Confidence Intervals
for Variance and
Standard Deviation
The Chi-Square
Chi Square Distribution
The point estimate for σ2 is s2, and the point estimate for σ
is s. s2 is the most unbiased estimate for σ2.
You can use the chi-square distribution to construct a
confidence interval for the variance and standard
deviation.
If the random variable x has a normal distribution, then
the distribution of
s
χ 2 = (n − 1)
2
σ
2
forms a chi-square
chi square distribution for samples of any size
n > 1.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
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1
The Chi-Square
Chi Square Distribution
Four properties of the chi-square distribution are as follows.
1. All chi-square values χ2 are greater than or equal to zero.
2. The chi-square distribution is a family of curves, each
determined by the degrees of freedom. To form a confidence
interval for σ2, use the χ2-distribution with degrees of
freedom. To form a confidence interval for σ2, use the
χ2-distribution with degrees of freedom equal to one less
than the sample size.
3. The area under each curve of the chi-square distribution
equals one.
4. Find the critical value zc that corresponds to the given level
of confidence.
5. Chi-square distributions are positively skewed.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
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Critical Values for X2
There are two critical values for each level of confidence.
The value χ2R represents the right-tail critical value and χ2L
represents the left-tail critical value.
1−
1 −c
2
X2R
X2
X2
X2L
Area to the right of X2R
(1 −2 c ) = 1 +2 c
Area to the right of X2L
1 −c
2
X2L
c
1 −c
2
X2R
The area between the left and
right critical values is c.
X2
Larson & Farber, Elementary Statistics: Picturing the World, 3e
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2
Critical Values for X2
Example:
Example
Find the critical values χ2R and χ2L for an 80% confidence
when the sample size is 18.
Because the sample size is 18, there are
d.f. = n – 1 = 18 – 1 = 17 degrees of freedom,
Area to the right of χ2R = 1 − c = 1 − 0.8 = 0.1
2
2
Area to the right of χ2L = 1 + c = 1 + 0.8 = 0.9
2
2
Use the Chi-square distribution table to find the
critical values.
Continued.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
5
Critical Values for X2
Example continued:
Appendix B: Table 6: χ2-Distribution
Degrees of
freedom
1
2
3
0.010
0.072
0.975
- 0.001
0.020 0.051
0.115 0.216
α
0.95
0.004
0.103
0.352
5.142
5.697
6.265
5.812
6.408
7.015
7.962 9.312 23.542 26.296
8.672 10.085 24.769 27.587
9.390 10.865 25.989 28.869
0.995
16
17
18
0.99
6.908
7.564
8.231
0.90
0.016
0.211
0.584
0.10
2.706
4.605
6.251
0.05
3.841
5.991
7.815
χ2R = 24.769
χ2L = 10.085
Larson & Farber, Elementary Statistics: Picturing the World, 3e
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3
Confidence Intervals for σ2 and σ
A c-confidence interval for a population variance and
standard deviation is as follows.
Confidence Interval for σ2:
(n − 1)s 2
X R2
<σ2 <
(n − 1)s 2
X L2
Confidence Interval for σ:
(n − 1)s 2
2
XR
<σ <
(n − 1)s 2
X L2
The probability that the confidence intervals contain
σ2 or σ is c.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
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Confidence Intervals for σ2 and σ
Constructing a Confidence Interval for a Variance and a Standard
Deviation
In Words
In Symbols
1. Verify that the population has a
normal distribution.
2. Identify the sample statistic n and
the degrees of freedom.
3. Find the point estimate s2.
4. Find the critical value χ2R and χ2L
that correspond to the given level of
confidence.
d.f. = n − 1
2
s 2 = ∑(x − x )
n −1
Use Table 6 in
Appendix B.
Continued.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
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4
Confidence Intervals for σ2 and σ
Constructing a Confidence Interval for a Variance and a Standard
Deviation
In Words
In Symbols
5. Find the left and right endpoints
and form the confidence interval.
6. Find the confidence interval for
the population standard deviation
by taking the square root of each
endpoint.
(n − 1)s 2
2
XR
(n − 1)s 2
2
XR
<σ2 <
<σ <
(n − 1)s 2
X L2
(n − 1)s 2
X L2
Larson & Farber, Elementary Statistics: Picturing the World, 3e
9
Constructing a Confidence Interval
Example:
You randomly select and weigh 41 samples of 16-ounce bags of
potato chips. The sample standard deviation is 0.05 ounce.
Assuming the weights are normally distributed, construct a 90%
confidence interval for the population standard deviation.
d.f. = n – 1 = 41 – 1 = 40 degrees of freedom,
Area to the right of χ2R = 1 − c = 1 − 0.9 = 0.05
2
2
Area to the right of χ2L = 1 + c = 1 + 0.9 = 0.95
2
2
The critical values are χ2R = 55.758 and χ2L = 26.509.
Continued.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
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5
Constructing a Confidence Interval
Example continued:
χ2L = 26.509
χ2R = 55.758
Left endpoint = ?
(n − 1)s 2
χ R2
Right endpoint = ?
