Section 8.2
Testing the Difference Between Means
(Small Independent Samples)
(Independent Samples,
σ1 and σ2 Unknown)
Section 8.2 Objectives
• Perform a t-test for the difference between two
population means μ1 and μ2 using independent
samples with population standard deviations
unknown.
Two Sample t-Test for the Difference
Between Means
These conditions are necessary to use a t-test for the
difference between means for independent samples.
1. The samples must be randomly selected.
2. The samples must be independent.
3. Each population must have a normal distribution -oreach sample size is at least 30.
4. The population standard deviations (σ1, σ2) are
UNKNOWN
Two Sample t-Test for the Difference
Between Means
• The test statistic is x1  x2.
• The standardized test statistic is
x1  x2  1  2

t
.
sx  x
1
2
• The standard error and the degrees of freedom of the
sampling distribution depend on whether the
population variances  12 and  22 are equal.
Two Sample t-Test for the Difference
Between Means
• Variances are equal
 Information from the two samples is combined to
calculate a pooled estimate of the standard deviation
ˆ .
n1  1 s12  n2  1 s22

ˆ 
n1  n2  2
 The standard error for the sampling distribution of
x1  x2 is
sx  x
1
2
1 1
 ˆ 

n1 n2
 d.f.= n1 + n2 – 2
Two Sample t-Test for the Difference
Between Means
• Variances are not equal
 If the population variances are not equal, then the
standard error is
s12 s22
sx  x 
 .
n1 n2
 d.f = smaller of n1 – 1 and n2 – 1
1
2
Normal or t-Distribution?
Both populations
normally distributed or
are both sample sizes at
least 30?
No
You cannot use the ztest or the t-test. (!)
Use the t-test
with
Yes
Are both population
standard deviations
(σ1 and σ2 )
known?
Yes
Use the z-test.
No
Are the population
variances equal?
Yes
sx  x  ˆ
1
2
d.f = n1 + n2 – 2.
No
Use the t-test with
sx  x 
1
1 1

n1 n2
2
s12 s22

n1 n2
d.f = smaller of n1 – 1 or n2 – 1.
Two-Sample t-Test for the Difference
Between Means (Independent Samples,
σ1,σ2 Unknown)
In Words
In Symbols
1. State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
2. Specify the level of significance.
3. Determine the degrees of
freedom.
State H0 and Ha.
Identify α.
d.f. = n1+ n2 – 2 (equal σ)
or
d.f. = smaller of n1 – 1 or
n2 – 1. (unequal σ)
4. Determine the critical value(s).
Use Table 5 in
Appendix B.
Two-Sample t-Test for the Difference
Between Means (Small Independent
Samples)
In Words
In Symbols
5. Determine the rejection region(s).
6. Find the standardized test statistic
and sketch the sampling
distribution.
x1  x2 1  2

t
7. Make a decision to reject or fail
to reject the null hypothesis.
If t is in the rejection
region, reject H0.
Otherwise, fail to
reject H0.
8. Interpret the decision in the
context of the original claim.
sx x
1
2
Example: Two-Sample t-Test for the
Difference Between Means
The results of a state mathematics test for random samples
of students taught by two different teachers at the same
school are shown below. Can you conclude that there is a
difference in the mean mathematics test scores for the
students of the two teachers? Use α = 0.10. Assume the
populations are normally distributed, and the population
variances are not equal and they are unknown .
Teacher 1
Teacher 2
x1  473
x2  459
s1 = 39.7
s2 = 24.5
n1 = 8
n2 = 18
Solution: Two-Sample t-Test for the
Difference Between Means
•
•
•
•
•
H0: μ 1 = μ 2
Ha: μ1 ≠ μ2 (claim)
α  0.10
d.f. = 8 – 1 = 7
Rejection Region:
t ≈ 0.922
• Test Statistic:
t
(473  459)  0
 0.922
39.7 2 24.52

8
18
• Decision: Fail to Reject H0 .
At the 10% level of significance,
there is not enough evidence to
support the claim that the mean
mathematics test scores for the
students of the two teachers are
different.
Example: Two-Sample t-Test for the
Difference Between Means
A manufacturer claims that the mean calling range (in feet) of
its 2.4-GHz cordless telephone is greater than that of its
leading competitor. You perform a study using 14 randomly
selected phones from the manufacturer and 16 randomly
selected similar phones from its competitor. The results are
shown below. At α = 0.05, can you support the manufacturer’s
claim? Assume the populations are normally distributed and
the population variances are equal (and unknown).
Manufacturer
Competitor
x1  1275ft
x2  1250 ft
s1 = 45 ft
s2 = 30 ft
n1 = 14
n2 = 16
Solution: Two-Sample t-Test for the
Difference Between Means
•
•
•
•
•
H0: μ 1 ≤ μ 2
Ha: μ1 > μ2 (claim)
α = 0.05
d.f. = 14 + 16 – 2 = 28
Rejection Region:
0.05
0
1.701
t
Solution: Two-Sample t-Test for the
Difference Between Means
• Test Statistic:
sx1  x2 
 n1  1 s12   n2  1 s2 2 
n1  n2  2
14  1 45  16  1 30 
2

