Transcript Chapter 8

Section 8.5
Testing a claim about a mean
(σ unknown)
Objective
For a population with mean µ (with σ unknown),
use a sample to test a claim about the mean.
Testing a mean (when σ known) uses the
t-distribution
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Notation
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Requirements
(1) The population standard deviation
σ is unknown
(2) One or both of the following:
The population is normally distributed
or
The sample size n > 30
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Test Statistic
Denoted t (as in t-score) since
the test uses the t-distribution.
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Example 1
People have died in boat accidents because an obsolete
estimate of the mean weight (of 166.3 lb.) was used.
A random sample of n = 40 men yielded the mean
x = 172.55 lb. and standard deviation s = 26.33 lb.
Do not assume the population standard deviation 
is known.
Test the claim that men have a mean weight greater
than 166.3 lb. using 90% confidence.
What we know:
µ0 = 166.3 n = 40 x = 172.55
Claim: µ > 166.3 using α = 0.1
s = 26.33
Note: Conditions for performing test are satisfied since n >30
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Example 1 Using Critical Regions
What we know:
µ0 = 166.3 n = 40 x = 172.55
Claim: µ > 166.3 using α = 0.1
s = 26.33
H0 : µ = 166.3
H1 : µ > 166.3 right-tailed test
Test statistic:
tα = 1.304
t = 1.501
Critical value:
(df = 39)
t in critical region
Initial Conclusion: Since t in critical region, Reject H0
Final Conclusion: Accept the claim that the mean weight
is greater than 166.3 lb.
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Calculating P-value for a Mean
(σ unknown)
Stat → T statistics → One sample → with summary
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Calculating P-value for a Mean
(σ unknown)
Enter the
Sample mean (x)
Sample std. dev. (s)
Sample size (n)
Then hit Next
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Calculating P-value for a Mean
(σ unknown)
Select
Enter the
Select
Hypothesis Test
Null:mean (µ0)
Alternative (“<“, “>”, or “≠”)
Then hit Calculate
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Calculating P-value for a Mean
(σ unknown)
The resulting table shows both the
test statistic (t) and the P-value
Test statistic (t)
P-value
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Example 1 Using the P-value
What we know:
H0 : µ = 166.3
H1 : µ > 166.3
Using
StatCrunch
µ0 = 166.3 n = 40 x = 172.55
Claim: µ > 166.3 using α = 0.1
s = 26.33
Stat → T statistics→ One sample → With summary
Sample mean: 172.55
Sample std. dev.: 37.8
Sample size:
40
● Hypothesis Test
Null: proportion=
Alternative
166.3
>
P-value = 0.0707
Initial Conclusion: Since P-value < α, Reject H0
Final Conclusion: Accept the claim that the mean weight
is greater than 166.3 lb.
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P-Values
A useful interpretation of the P-value: it is
observed level of significance
Thus, the value 1 – P-value is interpreted as
observed level of confidence
Recall: “Confidence Level” = 1 – “Significance Level”
Note: Only useful if we reject H0
If H0 accepted, the observed significance and
confidence are not useful.
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P-Values
From Example 1:
P-value = 0.0707
1 – P-value = 0.9293
Thus, we can say conclude the following:
The claim holds under 0.0707 significance.
or equivalently…
We are 92.93% confident the claim holds
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Example 2
Loaded Die
When a fair die (with equally likely outcomes 1-6) is
rolled many times, the mean valued rolled should be 3.5
Your suspicious a die being used at a casino is loaded
(that is, it’s mean is a value other than 3.5)
You record the values for 100 rolls and end up with a
mean of 3.87 and standard deviation 1.31
Using a confidence level of 99%, does the claim that the
dice are loaded?
What we know:
µ0 = 3.5
n = 100
x = 3.87
Claim: µ ≠ 3.5 using α = 0.01
s = 1.31
Note: Conditions for performing test are satisfied since n >30
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Example 2 Using Critical Regions
What we know:
µ0 = 3.5
n = 100
Claim: µ ≠ 3.5 using
x = 3.87
α = 0.01
s = 1.31
H0 : µ = 3.5
H1 : µ ≠ 3.5 two-tailed test
Test statistic:
zα = -2.626
zα = 2.626
z = 3.058
Critical value:
(df = 99)
t in critical region
Initial Conclusion: Since P-value < α, Reject H0
Final Conclusion: Accept the claim the die is loaded.