•
(41 − 1)(0.05)2
=
55.758
≈ 0.04
•
(n − 1)s 2
χ L2
(41 − 1)(0.05)2
=
26.509
≈ 0.06
0.04 < σ < 0.06
With 90% confidence we can say that the population
standard deviation is between 0.04 and 0.06 ounces.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
11
Hypothesis Testing
for Variance and
Standard Deviation
6
Critical Values for the χ2-Test
Test
Finding Critical Values for the χ2-Distribution
1. Specify the level of significance α.
2. Determine the degrees of freedom d.f. = n – 1.
3. The critical values for the χ2-distribution are found in Table
6 of Appendix B. To find the critical value(s) for a
a. right-tailed test, use the value that corresponds to
d.f. and α.
b. left-tailed test, use the value that corresponds to d.f.
and 1 – α.
c. two-tailed test, use the values that corresponds to
1
d.f. and 12 α and d.f. and 1 – 2 α.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
13
Finding Critical Values for the χ2
Example:
Find the critical value for a left-tailed test when n = 19
and α = 0.05.
There are 18 d.f. The area to the right of the critical
value is 1 – α = 1 – 0.05 = 0.95.
From Table 6, the critical value is χ20 = 9.390.
Example:
Find the critical value for a two-tailed test when n = 26
and α = 0.01.
There are 25 d.f. The areas to the right of the critical
1
1
values are 2 α = 0.005 and 1 – 2 α = 0.995.
From Table 6, the critical values are χ2L = 10.520 and
χ2R = 46.928.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
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7
The Chi-Square
Chi Square Test
The χ2-test for a variance or standard deviation is a
statistical test for a population variance or standard
deviation. The χ2-test can be used when the population is
normal.
The test statistic is s2 and the standardized test statistic
s
χ 2 = (n − 1)
2
σ
2
follows a chi-square distribution with degrees of freedom
d.f. = n – 1.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
15
The Chi-Square
Chi Square Test
Using the χ2-Test for a Variance or Standard Deviation
In Words
In Symbols
1. State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
State H0 and Ha.
2. Specify the level of significance.
Identify α.
3. Determine the degrees of freedom
and sketch the sampling
distribution.
d.f. = n – 1
4. Determine any critical values.
Use Table 6 in
Appendix B.
Continued.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
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8
The Chi-Square
Chi Square Test
Using the χ2-Test for a Variance or Standard Deviation
In Words
In Symbols
5. Determine any rejection regions.
2
6. Find the standardized test
statistic.
s
χ 2 = (n − 1)
σ2
7. Make a decision to reject or fail to
reject the null hypothesis.
If χ2 is in the
rejection region,
reject H0. Otherwise,
fail to reject H0.
8. Interpret the decision in the
context of the original claim.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
17
Hypothesis Test for Standard Deviation
Example:
Example
A college professor claims that the standard deviation
for students taking a statistics test is less than 30. 10
tests are randomly selected and the standard deviation
is found to be 28.8. Test this professor’s claim at the
α = 0.01 level.
H0: σ ≥ 30
Ha: σ < 30
(Claim)
This is a left-tailed test with d.f.= 9 and α = 0.01.
α = 0.01
X2
Continued.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
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9
Hypothesis Test for Standard Deviation
Example continued:
A college professor claims that the standard deviation
for students taking a statistics test is less than 30. 10
tests are randomly selected and the standard deviation
is found to be 28.8. Test this professor’s claim at the
α = 0.01 level.
Ha: σ < 30 (Claim)
H0: σ ≥ 30
χ20 = 2.088
2
2
s = (10 − 1)(28.8)
χ 2 = (n − 1)
σ2
302
α = 0.01
≈ 8.29
X20 = 2.088
X
2
Fail to reject H0.
At the 1% level of significance, there is not enough evidence
to support the professor’s claim.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
19
Hypothesis Test for Variance
Example:
Example
A local balloon company claims that the variance for the
time its helium balloons will stay afloat is 5 hours. A
disgruntled customer wants to test this claim. She
randomly selects 23 customers and finds that the variance
of the sample is 4.5 seconds. At α = 0.05, does she have
enough evidence to reject the company’s claim?
H0: σ2 = 5 (Claim)
Ha: σ2 ≠ 5
This is a two-tailed test with d.f.= 22 and α = 0.05.
1
α = 0.025
2
1
α = 0.025
2
X2L
X2R
X2
Larson & Farber, Elementary Statistics: Picturing the World, 3e
Continued.
20
10
Hypothesis Test for Variance
Example continued:
continued
A local balloon company claims that the variance for the
time its helium balloons will stay afloat is 5 hours. A
disgruntled customer wants to test this claim. She
randomly selects 23 customers and finds that the variance
of the sample is 4.5 seconds. At α = 0.05, does she have
enough evidence to reject the company’s claim?
Ha: σ2 ≠ 5
H0: σ2 = 5 (Claim)
The critical values are χ2L = 10.982 and χ2R = 36.781.
1
α = 0.025
2
1
α = 0.025
2
X2
Continued.
36.781
10.982
Larson & Farber, Elementary Statistics: Picturing the World, 3e
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Hypothesis Test for Variance
Example continued:
continued
A local balloon company claims that the variance for the
time one of its helium balloons will stay afloat is 5 hours. A
disgruntled customer wants to test this claim. She
randomly selects 23 customers and finds that the variance
of the sample is 4.5 seconds. At α = 0.05, does she have
enough evidence to reject the company’s claim?
Ha: σ2 ≠ 5
H0: σ2 = 5 (Claim)
2
s (23 − 1)(4.5)
χ 2 = (n − 1)
=
= 19.8 Fail to reject H0.
5
σ2
19.8
X2
10.982
36.781
At α = 0.05, there is not enough
evidence to reject the claim that the
variance of the float time is 5 hours.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
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