This chapter discusses methods for forming confidence intervals and conducting hypothesis tests to compare two population parameters, such as means, proportions, or variances. It covers topics like confidence intervals for the difference between two independent population means when the variances are known or unknown, confidence intervals for dependent sample means from before-after studies, and confidence intervals for comparing two independent population proportions. Examples are provided to demonstrate how to calculate confidence intervals for differences in means using pooled variances and how to form confidence intervals to compare proportions from two populations.

1. Chap 10-1
Chapter 10
INFERENCE: TWO POPULATIONS
BUSINESS STATISTICS

2. Chap 10-2
Chapter Contents
This chapter includes two main parts:
1. Estimation for the difference of two population
parameters
2. Hypothesis Testing for the difference of two population
parameters

3. Chap 10-3
Part 1: Goals
After completing this part, you should be able to:
§ Form confidence intervals for the mean difference from dependent
samples
§ Form confidence intervals for the difference between two
independent population means (standard deviations known or
unknown)
§ Compute confidence interval limits for the difference between two
independent population proportions
§ Create confidence intervals for a population variance
§ Find chi-square values from the chi-square distribution table
§ Determine the required sample size to estimate a mean or
proportion within a specified margin of error

4. Chap 10-4
Estimation: Two populations
Chapter Topics
Population
Means,
Independent
Samples
Population
Means,
Dependent
Samples
Population
Variance
Group 1 vs.
independent
Group 2
Same group
before vs. after
treatment
Variance of a
normal distribution
Examples:
Population
Proportions
Proportion 1 vs.
Proportion 2

5. Chap 10-5
Dependent Samples
Tests Means of 2 Related Populations
§ Paired or matched samples
§ Repeated measures (before/after)
§ Use difference between paired values:
§ Eliminates Variation Among Subjects
§ Assumptions:
§ Both Populations Are Normally Distributed
Dependent
samples
di = xi - yi

6. Chap 10-6
Mean Difference
The ith paired difference is di , where
di = xi - yi
The point estimate for
the population mean
paired difference is d :
n
d
d
n
1
i
i
å
=
=
n is the number of matched pairs in the sample
1
n
)
d
(d
S
n
1
i
2
i
d
-
-
=
å
=
The sample
standard
deviation is:
Dependent
samples

7. Chap 10-7
Confidence Interval for
Mean Difference
The confidence interval for difference
between population means, μd , is
Where
n = the sample size
(number of matched pairs in the paired sample)
n
S
t
d
μ
n
S
t
d d
α/2
1,
n
d
d
α/2
1,
n -
- +
<
<
-
Dependent
samples

8. Chap 10-8
§ The margin of error is
§ tn-1,a/2 is the value from the Student’s t
distribution with (n – 1) degrees of freedom
for which
Confidence Interval for
Mean Difference
(continued)
2
α
)
t
P(t α/2
1,
n
1
n =
> -
-
n
s
t
ME d
α/2
1,
n-
=
Dependent
samples

9. Chap 10-9
§ Six people sign up for a weight loss program. You
collect the following data:
Paired Samples Example
Weight:
Person Before (x) After (y) Difference, di
1 136 125 11
2 205 195 10
3 157 150 7
4 138 140 - 2
5 175 165 10
6 166 160 6
42
d =
S di
n
4.82
1
n
)
d
(d
S
2
i
d
=
-
-
=
å
= 7.0

10. Chap 10-10
§ For a 95% confidence level, the appropriate t value is
tn-1,a/2 = t5,.025 = 2.571
§ The 95% confidence interval for the difference between
means, μd , is
12.06
μ
1.94
6
4.82
(2.571)
7
μ
6
4.82
(2.571)
7
n
S
t
d
μ
n
S
t
d
d
d
d
α/2
1,
n
d
d
α/2
1,
n
<
<
+
<
<
-
+
<
<
- -
-
Paired Samples Example
(continued)
Since this interval contains zero, we cannot be 95% confident, given this
limited data, that the weight loss program helps people lose weight

