Download Ch8-5

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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