Download STAT 215 Fall 2006 L20 TESTING HYPOTHESIS * A statistical

STAT 215
Fall 2006
L20
TESTING HYPOTHESIS
* A statistical hypothesis is a claim about the value
of a single population characteristic.
* Usually in any hypothesis testing problem,
there are two contradictory hypothesis
under consideration
* The claim or research hypothesis that
we wish to establish is called
the alternative hypothesis H1
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* The opposite statement, one that nullifies
the research hypothesis
H1, is called the null hypothesis H0
* A test procedure is a rule, based on the
sample data, for deciding whether to reject H0.
The test procedure is specified by
1).
a test statistic, a function of the sample data
on which the decision
(reject H0 or do not reject H0) is to be based.
2).
a rejection region, the set of all
test statistic values for which H0 will be rejected.
*
The null hypothsis H0 will be rejected
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if and only if the observed
or computed test statistic value falls in
the rejection region.
Example 1.
claims
Suppose a cigarette manufacturer
that the average nicotine content µ of a brand B
cigarettes is at most 1.5 mg.
So that, the alternative hypothesis
H1: µ < 1.5 and the null hypothesis is
H0 :
µ = 1.5.
3).
Errors in hypothesis testing
*
A Type I error consists of rejecting the
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null hypothesis H0 when it is true.
*
A Type II error involves not rejecting
the null hypothesis H0 when it is false.
4).
Level of significance
α = Probability of making a Type I error is called
the level of significance.
Usually, the specified level of significance α = .05
or .01
Power of a Test: 1 - P (Typer II error) = 1 - β
Problems:
1. Stated here are some claims or research
hypotheses that are to be substantiated by
sample data.
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In each case, identify the null hypothesis
H0 and the alternative hypothesis
H1 in terms of the population mean µ.
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Examples:
(a)
The average mathematics score of the
college-bound students in Milwaukee who
participated in the American College Testing
(ACT) program in 1996 was highter than 17.2.
(b)
The meantime for an airline passenger
to obtain his or her luggage, once luggage
starts coming out on the conveyer belt,
is less than 210 seconds.
(c) The content of fat in a name-brand chocolate
ice cream is more than 4%, the amount printed
on the label.
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(d)
The average weight of a brand of motors
is different from the manufaturer’s
target of 6 pounds.
LARGE SAMPLE CASE .
Assume that we have some population with unknown
mean = µ and sd = σ. Suppose X1, ..., Xn
is a random sample from this population with n
- sample
size. Let us consider the case when n ≥ 30.
To test the hypothesis H0 : µ = µ0 versus the
alternative H1 : µ < µ0, ( or µ > µ0
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or µ 6= µ0 ) consider the standartized random
variable
Z=
X̄ − µ0
√ ,
S/ n
here again
n
1 X
Xi
X̄ =
n i=1
and
S=
v
u
u
u
t
n
1 X
(Xi − X̄)2.
n − 1 i=1
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Definitions and statistical concepts:
1. Null Hypothesis H0 : µ = µ0 ,
the alternative hypothesis H1 : µ < µ0
( H1 : µ > µ0 or, for example,
H1 : µ 6= µ0)
2. Type I and Type II errors:
Type I error: if H0 is true and we reject H0
Type II error: if H1 is true and we accept H0
3. The Level of significance: α = P (Type I error)
4. The Power of a Test:
1 - P (Type II error) = 1 - β
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5. Rejection Region: Z ≤ −zα
( or Z ≥ zα ),
where Z is a test statistic:
X̄ − µ0
√
S/ n
and zα is the α -upper point of
Z=
standard normal distribution.
Our general aim in Hypothesis Testing is
to use statistics (Tests) that make α and β
as small as possible. These actions are
contradictory.
Instead, our general strategy: to fix α at some
specific level, for example, α = .05, .01, and to
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use the test that max the power of the Test.
When H0 is true and n ≥ 30 then statistic Z
has approximately standard normal distribution,
z-distribution.
From the corresponding Table we can determine
the critical
value zα :
P (Z ≥ zα ) = α.
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The following sequence of steps is
recommended in the Hypothesis testing analysis:
1. Identify the parameter of interest and
describe it in the context of the problem
situation;
2. Determine the null value and state the null
hypothesis H0;
3. State the appropriate alternative
hypothesis H1;
4. Give the formula for the computation
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the value of the Test statistic ( T or Z statistics);
5. State the rejection region R for the specified
significance level α ( with specified values zα
or tα );
6. Compute any necessary sample quantities
( sample mean, sample deviation),
substitute them into the formula for the test
statistics and compute that value;
7. Decide whether H0 should be rejected and
state this conclusion in the problem context.
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Testing of statistical hypotheses about
population mean µ.
Small samples
Assume that the population distribution is normal with unknown
mean = µ and sd = σ. Suppose X1, ..., Xn is a
random sample from this population with n ≤ 30.
To test the hypothesis H0 : µ = µ0 versus the
alternative H1 : µ > µ0, consider the standartized
random variable
T =
X̄ − µ0
√ ,
S/ n
here again
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n
1 X
Xi
X̄ =
n i=1
and
v
u
u
u
t
n
1 X
S=
(Xi − X̄)2.
n − 1 i=1
When H0 is true, T statistic has t-distribution
(Student’s distribution) with degree of freedom
(d.f.) = n - 1.
This knowledge allows us to construct
the rejection region R : T ≥ tα , such that
P (type I error) = P (H0 is rejected when it is true) = P (T ≥ tα ) =
Here α is the specified level of significance
(usually = .05 ; .01).
Remark. The test statistic is the same as in the
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large sample case but is labeled by T to emphasis
that its
distribution is t-distribution with d.f. n−1 rather
then
standard normal z - distribution.
Examples:
1. A random sample of size 20 from a normal
population has x̄ = 182 and s = 2.3. To test
H0 : µ = 181 against H1 : µ > 181 with α = .05.
The null hypothesis H0 : µ = 181, the alternative
hypothesis H1 : µ > 181. The test statistic
X − 181
√
S/ 20
Since the sample size n = 20, the statistic T
T =
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has t-distribution with d.f. = 19. The specified
level
of significance is α = .05. The upper .05 point
of t-distribution with d.f. = 19 is t.05 = 1.729,
so that the rejection region is R : T ≥ 1.729.
Now let us compare the calculated value of T
with
the value t.05 = 1.729, we get
X − 181 182 − 181
√
√
=
= 1.94 > 1.729
S/ 20
2.3/ 20
i.e. the value of T lies in the rejection region R.
T =
So that we reject the null hypothesis H0 in favor
of
the alternative hypothesis H1 : µ > 181.
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