11. Chap 10-11
Difference Between Two Means
Population means,
independent
samples
Goal: Form a confidence interval
for the difference between two
population means, μx – μy
x – y
§ Different data sources
§ Unrelated
§ Independent
§ Sample selected from one population has no effect on the
sample selected from the other population
§ The point estimate is the difference between the two
sample means:

12. Chap 10-12
Difference Between Two Means
Population means,
independent
samples
Confidence interval uses za/2
Confidence interval uses a value
from the Student’s t distribution
σx
2 and σy
2
assumed equal
σx
2 and σy
2 known
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal
(continued)

13. Chap 10-13
Population means,
independent
samples
σx
2 and σy
2 Known
Assumptions:
§ Samples are randomly and
independently drawn
§ both population distributions
are normal
§ Population variances are
known
*
σx
2 and σy
2 known
σx
2 and σy
2 unknown

14. Chap 10-14
Population means,
independent
samples
…and the random variable
has a standard normal distribution
When σx and σy are known and
both populations are normal, the
variance of X – Y is
y
2
y
x
2
x
2
Y
X
n
σ
n
σ
σ +
=
-
(continued)
*
Y
2
y
X
2
x
Y
X
n
σ
n
σ
)
μ
(μ
)
y
x
(
Z
+
-
-
-
=
σx
2 and σy
2 known
σx
2 and σy
2 unknown
σx
2 and σy
2 Known

15. Chap 10-15
Population means,
independent
samples
The confidence interval for
μx – μy is:
Confidence Interval,
σx
2 and σy
2 Known
*
y
2
Y
x
2
X
α/2
Y
X
y
2
Y
x
2
X
α/2
n
σ
n
σ
z
)
y
x
(
μ
μ
n
σ
n
σ
z
)
y
x
( +
+
-
<
-
<
+
-
-
σx
2 and σy
2 known
σx
2 and σy
2 unknown

16. Chap 10-16
Population means,
independent
samples
σx
2 and σy
2 Unknown,
Assumed Equal
Assumptions:
§ Samples are randomly and
independently drawn
§ Populations are normally
distributed
§ Population variances are
unknown but assumed equal
*
σx
2 and σy
2
assumed equal
σx
2 and σy
2 known
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal

17. Chap 10-17
Population means,
independent
samples
(continued)
Forming interval
estimates:
§ The population variances
are assumed equal, so use
the two sample standard
deviations and pool them to
estimate σ
§ use a t value with
(nx + ny – 2) degrees of
freedom
*
σx
2 and σy
2
assumed equal
σx
2 and σy
2 known
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal
σx2 and σy
2 Unknown,
Assumed Equal

18. Chap 10-18
Population means,
independent
samples
The pooled variance is
(continued)
* 2
n
n
1)s
(n
1)s
(n
s
y
x
2
y
y
2
x
x
2
p
-
+
-
+
-
=
σx
2 and σy
2
assumed equal
σx
2 and σy
2 known
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal
σx
2 and σy
2 Unknown,
Assumed Equal

19. Chap 10-19
The confidence interval for
μ1 – μ2 is:
Where
*
Confidence Interval,
σx
2 and σy
2 Unknown, Equal
σx
2 and σy
2
assumed equal
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal
y
2
p
x
2
p
α/2
2,
n
n
Y
X
y
2
p
x
2
p
α/2
2,
n
n
n
s
n
s
t
)
y
x
(
μ
μ
n
s
n
s
t
)
y
x
( y
x
y
x
+
+
-
<
-
<
+
-
- -
+
-
+
2
n
n
1)s
(n
1)s
(n
s
y
x
2
y
y
2
x
x
2
p
-
+
-
+
-
=