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Example 2 Using the P-value
What we know:
H0 : µ = 3.5
H1 : µ ≠ 3.5
Using
StatCrunch
µ0 = 3.5
n = 100
Claim: µ ≠ 3.5 using
x = 3.87
α = 0.01
s = 1.31
Stat → T statistics→ One sample → With summary
● Hypothesis Test
Sample mean:
3.87
Sample std. dev.:
1.31
Null: proportion=
Sample size:
100
Alternative
3.5
≠
P-value = 0.0057
Initial Conclusion: Since P-value < α, Reject H0
Final Conclusion: Accept the claim the die is loaded.
We are 99.43% confidence the die are loaded
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Section 8.6
Testing a claim about a
standard deviation
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
c2-distribution
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Notation
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Notation
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Requirements
(1) The sample is a simple random sample
(2) The population is normally distributed
Very strict condition!!!
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Test Statistic
Denoted c2 (as in c2-score) since
the test uses the c2 -distribution.
n
Sample size
s
Sample standard deviation
σ0
Claimed standard deviation
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Critical Values
Right-tailed test “>“
Needs one critical value (right tail)
Use StatCrunch: Chi-Squared Calculator
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Critical Values
Left-tailed test
“<”
Needs one critical value (left tail)
Use StatCrunch: Chi-Squared Calculator
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Critical Values
Two-tailed test
“≠“
Needs two critical values (right and left tail)
Use StatCrunch: Chi-Squared Calculator
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Example 1
Problem 14, pg 449
Statisics Test Scores
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:
σ0 = 14.1
n = 27
Claim: σ < 14.1
s = 9.3
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:
σ0 = 14.1 n = 27 s = 9.3
Claim: σ < 14.1 using α = 0.01
H0 : σ = 14.1
H1 : σ < 14.1 Left-tailed
Test statistic:
c2L = 12.20
Critical value:
(df = 26)
c2 = 11.31
c2 in critical region
Initial Conclusion: Since c2 in critical region, Reject H0
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 (s2)
Sample size (n)
NOTE: Must use Variance
s2 = 9.32 = 86.49
Then hit Next
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Calculating P-value for a Variance
Select
Enter the
Select
Hypothesis Test
Null:variance (σ02)
Alternative (“<“, “>”, or “≠”)
σ02 = 14.12 = 198.81
Then hit Calculate
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Calculating P-value for a Variance
The resulting table shows both the
test statistic (c2) and the P-value
Test statistic (c2)
P-value
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Example 1 Using Critical Regions
What we know:
H0 : σ2 = 198.81
H1 :
σ2
< 198.81
Using
StatCrunch
s2 = 86.49
σ02 = 198.81
σ0 = 14.1 n = 27 s = 9.3
Claim: σ < 14.1 using α = 0.01
Stat → Variance → One sample → With summary
Sample variance:
Sample size:
86.49
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● Hypothesis Test
Null: proportion=
Alternative
198.81
<
P-value = 0.0056
Initial Conclusion: Since P-value < α (α = 0.01), Reject H0
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
Problem 17, pg 449
BMI for Miss America
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:
σ0 = 1.34
n = 10
Claim: σ ≠ 1.34
s = 1.186
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:
σ0 = 1.34
n = 10
Claim: σ ≠ 1.34
s = 1.186
using
α = 0.01
H0 : σ = 1.34
H1 : σ ≠ 1.34 two-tailed
0.005
Test statistic:
c2L = 2.088
Critical values:
(df = 26)
c2R = 26.67
c2 = 7.053
c2 not in critical region
Initial Conclusion: Since c2 not in critical region, Accept H0
Final Conclusion: Reject the claim recent winners have a
different variations than in the 20s and 30s
Since H0 accepted, the observed significance isn’t useful.
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Example 2 Using P-value
What we know:
σ0 = 1.34
n = 10
Claim: σ ≠ 1.34
H0 : σ2 = 1.796
H1 :
σ2
< 1.796
Using
StatCrunch
s2 = 1.407
σ02 = 1.796
s = 1.186
using
α = 0.01
Stat → Variance → One sample → With summary
Sample variance:
Sample size:
1.407
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● Hypothesis Test
Null: proportion=
1.796
Alternative
<
P-value = 0.509
Initial Conclusion: Since P-value ≥ α (α = 0.01), Accept H0
Final Conclusion: Reject the claim recent winners have a
different variations than in the 20s and 30s
Since H0 accepted, the observed significance isn’t useful.
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