Post on 16-Dec-2015
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Objective
Compare the proportions of two independent means using two samples from each population.
Hypothesis Tests and Confidence Intervals of two proportions use the t-distribution
Section 9.3Inferences About Two Means
(Independent)
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Definitions
Two samples are independent if the sample values selected from one population are not related to or somehow paired or matched with the sample values from the other population
Examples:
Flipping two coins (Independent)
Drawing two cards (not independent)
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Notation
μ1 First population mean
σ1 First population standard deviation
n1 First sample size
x1First sample mean
s1 First sample standard deviation
First Population
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Notation
μ2 Second population mean
σ2 Second population standard deviation
n2 Second sample size
x2Second sample mean
s2 Second sample standard deviation
Second Population
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(1) Have two independent random samples
(2) σ1 and σ2 are unknown and no assumption is made about their equality
(3)Either or both the following holds:Both sample sizes are large (n1>30, n2>30) or
Both populations have normal distributions
Requirements
All requirements must be satisfied to make a Hypothesis Test or to find a Confidence Interval
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Tests for Two Independent Means
The goal is to compare the two Means
H0 : μ1 =
μ
2
H1 : μ1 ≠ μ
2
Two tailed Left tailed Right tailed
Note: We only test the relation between μ1 and μ2
(not the actual numerical values)
H0 : μ1 =
μ
2
H1 : μ1 < μ
2
H0 : μ1 =
μ
2
H1 : μ1 > μ
2
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t x
1 x
2 1
2 s
12
n1
s
22
n2
Note: 1 –
2 =0 according to H0
Degrees of freedom: df = smaller of n1 – 1 and n2 – 1.
Finding the Test Statistic
This equation is an altered form of the test statistic for a single mean when σ unknown (see Ch. 8-5)
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Test Statistic
Note: Hypothesis Tests are done in same way as in Ch.8 (but with different test statistics)
Degrees of freedom df = min(n1 – 1, n2 – 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
• Find the Test Statistic
• Find the Degrees of Freedom
• Find the Critical Value(s)
• State the Initial Conclusion and Final Conclusion
Note: Same process as in Chapter 8
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A headline in USA Today proclaimed that “Men, women are equal talkers.” That headline referred to a study of the numbers of words that men and women spoke in a day.Use a 0.05 significance level to test the claim that men and women speak the same mean number of words in a day.
Example 1
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H0 : µ1 = µ2
H1 : µ1 ≠ µ2
Two-TailedH0 = Claim t = 7.602
tα/2 = 1.97
t-dist.df = 185
Test Statistic
Critical Value
Initial Conclusion: Since t is not in the critical region, accept H0
Final Conclusion: We accept the claim that men and women speak the same average number of words a day.
-tα/2 = -1.97
Example 1 n1 = 186 n2 = 210 α = 0.05
x1 = 15668.5 x2 = 16215.0 Claim: μ1 = μ2
s1 = 8632.5 s2 = 7301.2
Degrees of Freedom
df = min(n1 – 1, n2 – 1) = min(185, 209) = 185
tα/2 = t0.025 = 1.97 (Using StatCrunch)
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H0 : µ1 = µ2
H1 : µ1 ≠ µ2
Two-TailedH0 = Claim
Initial Conclusion: Since P-value > α (0.05), accept H0
Final Conclusion: We accept the claim that men and women speak the same average number of words a day.
Example 1 n1 = 186 n2 = 210 α = 0.05
x1 = 15668.5 x2 = 16215.0 Claim: μ1 = μ2
s1 = 8632.5 s2 = 7301.2
Stat → T statistics → Two sample → With summary
Null: prop. diff.=Alternative
Sample 1: MeanStd. Dev.
Size
Sample 2: MeanStd. Dev.