20. Chap 10-20
Pooled Variance Example
You are testing two computer processors for speed.
Form a confidence interval for the difference in CPU
speed. You collect the following speed data (in Mhz):
CPUx CPUy
Number Tested 17 14
Sample mean 3004 2538
Sample std dev 74 56
Assume both populations are
normal with equal variances,
and use 95% confidence

21. Chap 10-21
Calculating the Pooled Variance
( ) ( ) ( ) ( ) 4427.03
1)
14
1)
-
(17
56
1
14
74
1
17
1)
n
(n
S
1
n
S
1
n
S
2
2
y
2
y
y
2
x
x
2
p =
-
+
-
+
-
=
-
+
-
-
+
-
=
(
(
)
1
x
The pooled variance is:
The t value for a 95% confidence interval is:
2.045
t
t 0.025
,
29
α/2
,
2
n
n y
x
=
=
-
+

22. Chap 10-22
Calculating the Confidence Limits
§ The 95% confidence interval is
y
2
p
x
2
p
α/2
2,
n
n
Y
X
y
2
p
x
2
p
α/2
2,
n
n
n
s
n
s
t
)
y
x
(
μ
μ
n
s
n
s
t
)
y
x
( y
x
y
x
+
+
-
<
-
<
+
-
- -
+
-
+
14
4427.03
17
4427.03
(2.054)
2538)
(3004
μ
μ
14
4427.03
17
4427.03
(2.054)
2538)
(3004 Y
X +
+
-
<
-
<
+
-
-
515.31
μ
μ
416.69 Y
X <
-
<
We are 95% confident that the mean difference in
CPU speed is between 416.69 and 515.31 Mhz.

23. Chap 10-23
Population means,
independent
samples
σx
2 and σy
2 Unknown,
Assumed Unequal
Assumptions:
§ Samples are randomly and
independently drawn
§ Populations are normally
distributed
§ Population variances are
unknown and assumed
unequal
*
σx
2 and σy
2
assumed equal
σx
2 and σy
2 known
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal

24. Chap 10-24
Population means,
independent
samples
σx
2 and σy
2 Unknown,
Assumed Unequal
(continued)
Forming interval estimates:
§ The population variances are
assumed unequal, so a pooled
variance is not appropriate
§ use a t value with n degrees
of freedom, where
σx
2 and σy
2 known
σx
2 and σy
2 unknown
*
σx
2 and σy
2
assumed equal
σx
2 and σy
2
assumed unequal
1)
/(n
n
s
1)
/(n
n
s
)
n
s
(
)
n
s
(
y
2
y
2
y
x
2
x
2
x
2
y
2
y
x
2
x
-
÷
÷
ø
ö
ç
ç
è
æ
+
-
÷
÷
ø
ö
ç
ç
è
æ
ú
ú
û
ù
ê
ê
ë
é
+
=
v

25. Chap 10-25
The confidence interval for
μ1 – μ2 is:
*
Confidence Interval,
σx
2 and σy
2 Unknown, Unequal
σx
2 and σy
2
assumed equal
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal
y
2
y
x
2
x
α/2
,
Y
X
y
2
y
x
2
x
α/2
,
n
s
n
s
t
)
y
x
(
μ
μ
n
s
n
s
t
)
y
x
( +
+
-
<
-
<
+
-
- n
n
1)
/(n
n
s
1)
/(n
n
s
)
n
s
(
)
n
s
(
y
2
y
2
y
x
2
x
2
x
2
y
2
y
x
2
x
-
÷
÷
ø
ö
ç
ç
è
æ
+
-
÷
÷
ø
ö
ç
ç
è
æ
ú
ú
û
ù
ê
ê
ë
é
+
=
v
Where

26. Chap 10-26
Two Population Proportions
Goal: Form a confidence interval for
the difference between two
population proportions, Px – Py
The point estimate for
the difference is
Population
proportions
Assumptions:
Both sample sizes are large (generally at
least 40 observations in each sample)
y
x p
p ˆ
ˆ -

