Confidence Intervals for the Variance of a Normal
Population
Confidence Intervals for the Variance of a Normal
Population
Theorem
Let X1 , X2 , . . . , Xn be a random sample from a distribution with
mean µ and variance σ 2 . Then the random variable
P
(Xi − X )2
(n − 1)S 2
=
σ2
σ2
has s chi-squared (χ2 ) probability distribution with n − 1 degrees
of freedom (df).
Confidence Intervals for the Variance of a Normal
Population
Confidence Intervals for the Variance of a Normal
Population
Proposition
A 100(1 − α)% confidence interval for the variance σ 2 of a
normal population has lower limit
(n − 1)s 2 /χ2α ,n−1
2
and upper limit
(n − 1)s 2 /χ21− α ,n−1
2
A confidence interval for σ has lower and upper limits that are
the square roots of the corresponding limits in the interval for σ 2 .
Hypotheses and Test Procedures
Hypotheses and Test Procedures
Definition
The null hypothesis, denoted by H0 , is the claim that is initially
assumed to be true (the “prior belief” claim). The alternative
hypothesis, denoted by Ha , is the assertion that is contradictory
to H0 .
The null hypothesis will be rejected in favor of the alternative
hypothesis only if sample evidence suggests that H0 is false. If the
sample does not strongly contradict H0 , we will continue to believe
in the truth of the null hypothesis.
Hypotheses and Test Procedures
Definition
The null hypothesis, denoted by H0 , is the claim that is initially
assumed to be true (the “prior belief” claim). The alternative
hypothesis, denoted by Ha , is the assertion that is contradictory
to H0 .
The null hypothesis will be rejected in favor of the alternative
hypothesis only if sample evidence suggests that H0 is false. If the
sample does not strongly contradict H0 , we will continue to believe
in the truth of the null hypothesis.
Remark:
There are only two possible conclusions from a hypothesis testing:
reject H0 or fail to reject H0 ; or corresponding accept
Ha or fail to accept Ha .
We NEVER reject Ha .
Hypotheses and Test Procedures
Hypotheses and Test Procedures
A test procedure is specified by the following:
1. A test statistic, a function of the sample data on which the
decision (reject H0 or do not rejece H0 ) is to be based;
2. A rejection region, the set of all test statistic values for
which H0 will be rejected.
The null hypothesis will then be rejected if and only if the observed
or computed test statistic value falls in the rejection region.
Hypotheses and Test Procedures
A test procedure is specified by the following:
1. A test statistic, a function of the sample data on which the
decision (reject H0 or do not rejece H0 ) is to be based;
2. A rejection region, the set of all test statistic values for
which H0 will be rejected.
The null hypothesis will then be rejected if and only if the observed
or computed test statistic value falls in the rejection region.
Remark: The rejection region is more subjective.
Hypotheses and Test Procedures
Hypotheses and Test Procedures
Definition
A type I error consists of rejecting the null hypothesis H0 when it
is true, traditionally denoted the probability of making type I error
by α.
A type II error involves not rejecting H0 when it is false,
traditionally denoted the probability of making type I error by β.
Mathematically,
α = P(Type I Error) = P(reject H0 | H0 )
β = P(Type II Error) = P(fail to reject H0 | Ha )
Hypotheses and Test Procedures
Definition
A type I error consists of rejecting the null hypothesis H0 when it
is true, traditionally denoted the probability of making type I error
by α.
A type II error involves not rejecting H0 when it is false,
traditionally denoted the probability of making type I error by β.
Mathematically,
α = P(Type I Error) = P(reject H0 | H0 )
β = P(Type II Error) = P(fail to reject H0 | Ha )
Remark:
There are no error-free test procedures unless we have the
information for the whole population.
Therefore, we look for those test for which the probability of
making either type of error is small.
Test about a Population Mean
Test about a Population Mean
Example:
To determine whether the pipe welds in a nuclear power plant
meet specifications, a random sample of 10 welds is selected, and
tests are conducted on each weld in the sample. The sample data
is recorded as follows
101.9 100.4 101.2 100.9 101.7
with X = 101.10.
101.5 100.9 100.1 101.6 100.8
It is known that the weld strength is normally distributed with
mean µ and standard deviation σ = 2. If the specifications state
that the mean strength should be equal to 100 lb/in2 , shall we
accept that the pipe welds meet the specifications with significance
level .05?
Test about a Population Mean
Test about a Population Mean
1. Parameter of interest:
µ = population average strength.
Test about a Population Mean
1. Parameter of interest:
µ = population average strength.
2. Null hypothesis:
H0 : µ = µ0 = 100.
Test about a Population Mean
1. Parameter of interest:
µ = population average strength.
2. Null hypothesis:
H0 : µ = µ0 = 100.
3. Alternative hypothesis:
Ha : µ 6= 100.
Test about a Population Mean
1.
2.
3.
4.
Parameter of interest:
µ = population average strength.
Null hypothesis:
H0 : µ = µ0 = 100.
Alternative hypothesis:
Ha : µ 6= 100.
Test statistic value:
z=
x̄ − µ0
x̄ − 100
√ =
√
σ/ n
2/ n
Test about a Population Mean
1.
2.
3.
4.
Parameter of interest:
µ = population average strength.
Null hypothesis:
H0 : µ = µ0 = 100.
Alternative hypothesis:
Ha : µ 6= 100.
Test statistic value:
z=
x̄ − µ0
x̄ − 100
√ =
√
σ/ n
2/ n
5. Rejection region: z ≥ z.025 or z ≤ −z.025 , where z.025 = 1.96.
Test about a Population Mean
1.
2.
3.
4.
Parameter of interest:
µ = population average strength.
Null hypothesis:
H0 : µ = µ0 = 100.
Alternative hypothesis:
Ha : µ 6= 100.
Test statistic value:
z=
x̄ − µ0
x̄ − 100
√ =
√
σ/ n
2/ n
5. Rejection region: z ≥ z.025 or z ≤ −z.025 , where z.025 = 1.96.
6. Substituting n = 10 and x̄ = 101.10,
z=
101.10 − 100
√
= 1.74
2/ 10
Test about a Population Mean
1.
2.
3.
4.
Parameter of interest:
µ = population average strength.
Null hypothesis:
H0 : µ = µ0 = 100.
Alternative hypothesis:
Ha : µ 6= 100.
Test statistic value:
z=
x̄ − µ0
x̄ − 100
√ =
√
σ/ n
2/ n
5. Rejection region: z ≥ z.025 or z ≤ −z.025 , where z.025 = 1.96.
6. Substituting n = 10 and x̄ = 101.10,
z=
101.10 − 100
√
= 1.74
2/ 10
7. Since −1.96 < 1.74 < 1.96, i.e., the value of the test statistic
does not fall in the rejection region (−∞, 1.96) ∪ (1.96, ∞),
we can not reject H0 at significance level .05.
Test about a Population Mean
Test about a Population Mean
Test for Population Mean of A Normal Population with Known
σ
Null hypothesis:
Test statistic value
H0 : µ = µ0
x̄−µ
√0
z = σ/
n
Alternative Hypothesis
Ha : µ > µ0
Ha : µ < µ0
Ha : µ 6= µ0
Rejection Region for Level α Test
z ≥ zα (upper-tailed test)
z ≤ −zα (lower-tailed test)
z ≥ zα/2 or z ≤ −zα/2 (two-tailed test)
Test about a Population Mean
Test about a Population Mean
Example:
To determine whether the pipe welds in a nuclear power plant
meet specifications, a random sample of 10 welds is selected, and
tests are conducted on each weld in the sample. The sample data
is recorded as follows
101.9 100.4 101.2 100.9 101.7
with X = 101.10.
101.5 100.9 100.1 101.6 100.8
It is known that the weld strength is normally distributed with
mean µ and standard deviation σ = 2. If the specifications state
that the mean strength should be equal to 100 lb/in2 , shall we
accept that the pipe welds meet the specifications with significance
level .05? What is the probability of making type II error then?
Test about a Population Mean
Test about a Population Mean
H0 : µ = µ0 v.s. Ha : µ > µ0
Then the rejection region for level α test is z ≥ zα , or equivalently
√
x̄ ≥ µ0 + zα · σ/ n.
Let µ0 denote a particular value of µ that is less than the null value
µ0 , then
β(µ0 ) = P(H0 is not rejected | µ = µ0 )
√
= P(X < µ0 + zα · σ/ n | µ = µ0 )
X − µ0
µ0 − µ 0
0
√ < zα +
√ |µ=µ
=P
σ/ n
σ/ n
µ0 − µ 0
√
= Φ zα +
σ/ n
Test about a Population Mean
Test about a Population Mean
Alternative Hypothesis
Ha : µ > µ0
Ha : µ < µ0
Ha : µ 6= µ0
Type II Error Probability β(µ0 ) for Level α Test
0
√
Φ zα + µσ/0 −µ
n
0
µ0 −µ
1 − Φ −zα + σ/√n
0
√
Φ zα/2 + µσ/0 −µ
−
Φ
−zα/2 +
n
0
µ0 −µ
√
σ/ n
Test about a Population Mean
Test about a Population Mean
Example:
To determine whether the pipe welds in a nuclear power plant
meet specifications, a random sample of n welds is selected, and
tests are conducted on each weld in the sample.
It is known that the weld strength is normally distributed with
mean µ and standard deviation σ = 2. And the specifications state
that the mean strength should be equal to 100 lb/in2 . To
construct a hypothesis with α = .05 and β = .1, how large should
n be?
Test about a Population Mean
Test about a Population Mean
The sample size n for which a level α test also has β(µ0 ) = β at
the alternative value µ0 is
h
i
σ(zα +z ) 2


