Transcript lect7-3

For testing a claim about the
mean of a single population
Assumptions
1) Sample is large (n > 30)
a) Central limit theorem applies
b) Can use normal distribution
2) Can use sample standard deviation s
as estimate for s if s is unknown
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Addison Wesley Longman
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Three Methods Discussed
1) Traditional method
2) P-value method
3) Confidence intervals
Note: These three methods are equivalent, I.e., they
will provide the same conclusions.
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Procedure
Figure 7-4
1. Identify the specific claim or hypothesis to be tested, and
put it in symbolic form.
2. Give the symbolic form that must be true when the
original claim is false.
3. Of the two symbolic expressions obtained so far, put the
one you plan to reject in the null hypothesis H0 (make the
formula with equality). H1 is the other statement.
Or, One simplified rule suggested in the textbook: let null
hypothesis H0 be the one that contains the condition of
equality. H1 is the other statement.
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4. Select the significant level a based on the seriousness of
a type I error. Make a small if the consequences of
rejecting a true H0 are severe. The values of 0.05 and 0.01
are very common.
5. Identify the statistic that is relevant to this test and its
sampling distribution.
6. Determine the test statistic, the critical values, and the
critical region. Draw a graph and include the test
statistic, critical value(s), and critical region.
7. Reject H0 if the test statistic is in the critical region. Fail
to reject H0 if the test statistic is not in the critical region.
8. Restate this previous decision in simple non-technical
terms. (See Figure 7-2)
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In a P-value method, procedure is the
same except for steps 6 and 7
Step 6: Find the P-value
Step 7: Report the P-value
Reject the null hypothesis if the P-value
is less than or equal to the significance level a
Fail to reject the null hypothesis if the
P-value is greater than the significance level a
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Testing Claims with
Confidence Intervals
• A confidence interval estimate of a population
parameter contains the likely values of that
parameter. We should therefore reject a claim
that the population parameter has a value that is
not included in the confidence interval.
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Testing Claims
with Confidence Intervals
Claim: mean body temperature = 98.6°,
where n = 106, x = 98.2° and s = 0.62°
 95% confidence interval of 106 body temperature
data (that is, 95% of samples would contain true
value µ )
 98.08º < µ < 98.32º
 98.6º is not in this interval
 Therefore it is very unlikely that µ = 98.6º
 Thus we reject claim µ = 98.6
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Underlying Rationale of
Hypotheses Testing
• When testing a claim, we make an assumption (null
hypothesis) that contains equality. We then compare
the assumption and the sample results and we form
one of the following conclusions:
 If the sample results can easily occur when the
assumption is true, we attribute the relatively small
discrepancy between the assumption and the sample
results to chance.
 If the sample results cannot easily occur when the
assumption is true, we explain the relatively large
discrepancy between the assumption and the sample by
concluding that the assumption is not true.
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Testing a Claim about a
Mean: Small Samples
Section 7-4
M A R I O F. T R I O L A
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Wesley
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Wesley LongmanLongman
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Figure 7-10 Choosing between the Normal and Student
t-Distributions when Testing a Claim about a Population Mean µ
Start
Use normal distribution with
Is
n > 30
?
x – µx
Z=
s/ n
Yes
(If s is unknown use s instead.)
No
Is the
distribution of
the population essentially
normal ? (Use a
histogram.)
No
Yes
Use non-parametric methods,
which don’t require a normal
distribution.
Use normal distribution with
Is s
known
?
No
Z=
x – µx
s/ n
(This case is rare.)
Use the Student t-distribution
with
x – µx
t = s/
n
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The distribution of sample means will be
NORMAL and you can use Table A-2 :
a) For any population distribution (shape)
where, n > 30 (use s for a if a is unknown)
b) For a normal population distribution where
1)
s is known (n
2)
s is unknown and n
any size), or
> 30 (use s for s)
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The Student t-distribution and
Table A-3 should be used when:
 Population distribution is essentially normal
 s is unknown
 n  30
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Test Statistic (TS)
for a Student t-distribution
x – µx
t=s n
Critical Values (CV)
Found in Table A-3
Formula card, back book cover, or Appendix
Degrees of freedom (df) = n – 1
Critical t values to the left of the mean are
negative
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Important Properties of the
Student t Distribution
1. The Student t distribution is different for different sample sizes (see
Figure 6-5 in Section 6-3).
2. The Student t distribution has the same general bell shape as the
normal distribution; its wider shape reflects the greater variability that
is expected with small samples.
3. The Student t distribution has a mean of t = 0 (just as the standard
normal distribution has a mean of z = 0).
4. The standard deviation of the Student t distribution varies with the
sample size and is greater than 1 (unlike the standard normal
distribution, which has a s = 1).
5. As the sample size n gets larger, the Student t distribution get closer to
the normal distribution. For values of n > 30, the differences are so
small that we can use the z values instead of developing a much larger
table of critical t values. (The values in the bottom row of Table A-3 are
equal to the corresponding critical z values from the standard normal
distribution.)
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All three methods
1) Traditional method
2) P-value method
3) Confidence intervals
and the testing procedure
Step 1 to Step 8
are still valid, except that the test
statistic (therefore corresponding
Table) is different.
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The larger Student t-distribution
values shows that with small
samples the sample evidence must
be more extreme before the
difference is significant.
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P-Value Method
Table A-3 includes only selected values of a
Specific P-values usually cannot be found
Use Table to identify limits that contain the
P-value
Some calculators and computer programs
will find exact P-values
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Example
Conjecture: “the average starting salary for a
computer science gradate is $30,000 per year”.
For a randomly picked group of 25 computer
science graduates, their average starting salary
is $36,100 and the sample standard deviation
is $8,000.
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Example
Solution
Step 1: µ = 30k
Step 2: µ > 30k (if believe to be no less than 30k)
Step 3: H0: µ = 30k versus H1: µ > 30k
Step 4: Select a = 0.05 (significance level)
Step 5: The sample mean is relevant to this test and
its sampling distribution is t-distribution with
(25 - 1 ) = 24 degrees of freedom.
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(Step 6)
t-distribution (DF = 24)
Assume the
conjecture
is true!
Critical value =
Test Statistic:
t=
x – µx
S
/
n
1.71 * 8000/5 + 30000 = 32736
Fail to reject H0
30 K
( t = 0)
Reject H0
32.7 k
( t = 1.71 )
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(Step 7)
t-distribution (DF = 24)
Assume the
conjecture
is true!
Critical value =
Test Statistic:
t=
x – µx
s/
n
1.71 * 8000/5 + 30000 = 32736
Fail to reject H0
30 K
( t = 0)
Reject H0
Sample data: t = 3.8125
or
x = 36.1k
32.7 k
( t = 1.71 )
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Example
Step 8:
Conclusion: Based on the sample set, there is
sufficient evidence to warrant rejection of the
claim that “the average starting salary for a
computer science gradate is $30,000 per year”.
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(Step 6)
t-distribution (DF = 24)
Assume the
conjecture
is true!
Test Statistic:
x – µx
t=
S/
n
P-value = area
to the right of the
test statistic
30 K
t=
36.1 - 30
8/5
36.1 k
= 3.8125
P-value = .0004225
(by a computer program)
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(Step 7)
t-distribution (DF = 24)
P-value = area
to the right of the
test statistic
30 K
t=
36.1 - 30
8/5
36.1 k
= 3.8125
P-value = .0004225
P-value < 0.01
Highly statistically significant
(Very strong evidence against the null hypothesis)
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