Section 8.6
Objective
For a population with standard deviation σ , use a sample too test a claim about the standard deviation.
Tests of a standard deviation use the c
2 -distribution
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2

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(1) The sample is a simple random sample
(2) The population is normally distributed
Very strict condition!!!
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Denoted c
2 (as in c
2 -score) since the test uses the c
2 -distribution .
n Sample size s Sample standard deviation
σ
0
Claimed standard deviation
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Right-tailed test “>“
Needs one critical value ( right tail)
Use StatCrunch : Chi-Squared Calculator
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Left-tailed test “<”
Needs one critical value ( left tail)
Use StatCrunch : Chi-Squared Calculator
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Two-tailed test “ ≠ “
Needs two critical values ( right and left tail)
Use StatCrunch : Chi-Squared Calculator
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Example 1
Statisics Test Scores Problem 14, pg 449
Tests scores in the author’s previous statistic classes have followed a normal distribution with a standard deviation equal to 14.1. His current class has 27 tests scores with a standard deviation of 9.3.
Use a 0.01 significance level to test the claim that this class has less variation than the past classes.
What we know: σ
0
= 14.1 n = 27 s = 9.3
Claim: σ < 14.1 using α = 0.01
Note: Test conditions are satisfied since population is normally distributed
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Example 1 Using Critical Regions
What we know: σ
0
= 14.1 n = 27 s = 9.3
Claim: σ < 14.1 using α = 0.01
H
0
: σ = 14.1
H
1
: σ < 14.1
Left-tailed
Test statistic:
Critical value: ( df = 26) c
2
L
= 12.20
c
2
= 11.31
c
2 in critical region
Initial Conclusion: Since c
2 in critical region, Reject H
0
Final Conclusion: Accept the claim that the new class has less variance than the past classes
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Calculating P-value for a Variance
Stat → Variance → One sample → with summary
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Calculating P-value for a Variance
Enter the Sample variance ( s 2 )
Sample size ( n )
NOTE : Must use Variance s 2 = 9.3
2 = 86.49
Then hit Next
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Calculating P-value for a Variance
Select Hypothesis Test
Enter the Null:variance ( σ
0
2 )
Select Alternative
(“ < “, “ > ”, or “ ≠ ”)
σ
0
2 = 14.1
2 = 198.81
Then hit Calculate
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Calculating P-value for a Variance
The resulting table shows both the test statistic ( c
2 ) and the P-value
Test statistic ( c
2 )
P-value
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Example 1 Using Critical Regions
What we know: σ
0
= 14.1 n = 27 s = 9.3
Claim: σ < 14.1 using α = 0.01
H
0
: σ = 14.1
H
1
: σ < 14.1
Using
StatCrunch
Stat → Variance→ One sample → With summary
Sample variance: 86.49
● Hypothesis Test
Null : proportion=
Sample size:
27
Alternative
198.81
<
P-value = 0.0056
s 2 = 86.49
σ
0
2 = 198.81
Initial Conclusion: Since P-value < α ( α = 0.01), Reject H
0
Final Conclusion: Accept the claim that the new class has less variance than the past classes
We are 99.44% confident the claim holds
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Example 2
BMI for Miss America Problem 17, pg 449
Listed below are body mass indexes (BMI) for recent
Miss America winners. In the 1920s and 1930s, distribution of the BMIs formed a normal distribution with a standard deviation of 1.34.
Use a 0.01 significance level to test the claim that recent
Miss America winners appear to have variation that is different from that of the 1920s and 1930s.
19.5 20.3 19.6 20.2 17.8 17.9 19.1 18.8 17.6 16.8
Using StatCrunch : s = 1.1862172
What we know: σ
0
= 1.34 n = 10 s = 1.186
Claim: σ ≠ 1.34 using α = 0.01
Note: Test conditions are satisfied since population is normally distributed
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Example 2 Using Critical Regions
What we know: σ
0
= 1.34 n = 10 s = 1.186
Claim: σ ≠ 1.34 using α = 0.01
H
0
: σ = 1.34
H
1
: σ ≠ 1.34
two-tailed
Test statistic:
0.005
Critical values: ( df = 26) c
2
L
= 2.088
c
2
R
= 26.67
c
2
= 7.053
c
2 not in critical region
Initial Conclusion: Since c
2 not in critical region, Accept H
0
Final Conclusion: Reject the claim recent winners have a different variations than in the 20s and 30s
S ince H
0 accepted, the observed significance isn’t useful.
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Example 2 Using P-value
What we know: σ
0
= 1.34 n = 10 s = 1.186
Claim: σ ≠ 1.34 using α = 0.01
H
0
: σ 2 = 1.796
H
1
: σ 2 < 1.796
Using
StatCrunch
Stat → Variation → One sample → With summary
Sample variance: 1.407
● Hypothesis Test
Null : proportion=
Sample size:
10
Alternative
P-value = 0.509
s 2 = 1.407
σ
0
2 = 1.796
1.796
<
Initial Conclusion: Since P-value ≥ α ( α = 0.01), Accept H
0
Final Conclusion: Reject the claim recent winners have a different variations than in the 20s and 30s
S ince H
0 accepted, the observed significance isn’t useful.
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