for a one-tailed (upper or lower) test
 µ0 −µ0β
n = σ(z +z ) 2

β
α/2

for a two-tailed test (an approximate solution)

µ0 −µ0
Test about a Population Mean
Test about a Population Mean
Example:
To determine whether the pipe welds in a nuclear power plant
meet specifications, a random sample of 50 welds is selected, and
tests are conducted on each weld in the sample. The average
strength of this sample is X = 101.10, and the standard deviation
is s = 2. If the specifications state that the mean strength should
exceed 100 lb/in2 , shall we accept that the pipe welds meet the
specifications with significance level .05?
Test about a Population Mean
Test about a Population Mean
Example:
To determine whether the pipe welds in a nuclear power plant
meet specifications, a random sample of 10 welds is selected, and
tests are conducted on each weld in the sample. The sample data
is recorded as follows
101.9 100.4 101.2 100.9 101.7
with X = 101.10 and
101.5 100.9 100.1 101.6 100.8
s = .585.
It is known that the weld strength is normally distributed with
mean µ. If the specifications state that the mean strength should
exceed 100 lb/in2 , shall we accept that the pipe welds meet the
specifications with significance level .05?
Test about a Population Mean
Test about a Population Mean
Test for Population Mean of A Normal Population with
Unknown σ
Null hypothesis:
Test statistic value
H0 : µ = µ0
√0
t = x̄−µ
s/ n
Alternative Hypothesis
Ha : µ > µ0
Ha : µ < µ0
Ha : µ 6= µ0
Rejection Region for Level α Test
t ≥ tα,n−1 (upper-tailed test)
t ≤ −tα,n−1 (lower-tailed test)
t ≥ tα/2,n−1 or t ≤ −tα/2,n−1 (two-tailed test)