2. Chapter 8 Inferences from Two Samples
8-1 Overview
8-2 Inferences about Two Proportions
8-3 Inferences about Two Means: Independent Samples
8-4 Inferences about Matched Pairs
8-5 Comparing Variation in Two Samples

3. Overview and Inferences about Two Proportions
Section 8-1 & 8-2
Overview and Inferences about Two Proportions
Created by Erin Hodgess, Houston, Texas

4. Overview
There are many important and meaningful situations in which it becomes necessary to compare two sets of sample data.
page 438 of text
Examples in the discussion
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5. Inferences about Two Proportions
Assumptions
1. We have proportions from two     independent simple random samples.
2. For both samples, the conditions np  5 and nq  5 are satisfied.
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6. Notation for Two Proportions
For population 1, we let:
p1 = population proportion
n1 = size of the sample
x1 = number of successes in the sample
p1 = x1 (the sample proportion)
q1 = 1 – p1 ^ n1
population 2.
The corresponding meanings are attached
to p2, n2 , x2 , p2. and q2 , which come from ^
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7. Pooled Estimate of p1 and p2
The pooled estimate of p1 and p2   is denoted by p. = p
n1 + n2
x1 + x2
q = 1 – p
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8. Test Statistic for Two Proportions
For H0: p1 = p2 , H0: p1  p2 , H0: p1 p2
H1: p1  p2 , H1: p1 < p2 , H1: p 1> p2 +
z =
( p1 – p2 ) –
( p1 – p2 ) ^ n1 pq n2
Example given at the bottom of page
452

9. Test Statistic for Two Proportions
For H0: p1 = p2 , H0: p1  p2 , H0: p1 p2
H1: p1  p2 , H1: p1 < p2 , H1: p 1> p2
where p1 – p 2 = (assumed in the null hypothesis) = p1 ^ x1 n1 p2 x2 n2
and
Example given at the bottom of page
and
q = 1 – p
n1 + n2
p =
x1 + x2
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10. Example: For the sample data listed in Table 8-1, use a 0
Example: For the sample data listed in Table 8-1, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped.
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11. Example: For the sample data listed in Table 8-1, use a 0
Example: For the sample data listed in Table 8-1, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped.
200 n1
n1= 200
x1 = 24
p1 = x1 = 24 = 0.120 ^ n2
n2 = 1400
x2 = 147
p2 = x2 = 147 = 0.105
1400
H0: p1 = p2, H1: p1 > p2
p = x1 + x2 = =
n1 + n
q = 1 – =
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12. Example: For the sample data listed in Table 8-1, use a 0
Example: For the sample data listed in Table 8-1, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped.
200 n1
n1= 200
x1 = 24
p1 = x1 = 24 = 0.120 ^ n2
n2 = 1400
x2 = 147
p2 = x2 = 147 = 0.105
1400
z = (0.120 – 0.105) – 0
( )( ) + ( )( )
z = 0.64
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13. Example: For the sample data listed in Table 8-1, use a 0
Example: For the sample data listed in Table 8-1, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped.
200 n1
n1= 200
x1 = 24
p1 = x1 = 24 = 0.120 ^ n2
n2 = 1400
x2 = 147
p2 = x2 = 147 = 0.105
1400
z = 0.64
This is a right-tailed test, so the P-value is
the area to the right of the test statistic z = The P-value is
Because the P-value of is greater than the significance level of  = 0.05, we fail to reject the null hypothesis.
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14. Example: For the sample data listed in Table 8-1, use a 0
Example: For the sample data listed in Table 8-1, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped.
200 n1
n1= 200
x1 = 24
p1 = x1 = 24 = 0.120 ^ n2
n2 = 1400
x2 = 147
p2 = x2 = 147 = 0.105
1400
z = 0.64
Because we fail to reject the null hypothesis, we conclude that there is not sufficient evidence to support the claim that the proportion of black drivers stopped by police is greater than that for white drivers. This does not mean that racial profiling has been disproved. The evidence might be strong enough with more data.
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15. Example: For the sample data listed in Table 8-1, use a 0
Example: For the sample data listed in Table 8-1, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped.
200 n1
n1= 200
x1 = 24
p1 = x1 = 24 = 0.120 ^ n2
n2 = 1400
x2 = 147
p2 = x2 = 147 = 0.105
1400
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16. Confidence Interval Estimate of p1 - p2
( p1 – p2 ) – E < ( p1 – p2 ) < ( p1 – p2 ) + E ^ n1 n2
p1 q1
p2 q2 + ^
where E = z
Example at the bottom of page 463
Rationale for the procedures of this section on page
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17. Example: For the sample data listed in Table 8-1, find a 90% confidence interval estimate of the difference between the two population proportions. ^ ^ ^ ^
n1= 200
x1 = 24
p1 = x1 = 24 = 0.120
E = z
p1 q1
p2 q2 + n1 n2 ^
(.12)(.88)+
(0.105)(0.895)
E = 1.645 n1
200
200
1400
n2 = 1400
x2 = 147
p2 = x2 = 147 = 0.105
E = 0.400 ^ n2
1400
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18. (0.120 – 0.105) – 0.040 < ( p1– p2) < (0.120 – 0.105) + 0.040
Example: For the sample data listed in Table 8-1, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped.
n1= 200
x1 = 24
p1 = x1 = 24 = 0.120
(0.120 – 0.105) – < ( p1– p2) < (0.120 – 0.105)
–0.025 < ( p1– p2) < 0.055 ^ n1
200
n2 = 1400
x2 = 147
p2 = x2 = 147 = 0.105 ^ n2
1400
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