Download - Level of Significance
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Level of Significance
• α is a predetermined value by convention usually 0.05
• α = 0.05 corresponds to the 95% confidence level
• We are accepting the risk that out of 100 samples, we would reject a true null hypothesis five times
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Sampling Distribution Of Means
• A sampling distribution of means is the relative frequency distribution of the means of all possible samples of size n that could be selected from the population.
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One Sample Test
• Compares mean of a sample to known population mean– Z-test– T-test
This lecture focuses onone sample t-test
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The One Sample t – Test
Testing statistical hypothesis about µ when σ is not known OR sample size is
small
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An Example Problem
• Suppose that Dr. Tate learns from a national survey that the average undergraduate student in the United States spends 6.75 hours each week on the Internet – composing and reading e-mail, exploring the Web and constructing home pages. Dr. Tate is interested in knowing how Internet use among students at George Mason University compares with this national average.
• Dr. Tate randomly selects a sample of only 10 students. Each student is asked to report the number of hours he or she spends on the Internet in a typical week during the academic year.
Populaon mean
Small sample
Population variance is unknown & estimated from sample
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Steps in Test of Hypothesis
1. Determine the appropriate test 2. Establish the level of significance:α3. Determine whether to use a one tail or two
tail test4. Calculate the test statistic5. Determine the degree of freedom6. Compare computed test statistic against a
tabled value
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1. Determine the appropriate test
If sample size is more than 30 use z-test If sample size is less than 30 use t-test
Sample size of 10
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2. Establish Level of Significance
• α is a predetermined value
• The convention• α = .05
• α = .01
• α = .001
• In this example, assume α = 0.05
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3. Determine Whether to Use a One or Two Tailed Test
• H0 :µ = 6.75
• Ha :µ ≠ 6.75 A two tailedtest because it
can be either larger
or smaller
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4. Calculating Test Statistics
StudentNumber of Hours(X)
A 6 -3.9 15.21B 9 -0.9 0.81C 12 2.1 4.41D 3 -6.9 47.61E 11 1.1 1.21F 10 0.1 0.01G 18 8.1 56.61H 9 -0.9 0.81I 13 3.1 9.61J 8 -1.9 3.61
Σ=148.90
2)( XX )( XX 2)( XX
90.9X
StudentNumber of Hours(X)
A 6 -3.9 15.21B 9 -0.9 0.81C 12 2.1 4.41D 3 -6.9 47.61E 11 1.1 1.21F 10 0.1 0.01G 18 8.1 56.61H 9 -0.9 0.81I 13 3.1 9.61J 8 -1.9 3.61
Σ=148.90
2)( XX )( XX 2)( XX
90.9X
Sample mean
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4. Calculating Test Statistics
StudentNumber of Hours(X)
A 6 -3.9 15.21B 9 -0.9 0.81C 12 2.1 4.41D 3 -6.9 47.61E 11 1.1 1.21F 10 0.1 0.01G 18 8.1 56.61H 9 -0.9 0.81I 13 3.1 9.61J 8 -1.9 3.61
Σ=148.90
2)( XX )( XX 2)( XX
90.9X
StudentNumber of Hours(X)
A 6 -3.9 15.21B 9 -0.9 0.81C 12 2.1 4.41D 3 -6.9 47.61E 11 1.1 1.21F 10 0.1 0.01G 18 8.1 56.61H 9 -0.9 0.81I 13 3.1 9.61J 8 -1.9 3.61
Σ=148.90
2)( XX )( XX 2)( XX
90.9X
Deviationfrom sample
mean
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StudentNumber of Hours(X)
A 6 -3.9 15.21B 9 -0.9 0.81C 12 2.1 4.41D 3 -6.9 47.61E 11 1.1 1.21F 10 0.1 0.01G 18 8.1 56.61H 9 -0.9 0.81I 13 3.1 9.61J 8 -1.9 3.61
Σ=148.90
)( XX
4. Calculating Test Statistics
Squared deviation
from samplemean
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4. Calculating Test Statistics
Standard deviation
of observations
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4. Calculating Test Statistics
Calculated
t value
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4. Calculating Test Statistics
Standard deviationof sample means
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4. Calculating Test Statistics
Calculated t
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5. Determine Degrees of Freedom
• Degrees of freedom, df, is value indicating the number of independent pieces of information a sample can provide for purposes of statistical inference.
• Df = Sample size – Number of parameters estimated
• Df is n-1 for one sample test of mean because the population variance is estimated from the sample
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Degrees of Freedom• Suppose you have a sample of three observations:
2 -1 1
2 -1 1
5 +2 4
-------- ------ ------
Σ=0 6
-------- ----------------2)( XX )( XX X
3X
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Degrees of Freedom Continued
• For your sample scores, you have only two independent pieces of information, or degrees of freedom, on which to base your estimates of S and xS
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6. Compare the Computed Test Statistic Against a Tabled Value
• α = .05
• Df = n-1 = 9
• Therefore, reject H0
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Decision Rule for t-Scores
If |tc| > |tα| Reject H0
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Decision Rule for P-values
If p value < α Reject H0
Pvalue is one minusprobability of observing
the t-value calculated from our sample
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Example of Decision Rules
• In terms of t score:
|tc = 2.449| > |tα= 2.262| Reject H0
• In terms of p-value:
If p value = .037 < α = .05 Reject H0
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Constructing a Confidence Interval for µ
Sam
ple m
ean
Standard deviation of sample means
Critical t value
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Constructing a Confidence Interval for µ for the Example
• Sample mean is 9.90
• Critical t value is 2.262
• Standard deviation of sample means is 1.29
• 9.90 + 2.262 * 1.29
• The estimated interval goes from 6.98 to 12.84