Size
● Hypothesis Test
P-value = 0.4998
15668.5
0
≠ Using StatCrunch 8632.5
18616215.07301.2
210 (No pooled variance)
(Be sure to not use pooled variance)
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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 = ( (x1–x2) – E, (x1–x2) + E )
Where
df = min(n1–1, n2–1)
2 2
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df = min(n1–1, n2–1) = min(185, 210) = 185
tα/2 = t0.1/2 = t0.05 = 1.973
x1 - x2 = 15668.5 – 16215.0 = -546.5
(x1 - x2) + E = -546.5 + 1596.17 = 1049.67
(x1 - x2) – E = -546.5 – 1596.17 = -2142.67
Use the same sample data in Example 1 to construct a 95% Confidence Interval Estimate of the difference between the two population proportions (µ1–µ2)
Example 2
n1 = 186 n2 = 210
x1 = 15668.5 x2 = 16215.0
s1 = 8632.5 s2 = 7301.2
df = min(n1–1, n2–1) = min(185, 210) = 185
tα/2 = t0.05/2 = t0.025 = 1.973
x1 - x2 = 15668.5 – 16215.0 = -546.5
CI = (-2142.7, 1049.7)
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Use the same sample data in Example 1 to construct a 95% Confidence Interval Estimate of the difference between the two population proportions (µ1–µ2)
Example 2
CI = (-2137.4, 1044.4)
n1 = 186 n2 = 210
x1 = 15668.5 x2 = 16215.0
s1 = 8632.5 s2 = 7301.2
Stat → T statistics → Two sample → With summary
Level:
Sample 1: MeanStd. Dev.
Size
Sample 2: MeanStd. Dev.
Size
● Confidence Interval15668.5
0.95
Using StatCrunch
8632.5186
16215.07301.2
210
Note: slightly different because of rounding errors
(No pooled variance)
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Consider two different classes. The students in the first class are thought to generally be older than those in the second. The students’ ages for this semester are summed as follows:
(a) Use a 0.1 significance level to test the claim that the average age of students in the first class is greater than the average age of students in the second class.
(b) Construct a 90% confidence interval estimate of the difference in average ages.
Example 3
n1 = 93 n2 = 67
x1 = 21.2 x2 = 19.8
s1 = 2.42 s2 = 4.77
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t = 7.602tα/2 = 1.668Test Statistic
Critical Value
Degrees of Freedom
df = min(n1 – 1, n2 – 1) = min(92, 66) = 66
tα/2 = t0.05 = 1.668 (Using StatCrunch)
H0 : µ1 = µ2
H1 : µ1 > µ2
Right-TailedH1 = Claim
n1 = 93 n2 = 67α = 0.1
x1 = 21.2 x2 = 19.8 Claim: µ1 > µ2
s1 = 2.42 s2 = 4.77
t-dist.df = 66
Example 3a
Initial Conclusion: Since t is in the critical region, reject H0
Final Conclusion: We accept the claim that the average age of students
in the first class is greater than that in the second.
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H0 : µ1 = µ2
H1 : µ1 > µ2
Right-TailedH1 = Claim
Example 3a n1 = 93 n2 = 67α = 0.1
x1 = 21.2 x2 = 19.8 Claim: µ1 > µ2
s1 = 2.42 s2 = 4.77
Stat → T statistics → Two sample → With summary
Null: prop. diff.=Alternative
Sample 1: MeanStd. Dev.
Size
Sample 2: MeanStd. Dev.
Size
● Hypothesis Test
P-value = 0.0299
21.2
0
≠
2.4293
19.84.7767 (No pooled variance)
Using StatCrunch(Be sure to not use pooled variance)
Initial Conclusion: Since P-value < α (0.1), reject H0
Final Conclusion: We accept the claim that the average age of students
in the first class is greater than that in the second.
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df = min(n1–1, n2–1) = min(92, 66) = 66
tα/2 = t0.1/2 = t0.05 = 1.668
x1 - x2 = 21.2 – 19.8 = 1.4
(x1 - x2) + E = 1.4 + 1.058 = 2.458
(x1 - x2) – E = 1.4 – 1.058 = 0.342
CI = (0.34, 2.46)
n1 = 93 n2 = 67α = 0.1
x1 = 21.2 x2 = 19.8
s1 = 2.42 s2 = 4.77
Example 3b
mp
(90% Confidence Interval)
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Example 3b
Stat → T statistics → Two sample → With summary
Null: prop. diff.=Alternative
Sample 1: MeanStd. Dev.
Size
Sample 2: MeanStd. Dev.
Size
● Hypothesis Test21.2
0
≠
2.4293
19.84.7767 (No pooled variance)
Using StatCrunch(Be sure to not use pooled variance)
CI = (0.35, 2.45)
n1 = 93 n2 = 67α = 0.1
x1 = 21.2 x2 = 19.8
s1 = 2.42 s2 = 4.77(90% Confidence Interval)