Section 8.5 Testing a claim about a mean ( σ unknown )
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Transcript of Section 8.5 Testing a claim about a mean ( σ unknown )
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Section 8.5Testing a claim about a mean
(σ unknown)
ObjectiveFor 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 distributedor
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 TestEnter 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-ValuesA 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-ValuesFrom 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 DieWhen 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