If the probability is not low enough, fail to reject H0.
H0 always contains the = condition
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Decision
Errors and Level of Significance
Actual Truth of H0
H0 True
Do not
reject H0
Reject H0
Correct
Decision
Type I
Error
H0 False
Types of Hypothesis Tests
Ha is more probable
Right-tail test
Type II
Error
Correct
Decision
Ha is more probable
Type I Error: Null hypothesis is true but reject it.
One-tail test
Left-tail test
Level of significance,
(e.g. 0.01, 0.05, 0.10)
Maximum probability of committing a Type I Error.
Type II Error: Null hypothesis is false but accept it.
Probability of Type II Error 1- (power of test)
“Innocent until proven guilty”
Ha is more probable
Two-tail test
“Beyond reasonable doubt”
P-values
Finding P-values: 1-tail Test
The P-value is the probability of obtaining a sample
statistic with a value as or more extreme than the one
determined by the sample data.
The test statistic for a right-tail test is z = 1.56. Find the
P-value.
P-value = indicated area
Area in
left tail
Area in
right tail
Area in right tail
z = 1.56
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
The area to the right of z = 1.56 is 1 – .9406 = 0.0594.
The P-value is 0.0594.
z
For a two-tail test
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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.
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 arbitrary level of significance of the test.
If
,
reject the null hypothesis.
z = –2.63
If
fail to reject the null hypothesis.
The area to the left of z = –2.63 is 0.0043.
The P-value is 2(0.0043) = 0.0086.
Interpreting the Decision
Method I: Using P-values
Claim
The P-value of a hypothesis test is 0.0749. Make your
decision at the 0.05 level of significance.
If P = 0.0246, what is your decision if
1) Since
Claim is H0
. Since 0.0749 > 0.05,
, reject H0.
2) Since 0.0246 > 0.01, fail to reject H0.
Decision
Compare the P-value to
fail to reject H0.
Claim is Ha
Reject H0
There is enough
evidence to
reject the claim.
There is enough
evidence to
support the
claim.
Fail to
reject H0
There is not
enough
evidence to
reject the claim.
There is not
enough
evidence to
support the
claim.
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Steps in a Hypothesis Test
1. Write the null and alternative hypothesis.
Write H0 and Ha as mathematical statements (H0 always
contains the = symbol).
2. State the level of significance.
Maximum probability of rejecting the null hypothesis when
it is true. (Making a Type I error.)
Steps in a Hypothesis Test
4. Find the test statistic and standardize it.
Perform 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.
3. Identify the sampling distribution.
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.
6. Make your decision.
If the P-value is less than , reject H0.
If the P value is greater than , fail to reject H0.
7. Interpret your decision.
Section 7.2
Hypothesis Testing for
the Mean
(n ≥ 30)
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
The z-test is a statistical test for a population mean. The z-test
can be used:
(1) if the population is normal and  is known or
(2) when the sample size, n, is at least 30.
Test statistic is the sample mean
test statistic is z.
and the standardized
The z-Test for a Mean (P-value)
Example: 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.
When n ≥ 30, use s in place of
.
= 0.05
3. Determine the sampling distribution.
Since the sample size is at least 30, the sampling distribution is normal
(Central Limit Theorem, Chptr. 5).
4. Find the test statistic and standardize it.
n = 52
s = 10 (n>30)
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
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.
z = 1.44
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Critical Values
Method II: Rejection Regions
Sampling distribution for
Rejection Region
Rejection
region
Rejection
region
z0
z0
z
z0
Critical Value z0
Rejection region is the range of values for which the
null hypothesis is not probable. Always in the direction
of the alternative hypothesis. Its area is equal to .
A critical value separates rejection region from the nonrejection region.
Find z0 for a right-tail
test with = .05.
Find z0 for a left-tail
test with = .01.
z0 = –2.33
z0 = 1.645
Rejection
region
z0
Rejection
region
z0
Find –z0 and z0 for a two-tail test with  = .01
–z0 = –2.575
and z0 = 2.575
Using Critical Values to Make Test Decisions
1. Write the null and alternative hypothesis.
4. Find the critical value.
Write H0 and Ha as mathematical statements.
Remember H0 always contains the = symbol.
2. State the level of significance.
The maximum probability of rejecting the null hypothesis
when it is actually true (Type I Error.)
3. Identify the sampling distribution.
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.
Rejection Region
z0
5. Find the
rejection region.
Critical value separates
rejection region of the
sampling distribution from
the non-rejection region.
Area of the critical region
is equal to the level of
significance of the test.
6. Find the test statistic.
Perform the calculations to standardize your sample statistic.
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The z-Test for a Mean
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.
Example: 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.
Using the P-value of a Test to Compare
Areas
Since Ha contains the > symbol, this is a right-tail test.
Rejection
region
z0
4. Find the critical value.
1.645,  = 0.05, cum. area = 0.95
5. Find the rejection region.
6. Find the test statistic and standardize it.
n = 52
= 232 s = 10
z = 1.44
7. Make your decision.
Test statistic 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 230 mg of sodium in one serving of its cereal.
= 0.05
z0 = –1.645
Rejection area
0.05
z0
Test statistic
z = –1.23
P = 0.1093
z
For a P-value decision, compare areas.
fail to reject H0.
If
reject H0. If
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.
*Decision same, critical values & z-scores, P-values & areas*
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Section 7.3
The t Sampling Distribution
Find the critical value t0 for a left-tailed test given  = 0.01 and n = 18.
Hypothesis Testing for
the Mean
(n < 30)
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.
d.f. = 11 – 1 = 10
–t0 = –2.228 and t0 = 2.228
t0
Testing
–Small Sample
Example: A university says the mean number of classroom hours per
week for full-time faculty is 11.0. A random sample 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
= 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.
*If > 30 d.f. then ~normal*
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
2. State the level of significance
t0
= 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 university’s claim that
faculty spend a mean of 11 classroom hours.
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Test for Proportions
Section 7.4
p is the population proportion of successes. The
test statistic is
.
Hypothesis Testing for
Proportions
(the proportion of sample successes)
If
and
the sampling distribution for
is normal.
The standardized test statistic is:
Test for Proportions
Example: 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 1,036
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.
3. Determine the sampling distribution.
1036(.40) > 5 and 1036(.60) > 5. The sampling distribution is
normal.
Rejection
region
= 0.05
5. Find the rejection region.
1.645
6. Find the test statistic and standardize it.
n = 1036
2. State the level of significance.
4. Find the critical value.
x = 456
7. Make your decision.
z = 2.63 falls in the rejection region, so reject H0
What is p-value?
0.0043
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
Critical Values for
s2 is the test statistic for the
population variance. Its
sampling distribution is a 2
distribution with n – 1 d.f.
Hypothesis Testing for
Variance and
Standard Deviation
Find a 20 critical value for a left-tail test when n = 17 and
= 0.05.
20 = 7.962
= 0.01.
Find critical values 20 for a two-tailed test when n = 12,
2L = 2.603 and 2R = 26.757
The standardized test statistic is
Test for
Example: 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.
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.
2 = 8.2944 does not fall in the rejection region, so fail to reject H0
2. State the level of significance.
8. Interpret your decision.
= 0.01
3. Determine the sampling distribution.
There is not enough evidence to support the administrator’s claim
that the standard deviation is less than 30.
The sampling distribution is 2 with 10 – 1 = 9 d.f.
1- or 2-tailed test?
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