1 1

n1 n2
14  16  2
2
1 1


 13.8018
14 16
x1  x2    1  2  1275  1250   0

t

 1.811
sx1  x2
13.8018
Solution: Two-Sample t-Test for the
Difference Between Means
•
•
•
•
•
H0: μ 1 ≤ μ 2
Ha: μ1 > μ2 (claim)
α = 0.05
d.f. = 14 + 16 – 2 = 28
Rejection Region:
0.05
0
1.701
1.811
t
• Test Statistic:
t  1.811
• Decision: Reject H0 .
At the 5% level of significance,
there is enough evidence to
support the manufacturer’s
claim that its phone has a
greater mean calling range than
its competitor.
The maximal oxygen consumption is a way to
measure the fitness of an individual. It is the amount
of oxygen in milliliters a person uses per kilogram of
body weight per minute. A medical research center
claims that athletes have a greater mean maximal
oxygen consumption than non-athletes. The results
for samples from the 2 groups are below. At α =
0.05, can you support the research center’s claim?
(Assume population variances are equal, and
populations are normally distributed, σ’s unknown.)
Athletes
Non-athletes
𝑥1= 56 ml/kg/min
𝑥2 = 47 ml/kg/min
s1 = 4.9 ml/kg/min
s2 = 3.1 ml/kg/min
n1 = 23
n2 = 21
Test of Inference
I. Ho : µ1 ≤ µ2
Ha : µ1 > µ2 (claim)
III.
IV.
V.
VI.
II. α = 0.05
Test of Inference
I. Ho : µ1 ≤ µ2
Ha : µ1 > µ2 (claim)
III. t =
=
IV.
V.
VI.
𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 −𝑚𝑒𝑎𝑛
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟
9
720.4200
∗
42
=
II. α = 0.05
𝑥1 − 𝑥2 −(µ1 −µ2)
𝑠
𝑥1− 𝑥2
9
9
=
56−47 −0
22∗4.92+20∗3.12
1 1
∗
+
23+21−2
23 21
= 4.1416∗.3018 = 1.2499 = 7.200
.0911
Test of Inference
I. Ho : µ1 ≤ µ2
Ha: µ1 > µ2 (claim)
III. t =
=
𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 −𝑚𝑒𝑎𝑛
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟
9
720.4200
∗
42
=
II. α = 0.05
𝑥1 − 𝑥2 −(µ1 −µ2)
𝑠
𝑥1− 𝑥2
9
=
56−47 −0
22∗4.92+20∗3.12
1 1
∗
+
23+21−2
23 21
9
= 4.1416∗.3018 = 1.2499 = 7.200
.0911
IV. Define t0 (table): df=n1+n2-2 = 23+21-2=42
V.
VI.
t0 = 1.684
Test of Inference
I. Ho : µ1 ≤ µ2
II. α = 0.05
Ha : µ1 > µ2 (claim)
III. t =
=
𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 −𝑚𝑒𝑎𝑛
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟
9
720.4200
∗
42
=
𝑥1 − 𝑥2 −(µ1 −µ2)
𝑠
𝑥1− 𝑥2
9
=
56−47 −0
2
22∗4.9 +20∗3.1
23+21−2
∗
1 1
+
23 21
9
= 4.1416∗.3018 = 1.2499 = 7.200
.0911
IV. Define t0 (table): df=n1+n2-2 = 23+21-2=42
V. Define rejection region: reject if t > 1.684
VI.
2
t0 = 1.684
Test of Inference
I. Ho : µ1 ≤ µ2
II. α = 0.05
Ha : µ1 > µ2 (claim)
III. t =
=
𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 −𝑚𝑒𝑎𝑛
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟
9
720.4200
∗
42
=
𝑥1 − 𝑥2 −(µ1 −µ2)
𝑠
𝑥1− 𝑥2
9
=
56−47 −0
2
22∗4.9 +20∗3.1
23+21−2
2
∗
1 1
+
23 21
9
= 4.1416∗.3018 = 1.2499 = 7.200
.0911
IV. Define t0 (table): df=n1+n2-2 = 23+21-2=42
V. Define rejection region: reject if t > 1.684
VI. Decision : reject H0
t0 = 1.684
Test of Inference
I. Ho : µ1 ≤ µ2
Ha : µ1 > µ2 (claim)
III. t =
=
𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 −𝑚𝑒𝑎𝑛
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟
9
720.4200
∗
42
=
II. α = 0.05
𝑥1 − 𝑥2 −(µ1 −µ2)
𝑠
𝑥1− 𝑥2
9
=
56−47 −0
22∗4.92+20∗3.12
1 1
∗
+
23+21−2
23 21
9
= 4.1416∗.3018 = 1.2499 = 7.200
.0911
IV. Define t0 (table): df=n1+n2-2 = 23+21-2=42
t0 = 1.645
V. Define rejection region: reject if t > 1.645
VI. Decision : reject H0
VII. There is enough evidence, at the 5% significance level to
support the claim that athletes have a greater mean maximal
oxygen consumption than non-athletes.
Section 8.2 Summary
• Performed a t-test for the difference between two
means μ1 and μ2 from independent samples when the
population standard deviations are not known.