27. Chap 10-27
Two Population Proportions
Population
proportions
(continued)
§ The random variable
is approximately normally distributed
y
y
y
x
x
x
y
x
y
x
n
)
p
(1
p
n
)
p
(1
p
)
p
(p
)
p
p
(
Z
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
-
+
-
-
-
-
=

28. Chap 10-28
Confidence Interval for
Two Population Proportions
Population
proportions
The confidence limits for
Px – Py are:
y
y
y
x
x
x
y
x
n
)
p
(1
p
n
)
p
(1
p
Z
)
p
p
(
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ 2
/
-
+
-
±
- a

29. Chap 10-29
Example:
Two Population Proportions
Form a 90% confidence interval for the
difference between the proportion of
men and the proportion of women who
have college degrees.
§ In a random sample, 26 of 50 men and
28 of 40 women had an earned college
degree

30. Chap 10-30
Example:
Two Population Proportions
Men:
Women:
0.1012
40
0.70(0.30)
50
0.52(0.48)
n
)
p
(1
p
n
)
p
(1
p
y
y
y
x
x
x
=
+
=
-
+
- ˆ
ˆ
ˆ
ˆ
0.52
50
26
px =
=
ˆ
0.70
40
28
py =
=
ˆ
(continued)
For 90% confidence, Za/2 = 1.645

31. Chap 10-31
Example:
Two Population Proportions
The confidence limits are:
so the confidence interval is
-0.3465 < Px – Py < -0.0135
Since this interval does not contain zero we are 90% confident that the
two proportions are not equal
(continued)
(0.1012)
1.645
.70)
(.52
n
)
p
(1
p
n
)
p
(1
p
Z
)
p
p
(
y
y
y
x
x
x
α/2
y
x
±
-
=
-
+
-
±
-
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ

32. Chap 10-32
Confidence Intervals for the
Population Variance
Population
Variance
§ Goal: Form a confidence interval
for the population variance, σ2
§ The confidence interval is based on
the sample variance, s2
§ Assumed: the population is
normally distributed

33. Chap 10-33
Confidence Intervals for the
Population Variance
Population
Variance
The random variable
2
2
2
1
n
σ
1)s
(n-
=
-
c
follows a chi-square distribution
with (n – 1) degrees of freedom
(continued)
The chi-square value denotes the number for which
2
,
1
n a
c -
α
)
P( 2
α
,
1
n
2
1
n =
> -
- χ
χ

34. Chap 10-34
Confidence Intervals for the
Population Variance
Population
Variance
The (1 - a)% confidence interval
for the population variance is
2
/2
-
1
,
1
n
2
2
2
/2
,
1
n
2
1)s
(n
σ
1)s
(n
α
α χ
χ -
-
-
<
<
-
(continued)

35. Chap 10-35
Example
You are testing the speed of a computer processor. You
collect the following data (in Mhz):
CPUx
Sample size 17
Sample mean 3004
Sample std dev 74
Assume the population is normal.
Determine the 95% confidence interval for σx
2

36. Chap 10-36
Finding the Chi-square Values
§ n = 17 so the chi-square distribution has (n – 1) = 16
degrees of freedom
§ a = 0.05, so use the the chi-square values with area
0.025 in each tail:
probability
α/2 = .025
c2
16
c2
16
= 28.85
6.91
28.85
2
0.975
,
16
2
/2
-
1
,
1
n
2
0.025
,
16
2
/2
,
1
n
=
=
=
=
-
-
χ
χ
χ
χ
α
α
c2
16 = 6.91
probability
α/2 = .025

37. Chap 10-37
Calculating the Confidence Limits
§ The 95% confidence interval is
Converting to standard deviation, we are 95%
confident that the population standard deviation of
CPU speed is between 55.1 and 112.6 Mhz
2
/2
-
1
,
1
n
2
2
2
/2
,
1
n
2
1)s
(n
σ
1)s
(n
α
α χ
χ -
-
-
<
<
-
6.91
1)(74)
(17
σ
28.85
1)(74)
(17 2
2
2
-
<
<
-
12683
σ
3037 2
<
<

