1 section 8.5 testing a claim about a mean (σ unknown) objective for a population with mean µ...
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Section 8.5Testing 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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(1) The population standard deviation σ is unknown
(2) One or both of the following:
Requirements
The population is normally distributed
or
The sample size n > 30
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Test StatisticDenoted t (as in t-score) since the test uses the t-distribution.
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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 s = 26.33
Claim: µ > 166.3 using α = 0.1
Note: Conditions for performing test are satisfied since n >30
Example 1
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What we know: µ0 = 166.3 n = 40 x = 172.55 s = 26.33
Claim: µ > 166.3 using α = 0.1
H0 : µ = 166.3
H1 : µ > 166.3 right-tailed test
Initial Conclusion: Since t in critical region, Reject H0
Final Conclusion: Accept the claim that the mean weight is greater than 166.3 lb.
t in critical region(df = 39)
Using Critical RegionsExample 1
tα = 1.304
t = 1.501
Test statistic:
Critical value:
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Stat → T statistics → One sample → with summary
Calculating P-value for a Mean(σ unknown)
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Then hit Next
Enter the Sample mean (x)
Sample std. dev. (s)
Sample size (n)
Calculating P-value for a Mean(σ unknown)
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Then hit Calculate
Select Hypothesis Test
Enter the Null:mean (µ0)
Select Alternative (“<“, “>”, or “≠”)
Calculating P-value for a Mean(σ unknown)
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Test statistic (t)
P-value
Calculating P-value for a Mean(σ unknown)
The resulting table shows both the test statistic (t) and the P-value
Initial Conclusion
Since P-value < α (α = 0.1), reject H0
Final Conclusion
Accept the claim the mean weight greater than 166.3 Ib
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Using StatCrunch
Using the P-valueExample 1
Stat → T statistics→ One sample → With summary
Null: proportion=
Alternative
Sample mean:
Sample std. dev.:
Sample size:
● Hypothesis Test172.55
37.8
40
166.3
>
P-value = 0.0707
What we know: µ0 = 166.3 n = 40 x = 172.55 s = 26.33
Claim: µ > 166.3 using α = 0.1
Initial Conclusion: Since P-value < α, Reject H0
Final Conclusion: Accept the claim that the mean weight is greater than 166.3 lb.
H0 : µ = 166.3
H1 : µ > 166.3
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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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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 s = 1.31
Claim: µ ≠ 3.5 using α = 0.01
Note: Conditions for performing test are satisfied since n >30
Example 2
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H0 : µ = 3.5
H1 : µ ≠ 3.5
What we know: µ0 = 3.5 n = 100 x = 3.87 s = 1.31
Claim: µ ≠ 3.5 using α = 0.01
two-tailed test
Example 2
t in critical region(df = 99)
Test statistic:
Critical value:z = 3.058
zα = -2.626 zα = 2.626
Using Critical Regions
Initial Conclusion: Since P-value < α, Reject H0
Final Conclusion: Accept the claim the die is loaded.
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Using StatCrunch
Using the P-valueExample 2
Null: proportion=
Alternative
Sample mean:
Sample std. dev.:
Sample size:
● Hypothesis Test3.87
1.31
100
3.5
≠
P-value = 0.0057
Initial Conclusion: Since P-value < α, Reject H0
Final Conclusion: Accept the claim the die is loaded.
H0 : µ = 3.5
H1 : µ ≠ 3.5
What we know: µ0 = 3.5 n = 100 x = 3.87 s = 1.31
Claim: µ ≠ 3.5 using α = 0.01
We are 99.43% confidence the die are loaded
Stat → T statistics→ One sample → With summary
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Section 8.6Testing 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 2-distribution
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Notation
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Notation
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(1) The sample is a simple random sample
(2) The population is normally distributed
Very strict condition!!!
Requirements
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Test StatisticDenoted 2 (as in 2-score) since the test uses the 2 -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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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.
Example 1
What we know: σ0 = 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
Problem 14, pg 449
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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
Using Critical RegionsExample 1
2 in critical region
(df = 26)
Initial Conclusion: Since 2 in critical region, Reject H0
Final Conclusion: Accept the claim that the new class has less variance than the past classes
2
2L
Test statistic:
Critical value:
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Calculating P-value for a Variance
Stat → Variance → One sample → with summary
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Then hit Next
Enter the Sample variance (s2)
Sample size (n)
Calculating P-value for a Variance
s2 = 9.32 = 86.49
NOTE: Must use Variance
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Then hit Calculate
Select Hypothesis Test
Enter the Null:variance (σ02)
Select Alternative (“<“, “>”, or “≠”)
Calculating P-value for a Variance
σ02 = 14.12 = 198.81
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Test statistic (2)
P-value
The resulting table shows both the test statistic (2) and the P-value
Calculating P-value for a Variance
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What we know: σ0 = 14.1 n = 27 s = 9.3
Claim: σ < 14.1 using α = 0.01
Using Critical RegionsExample 1
Using StatCrunch
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
Stat → Variance → One sample → With summary
Null: proportion=
Alternative
Sample variance:
Sample size:
86.49
27
198.81
<
● Hypothesis Test
P-value = 0.0056s2 = 86.49
σ02 = 198.81
H0 : σ2 = 198.81
H1 : σ2 < 198.81
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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.
Example 2
What we know: σ0 = 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
Problem 17, pg 449
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
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Using Critical RegionsExample 2
2 not in critical region(df = 26)
Test statistic:
Critical values: 2
0.005
2R
2L
Initial Conclusion: Since 2 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.
What we know: σ0 = 1.34 n = 10 s = 1.186
Claim: σ ≠ 1.34 using α = 0.01
H0 : σ = 1.34
H1 : σ ≠ 1.34 two-tailed
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Using P-valueExample 2
Using StatCrunch
Null: proportion=
Alternative
Sample variance:
Sample size:
1.407
10
1.796
<
● Hypothesis Test
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.
s2 = 1.407
σ02 = 1.796
What we know: σ0 = 1.34 n = 10 s = 1.186
Claim: σ ≠ 1.34 using α = 0.01
H0 : σ2 = 1.796
H1 : σ2 < 1.796
Stat → Variance → One sample → With summary