1. Testing the Difference Between Means (Dependent Samples)
Section 8.3
Testing the Difference Between Means (Dependent Samples)
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2. Section 8.3 Objectives
Perform a t-test to test the mean of the differences for a population of paired data
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3. t-Test for the Difference Between Means
To perform a two-sample hypothesis test with dependent samples, the difference between each data pair is first found:
d = x1 – x2 Difference between entries for a data pair
The test statistic is the mean of these differences.
Mean of the differences between paired data entries in the dependent samples
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4. t-Test for the Difference Between Means
Three conditions are required to conduct the test.
The samples must be randomly selected.
The samples must be dependent (paired).
Both populations must be normally distributed.
If these requirements are met, then the sampling distribution for is approximated by a t-distribution with n – 1 degrees of freedom, where n is the number of data pairs.
-t0 t0 μd
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5. Symbols used for the t-Test for μd
Symbol Description n
The number of pairs of data d
The difference between entries for a data pair, d = x1 – x2
The hypothesized mean of the differences of paired data in the population
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6. Symbols used for the t-Test for μd
Symbol Description
The mean of the differences between the paired data entries in the dependent samples sd
The standard deviation of the differences between the paired data entries in the dependent samples
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7. t-Test for the Difference Between Means
The test statistic is
The standardized test statistic is
The degrees of freedom are
d.f. = n – 1.
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8. t-Test for the Difference Between Means (Dependent Samples)
In Words In Symbols
State the claim mathematically and verbally. Identify the null and alternative hypotheses.
Specify the level of significance.
Determine the degrees of freedom.
Determine the critical value(s).
State H0 and Ha.
Identify α.
d.f. = n – 1
Use Table 5 in Appendix B if n > 29 use the last row (∞) .
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9. t-Test for the Difference Between Means (Dependent Samples)
In Words In Symbols
Determine the rejection region(s).
Calculate and
Find the standardized test statistic.
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10. t-Test for the Difference Between Means (Dependent Samples)
In Words In Symbols
Make a decision to reject or fail to reject the null hypothesis.
Interpret the decision in the context of the original claim.
If t is in the rejection region, reject H0. Otherwise, fail to reject H0.
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11. Example: t-Test for the Difference Between Means
A shoe manufacturer claims that athletes can increase their vertical jump heights using the manufacturer’s new Strength Shoes®. The vertical jump heights of eight randomly selected athletes are measured. After the athletes have used the Strength Shoes® for 8 months, their vertical jump heights are measured again. The vertical jump heights (in inches) for each athlete are shown in the table. Assuming the vertical jump heights are normally distributed, is there enough evidence to support the manufacturer’s claim at α = 0.10? (Adapted from Coaches Sports Publishing)
Athletes 1 2 3 4 5 6 7 8
Height (old) 24 22 25 28 35 32 30 27
Height (new) 26 29 33 34
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12. Solution: Two-Sample t-Test for the Difference Between Means
d = (jump height before shoes) – (jump height after shoes)
H0:
Ha:
α =
d.f. =
Rejection Region:
μd ≥ 0
μd < 0 (claim)
0.10
8 – 1 = 7
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13. Solution: Two-Sample t-Test for the Difference Between Means
d = (jump height before shoes) – (jump height after shoes)
Before
After d d2 24 26 –2 4 22 25 –3 9 28 29 –1 1 35 33 2 32 34 30 –5 27
Σ = –14
Σ = 56
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14. Solution: Two-Sample t-Test for the Difference Between Means
d = (jump height before shoes) – (jump height after shoes)
H0:
Ha:
α =
d.f. =
Rejection Region:
Test Statistic:
µd ≥ 0
μd < 0 (claim)
0.10
8 – 1 = 7
Decision:
Reject H0.
At the 10% level of significance, there is enough evidence to support the shoe manufacturer’s claim that athletes can increase their vertical jump heights using the new Strength Shoes®.
t ≈ –2.333
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15. To use the calculator for testing the difference between the
means of dependent samples, enter the data in two columns
and form a third Column ( 𝐿 3 ) in which you calculate the
difference for each pair ( 𝐿 1 − 𝐿 2 ). You can then perform
a one-sample t-test on the difference column ( 𝐿 3 ).