38. Chap 10-38
Sample Size Determination
For the
Mean
Determining
Sample Size
For the
Proportion

39. Chap 10-39
Margin of Error
§ The required sample size can be found to reach
a desired margin of error (ME) with a specified
level of confidence (1 - a)
§ The margin of error is also called sampling error
§ the amount of imprecision in the estimate of the
population parameter
§ the amount added and subtracted to the point
estimate to form the confidence interval

40. Chap 10-40
For the
Mean
Determining
Sample Size
n
σ
z
x α/2
±
n
σ
z
ME α/2
=
Margin of Error
(sampling error)
Sample Size Determination

41. Chap 10-41
For the
Mean
Determining
Sample Size
n
σ
z
ME α/2
=
(continued)
2
2
2
α/2
ME
σ
z
n =
Now solve
for n to get
Sample Size Determination

42. Chap 10-42
§ To determine the required sample size for the
mean, you must know:
§ The desired level of confidence (1 - a), which
determines the za/2 value
§ The acceptable margin of error (sampling error), ME
§ The standard deviation, σ
(continued)
Sample Size Determination

43. Chap 10-43
Required Sample Size Example
If s = 45, what sample size is needed to
estimate the mean within ± 5 with 90%
confidence?
(Always round up)
219.19
5
(45)
(1.645)
ME
σ
z
n 2
2
2
2
2
2
α/2
=
=
=
So the required sample size is n = 220

44. Chap 10-44
n
)
p
(1
p
z
p α/2
ˆ
ˆ
ˆ -
±
n
)
p
(1
p
z
ME α/2
ˆ
ˆ -
=
Determining
Sample Size
For the
Proportion
Margin of Error
(sampling error)
Sample Size Determination

45. Chap 10-45
Determining
Sample Size
For the
Proportion
2
2
α/2
ME
z
0.25
n =
Substitute
0.25 for
and solve for
n to get
(continued)
Sample Size Determination
n
)
p
(1
p
z
ME α/2
ˆ
ˆ -
=
cannot
be larger than
0.25, when =
0.5
)
p
(1
p ˆ
ˆ -
p̂
)
p
(1
p ˆ
ˆ -

46. Chap 10-46
§ The sample and population proportions, and P, are
generally not known (since no sample has been taken
yet)
§ P(1 – P) = 0.25 generates the largest possible margin
of error (so guarantees that the resulting sample size
will meet the desired level of confidence)
§ To determine the required sample size for the
proportion, you must know:
§ The desired level of confidence (1 - a), which determines the
critical za/2 value
§ The acceptable sampling error (margin of error), ME
§ Estimate P(1 – P) = 0.25
(continued)
Sample Size Determination
p̂

47. Chap 10-47
Required Sample Size Example
How large a sample would be necessary
to estimate the true proportion defective in
a large population within ±3%, with 95%
confidence?

48. Chap 10-48
Required Sample Size Example
Solution:
For 95% confidence, use z0.025 = 1.96
ME = 0.03
Estimate P(1 – P) = 0.25
So use n = 1068
(continued)
1067.11
(0.03)
6)
(0.25)(1.9
ME
z
0.25
n 2
2
2
2
α/2
=
=
=

49. Chap 10-49
Part 1: Summary
§ Compared two dependent samples (paired samples)
§ Formed confidence intervals for the paired difference
§ Compared two independent samples
§ Formed confidence intervals for the difference between two
means, population variance known, using z
§ Formed confidence intervals for the differences between two
means, population variance unknown, using t
§ Formed confidence intervals for the differences between two
population proportions
§ Formed confidence intervals for the population variance
using the chi-square distribution
§ Determined required sample size to meet confidence
and margin of error requirements

