section 9.4 inferences about two means (matched pairs)
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Section 9.4 Inferences About Two Means (Matched Pairs). Objective Compare of two matched-paired means using two samples from each population. Hypothesis Tests and Confidence Intervals of two dependent means use the t -distribution. Definition. - PowerPoint PPT PresentationTRANSCRIPT
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Objective
Compare of two matched-paired means using two samples from each population.
Hypothesis Tests and Confidence Intervals of two dependent means use the t-distribution
Section 9.4Inferences About Two Means
(Matched Pairs)
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Definition
Two samples are dependent if there is some relationship between the two samples so that each value in one sample is paired with a corresponding value in the other sample.
Two samples can be treated as the matched pairs of values.
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Examples
• Blood pressure of patients before they are given medicine and after they take it.
• Predicted temperature (by Weather Forecast) and the actual temperature.
• Heights of selected people in the morning and their heights by night time.
• Test scores of selected students in Calculus-I and their scores in Calculus-II.
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Example 1
First sample: weights of 5 students in April
Second sample: their weights in September
These weights make 5 matched pairs
Third line: differences between April weights and September weights (net change in weight for each student, separately)
In our calculations we only use differences (d), not the values in the two samples.
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Notation
d Individual difference between two matched paired values
μd Population mean for the difference of the two values.
n Number of paired values in sample
d Mean value of the differences in sample
sd Standard deviation of differences in sample
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(1) The sample data are dependent (i.e. they make matched pairs)
(2) Either or both the following holds:
The number of matched pairs is large (n>30) orThe differences have a normal distribution
Requirements
All requirements must be satisfied to make a Hypothesis Test or to find a Confidence Interval
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Tests for Two Dependent Means
Goal: Compare the mean of the differences
H0 : μd =
0
H1 : μd ≠ 0
Two tailed Left tailed Right tailed
H0 : μd =
0
H1 : μd < 0
H0 : μd =
0
H1 : μd > 0
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t = d – µdsdn
degrees of freedom: df = n – 1
Note: d
= 0 according to H0
Finding the Test Statistic
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Test Statistic
Note: Hypothesis Tests are done in same way as in Ch.8-5
Degrees of freedom df = n – 1
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Steps for Performing a Hypothesis Test on Two Independent Means
• Write what we know
• State H0 and H1
• Draw a diagram
• Calculate the Sample Stats
• Find the Test Statistic
• Find the Critical Value(s)
• State the Initial Conclusion and Final Conclusion
Note: Same process as in Chapter 8
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Example 1
Assume the differences in weight form a normal distribution.
Use a 0.05 significance level to test the claim that for the population of students, the mean change in weight from September to April is 0 kg (i.e. on average, there is no change)
Claim: μd = 0 using α = 0.05
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H0 : µd = 0
H1 : µd ≠ 0t = 0.186
tα/2 = 2.78
t-dist.df = 4
Test Statistic
Critical Value
Initial Conclusion: Since t is not in the critical region, accept H0
Final Conclusion: We accept the claim that mean change in weight from
September to April is 0 kg.
-tα/2 = -2.78
Example 1
tα/2 = t0.025 = 2.78 (Using StatCrunch, df = 4)
d Data: -1 -1 4 -2 1
Sample Stats
n = 5 d = 0.2 sd = 2.387
Use StatCrunch: Stat – Summary Stats – Columns
Two-TailedH0 = Claim
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H0 : µd = 0
H1 : µd ≠ 0Two-TailedH0 = Claim
Initial Conclusion: Since P-value is greater than α (0.05), accept H0
Final Conclusion: We accept the claim that mean change in weight from
September to April is 0 kg.
Example 1 d Data: -1 -1 4 -2 1Sample Stats
n = 5 d = 0.2 sd = 2.387
Use StatCrunch: Stat – Summary Stats – Columns
Null: proportion=
Alternative
Sample mean:
Sample std. dev.:
Sample size:
● Hypothesis Test0.2
2.387
5
0
≠
P-value = 0.8605
Stat → T statistics→ One sample → With summary
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Confidence Interval Estimate
We can observe how the two proportions relate by looking at the Confidence Interval Estimate of μ1–μ2
CI = ( d – E, d + E )
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Example 2 Find the 95% Confidence Interval Estimate of μd from the data in Example 1
Sample Stats
n = 5 d = 0.2 sd = 2.387
CI = (-2.8, 3.2)
tα/2 = t0.025 = 2.78 (Using StatCrunch, df = 4)