1. Section 9.3 Inferences About Two Means (Independent)
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

2. 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)
Text will use the wording ‘matched pairs’.
Example at bottom of page of Elementary Statistics, 10th Edition

3. Notation First Population μ1 First population mean
σ1 First population standard deviation
n1 First sample size
x1 First sample mean
s1 First sample standard deviation

4. Notation Second Population μ2 Second population mean
σ2 Second population standard deviation
n2 Second sample size
x2 Second sample mean
s2 Second sample standard deviation

5. Requirements (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
All requirements must be satisfied to make a Hypothesis Test or to find a Confidence Interval

6. Tests for Two Independent Means
The goal is to compare the two Means
H0 : μ1 = μ2
H1 : μ1 ≠ μ2
H0 : μ1 = μ2
H1 : μ1 < μ2
H0 : μ1 = μ2
H1 : μ1 > μ2
Two tailed
Left tailed
Right tailed
Example given at the bottom of page of Elementary Statistics, 10th Edition.
Note: We only test the relation between μ1 and μ2
(not the actual numerical values)

7. Finding the Test Statistic
Note: m1 – m2 =0 according to H0
Degrees of freedom: df = smaller of n1 – 1 and n2 – 1.
This equation is an altered form of the test statistic for a single mean when σ unknown (see Ch. 8-5)

8. Test Statistic Degrees of freedom df = min(n1 – 1, n2 – 1)
Degrees of freedom df = min(n1 – 1, n2 – 1)
Example given at the bottom of page of Elementary Statistics, 10th Edition.
Note: Hypothesis Tests are done in same way as in Ch.8 (but with different test statistics)

9. 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

10. Example 1
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.

11. Example 1 H0 : µ1 = µ2 H1 : µ1 ≠ µ2 n1 = 186 n2 = 210 α = 0.05
x1 = x2 = Claim: μ1 = μ2
s1 = s2 =
H0 : µ1 = µ2
H1 : µ1 ≠ µ2
Two-Tailed
H0 = Claim
t-dist.
df = 185
t = 7.602
-tα/2 = -1.97
tα/2 = 1.97
Test Statistic
Degrees of Freedom
df = min(n1 – 1, n2 – 1) = min(185, 209) = 185
Critical Value
tα/2 = t0.025 = (Using StatCrunch)
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.

12. Stat → T statistics → Two sample → With summary
Example 1
n1 = n2 = α = 0.05
x1 = x2 = Claim: μ1 = μ2
s1 = s2 =
H0 : µ1 = µ2
H1 : µ1 ≠ µ2
Two-Tailed
H0 = Claim
Stat → T statistics → Two sample → With summary
Null: prop. diff.=
Alternative
Sample 1: Mean
Std. Dev.
Size
Sample 2: Mean
● Hypothesis Test
Using StatCrunch
8632.5
186 ≠ 
(Be sure to not use pooled variance)
7301.2
210
(No pooled variance)
P-value =
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.

13. 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 ) 2 2
Example at the bottom of page 461 of Elementary Statistics, 10th Edition
Rationale for the procedures of this section on page of Elementary Statistics, 10th Edition
Where
df = min(n1–1, n2–1)

14. Example 2
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)
n1 = n2 = 210
x1 = x2 =
s1 = s2 =
df = min(n1–1, n2–1) = min(185, 210) = 185
tα/2 = t0.05/2 = t0.025 = 1.973
x1 - x2 = – =
df = min(n1–1, n2–1) = min(185, 210) = 185
tα/2 = t0.1/2 = t0.05 = 1.973
x1 - x2 = – =
(x1 - x2) + E = =
(x1 - x2) – E = – =
CI = ( , )

15. Stat → T statistics → Two sample → With summary
Example 2
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)
n1 = n2 = 210
x1 = x2 =
s1 = s2 =
Stat → T statistics → Two sample → With summary
Level:
Sample 1: Mean
Std. Dev.
Size
Sample 2: Mean
● Confidence Interval
8632.5
0.95
186
Using StatCrunch
7301.2
(No pooled variance)
210
CI = ( , )
Note: slightly different because of rounding errors

16. Example 3
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.
n1 = 93 n2 = 67
x1 = x2 = 19.8
s1 = s2 = 4.77

17. Example 3a H0 : µ1 = µ2 H1 : µ1 > µ2 n1 = 93 n2 = 67 α = 0.1
x1 = x2 = 19.8 Claim: µ1 > µ2
s1 = s2 = 4.77
H0 : µ1 = µ2
H1 : µ1 > µ2
Right-Tailed
H1 = Claim
t-dist.
df = 66
Test Statistic
tα/2 = 1.668
t = 7.602
Degrees of Freedom
df = min(n1 – 1, n2 – 1) = min(92, 66) = 66
Critical Value
tα/2 = t0.05 = (Using StatCrunch)
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.

18. Stat → T statistics → Two sample → With summary
Example 3a
n1 = 93 n2 = 67 α = 0.1
x1 = x2 = 19.8 Claim: µ1 > µ2
s1 = s2 = 4.77
H0 : µ1 = µ2
H1 : µ1 > µ2
Right-Tailed
H1 = Claim
Stat → T statistics → Two sample → With summary
Null: prop. diff.=
Alternative
Sample 1: Mean
Std. Dev.
Size
Sample 2: Mean
● Hypothesis Test
21.2
Using StatCrunch
2.42 93 ≠ 
19.8
(Be sure to not use pooled variance)
4.77 67
(No pooled variance)
P-value =
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.

19. Example 3b mp CI = (0.34, 2.46) n1 = 93 n2 = 67 α = 0.1
x1 = x2 = 19.8
s1 = s2 = 4.77
(90% Confidence Interval)
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 = = 2.458
(x1 - x2) – E = 1.4 – = 0.342 mp
CI = (0.34, 2.46)

20. Stat → T statistics → Two sample → With summary
Example 3b
n1 = 93 n2 = 67 α = 0.1
x1 = x2 = 19.8
s1 = s2 = 4.77
(90% Confidence Interval)
Stat → T statistics → Two sample → With summary
Null: prop. diff.=
Alternative
Sample 1: Mean
Std. Dev.
Size
Sample 2: Mean
● Hypothesis Test
21.2
Using StatCrunch
2.42 93 ≠ 
19.8
(Be sure to not use pooled variance)
4.77 67
(No pooled variance)
CI = (0.35, 2.45)