50. Chap 10-50
Hypothesis Testing: Two
populations
Part 2

51. Chap 10-51
Part 2: Goals
After completing this part, you should be able to:
§ Test hypotheses for the difference between two population means
§ Two means, matched pairs
§ Independent populations, population variances known
§ Independent populations, population variances unknown but
equal
§ Complete a hypothesis test for the difference between two
proportions (large samples)
§ Use the chi-square distribution for tests of the variance of a normal
distribution
§ Use the F table to find critical F values
§ Complete an F test for the equality of two variances

52. Chap 10-52
Two Sample Tests
Two Sample Tests
Population
Means,
Independent
Samples
Population
Means,
Matched
Pairs
Population
Variances
Group 1 vs.
independent
Group 2
Same group
before vs. after
treatment
Variance 1 vs.
Variance 2
Examples:
Population
Proportions
Proportion 1 vs.
Proportion 2
(Note similarities to part 1)

53. Chap 10-53
Matched Pairs
Tests Means of 2 Related Populations
§ Paired or matched samples
§ Repeated measures (before/after)
§ Use difference between paired values:
§ Assumptions:
§ Both Populations Are Normally Distributed
Matched
Pairs
di = xi - yi

54. Chap 10-54
The test statistic for the mean
difference is a t value, with
n – 1 degrees of freedom:
n
s
D
d
t
d
0
-
=
Test Statistic: Matched Pairs
Where
D0 = hypothesized mean difference
sd = sample standard dev. of differences
n = the sample size (number of pairs)
Matched
Pairs

55. Chap 10-55
Lower-tail test:
H0: μx – μy ³ 0
H1: μx – μy < 0
Upper-tail test:
H0: μx – μy ≤ 0
H1: μx – μy > 0
Two-tail test:
H0: μx – μy = 0
H1: μx – μy ≠ 0
Paired Samples
Decision Rules: Matched Pairs
a a/2 a/2
a
-ta -ta/2
ta ta/2
Reject H0 if t < -tn-1, a Reject H0 if t > tn-1, a Reject H0 if t < -tn-1 , a/2
or t > tn-1 , a/2
Where
n
s
D
d
t
d
0
-
=
has n - 1 d.f.

56. Chap 10-56
§ Assume you send your salespeople to a “customer
service” training workshop. Has the training made a
difference in the number of complaints? You collect
the following data:
Matched Pairs Example
Number of Complaints: (2) - (1)
Salesperson Before (1) After (2) Difference, di
C.B. 6 4 - 2
T.F. 20 6 -14
M.H. 3 2 - 1
R.K. 0 0 0
M.O. 4 0 - 4
-21
d =
S di
n
5.67
1
n
)
d
(d
S
2
i
d
=
-
-
=
å
= - 4.2

57. Chap 10-57
§ Has the training made a difference in the number of
complaints (at the a = 0.01 level)?
- 4.2
d =
1.66
5
5.67/
0
4.2
n
/
s
D
d
t
d
0
-
=
-
-
=
-
=
H0: μx – μy = 0
H1: μx – μy ¹ 0
Test Statistic:
Critical Value = ± 4.604
d.f. = n - 1 = 4
Reject
a/2
- 4.604 4.604
Decision: Do not reject H0
(t stat is not in the reject region)
Conclusion: There is not a
significant change in the
number of complaints.
Matched Pairs: Solution
Reject
a/2
- 1.66
a = .01

58. Chap 10-58
Difference Between Two Means
Population means,
independent
samples
Goal: Form a confidence interval
for the difference between two
population means, μx – μy
§ Different data sources
§ Unrelated
§ Independent
§ Sample selected from one population has no effect on the
sample selected from the other population

59. Chap 10-59
Difference Between Two Means
Population means,
independent
samples
Test statistic is a z value
Test statistic is a a value from the
Student’s t distribution
σx
2 and σy
2
assumed equal
σx
2 and σy
2 known
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal
(continued)