16. Testing the Difference Between Proportions
Section 8.4
Testing the Difference Between Proportions
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17. Section 8.4 Objectives
Perform a z-test for the difference between two population proportions p1 and p2
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18. Two-Sample z-Test for Proportions
Used to test the difference between two population proportions, p1 and p2.
Three conditions are required to conduct the test.
The samples must be randomly selected.
The samples must be independent.
The samples must be large enough to use a normal sampling distribution. That is, n1p1 ≥ 5, n1q1 ≥ 5, n2p2 ≥ 5, and n2q2 ≥ 5.
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19. Two-Sample z-Test for the Difference Between Proportions
If these conditions are met, then the sampling distribution for is a normal distribution.
Mean:
A weighted estimate of p1 and p2 can be found by using
Standard error:
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20. Two-Sample z-Test for the Difference Between Proportions
The test statistic is
The standardized test statistic is
where
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21. Two-Sample z-Test for the Difference Between Proportions
In Words In Symbols
State the claim. Identify the null and alternative hypotheses.
Specify the level of significance.
Determine the critical value(s).
Determine the rejection region(s).
Find the weighted estimate of p1 and p2. Verify that
State H0 and Ha.
Identify α.
Use Table 4 in Appendix B.
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22. Two-Sample z-Test for the Difference Between Proportions
In Words In Symbols
Find the standardized test statistic and sketch the sampling distribution.
Make a decision to reject or fail to reject the null hypothesis.
Interpret the decision in the context of the original claim.
If z is in the rejection region, reject H0. Otherwise, fail to reject H0.
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23. Example: Two-Sample z-Test for the Difference Between Proportions
In a study of 150 randomly selected occupants in passenger cars and 200 randomly selected occupants in pickup trucks, 86% of occupants in passenger cars and 74% of occupants in pickup trucks wear seat belts. At α = 0.10 can you reject the claim that the proportion of occupants who wear seat belts is the same for passenger cars and pickup trucks? (Adapted from National Highway Traffic Safety Administration)
Solution:
1 = Passenger cars 2 = Pickup trucks
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24. Solution: Two-Sample z-Test for the Difference Between Means
Ha:
α =
n1= , n2 =
Rejection Region:
p1 = p2 (claim)
p1 ≠ p2
0.10
150
200
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25. Solution: Two-Sample z-Test for the Difference Between Means
Note:
𝑛 1 𝑝 = ≥ 𝑛 1 𝑞 = ≥5
𝑛 2 𝑝 = ≥ 𝑛 2 𝑝 = ≥5
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26. Solution: Two-Sample z-Test for the Difference Between Means
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27. Solution: Two-Sample z-Test for the Difference Between Means
Ha:
α =
n1= , n2 =
Rejection Region:
p1 = p2 (claim)
p1 ≠ p2
Test Statistic:
0.10
150
200
Decision:
Reject H0
At the 10% level of significance, there is enough evidence to reject the claim that the proportion of occupants who wear seat belts is the same for passenger cars and pickup trucks.
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28. Example: Two-Sample z-Test for the Difference Between Proportions
A medical research team conducted a study to test the effect of a cholesterol-reducing medication. At the end of the study, the researchers found that of the 4700 randomly selected subjects who took the medication, 301 died of heart disease. Of the 4300 randomly selected subjects who took a placebo, 357 died of heart disease. At α = 0.01 can you conclude that the death rate due to heart disease is lower for those who took the medication than for those who took the placebo? (Adapted from New England Journal of Medicine)
Solution:
1 = Medication 2 = Placebo
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29. Solution: Two-Sample z-Test for the Difference Between Means
Ha:
α =
n1= , n2 =
Rejection Region:
p1 ≥ p2
p1 < p2 (claim)
0.01
4700
4300 z
–2.33
0.01
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30. Solution: Two-Sample z-Test for the Difference Between Means
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31. Solution: Two-Sample z-Test for the Difference Between Means
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32. Solution: Two-Sample z-Test for the Difference Between Means
Ha:
α =
n1= , n2 =
Rejection Region:
p1 ≥ p2
p1 < p2 (claim)
Test Statistic:
0.01
4700
4300
Decision:
Reject H0
At the 1% level of significance, there is enough evidence to conclude that the death rate due to heart disease is lower for those who took the medication than for those who took the placebo. z
–2.33
0.01
–2.33
–3.46
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33. Section 8.4 Summary
Performed a z-test for the difference between two population proportions p1 and p2
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