60. Chap 10-60
Population means,
independent
samples
σx
2 and σy
2 Known
Assumptions:
§ Samples are randomly and
independently drawn
§ both population distributions
are normal
§ Population variances are
known
*
σx
2 and σy
2 known
σx
2 and σy
2 unknown

61. Chap 10-61
Population means,
independent
samples
…and the random variable
has a standard normal distribution
When σx
2 and σy
2 are known and
both populations are normal, the
variance of X – Y is
y
2
y
x
2
x
2
Y
X
n
σ
n
σ
σ +
=
-
(continued)
*
Y
2
y
X
2
x
Y
X
n
σ
n
σ
)
μ
(μ
)
y
x
(
Z
+
-
-
-
=
σx
2 and σy
2 known
σx
2 and σy
2 unknown
σx
2 and σy
2 Known

62. Chap 10-62
Population means,
independent
samples
Test Statistic,
σx
2 and σy
2 Known
*
σx
2 and σy
2 known
σx
2 and σy
2 unknown
( )
y
2
y
x
2
x
0
n
σ
n
σ
D
y
x
z
+
-
-
=
The test statistic for
μx – μy is:

63. Chap 10-63
Hypothesis Tests for
Two Population Means
Lower-tail test:
H0: μx ³ μy
H1: μx < μy
i.e.,
H0: μx – μy ³ 0
H1: μx – μy < 0
Upper-tail test:
H0: μx ≤ μy
H1: μx > μy
i.e.,
H0: μx – μy ≤ 0
H1: μx – μy > 0
Two-tail test:
H0: μx = μy
H1: μx ≠ μy
i.e.,
H0: μx – μy = 0
H1: μx – μy ≠ 0
Two Population Means, Independent Samples

64. Chap 10-64
Two Population Means, Independent
Samples, Variances Known
Lower-tail test:
H0: μx – μy ³ 0
H1: μx – μy < 0
Upper-tail test:
H0: μx – μy ≤ 0
H1: μx – μy > 0
Two-tail test:
H0: μx – μy = 0
H1: μx – μy ≠ 0
a a/2 a/2
a
-za -za/2
za za/2
Reject H0 if z < -za Reject H0 if z > za Reject H0 if z < -za/2
or z > za/2
Decision Rules

65. Chap 10-65
Population means,
independent
samples
σx
2 and σy
2 Unknown,
Assumed Equal
Assumptions:
§ Samples are randomly and
independently drawn
§ Populations are normally
distributed
§ Population variances are
unknown but assumed equal
*
σx
2 and σy
2
assumed equal
σx
2 and σy
2 known
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal

66. Chap 10-66
Population means,
independent
samples
(continued)
Forming interval
estimates:
§ The population variances
are assumed equal, so use
the two sample standard
deviations and pool them to
estimate σ
§ use a t value with
(nx + ny – 2) degrees of
freedom
*
σx
2 and σy
2
assumed equal
σx
2 and σy
2 known
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal
σx2 and σy
2 Unknown,
Assumed Equal

67. Chap 10-67
*
Test Statistic,
σx
2 and σy
2 Unknown, Equal
σx
2 and σy
2
assumed equal
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal
2
n
n
1)s
(n
1)s
(n
s
y
x
2
y
y
2
x
x
2
p
-
+
-
+
-
=
Where t has (n1 + n2 – 2) d.f.,
and
( ) ( )
÷
÷
ø
ö
ç
ç
è
æ
+
-
-
-
=
y
x
2
p
y
x
n
1
n
1
S
μ
μ
t
y
x
The test statistic for
μx – μy is:

68. Chap 10-68
Population means,
independent
samples
σx
2 and σy
2 Unknown,
Assumed Unequal
Assumptions:
§ Samples are randomly and
independently drawn
§ Populations are normally
distributed
§ Population variances are
unknown and assumed
unequal
*
σx
2 and σy
2
assumed equal
σx
2 and σy
2 known
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal

69. Chap 10-69
Population means,
independent
samples
σx
2 and σy
2 Unknown,
Assumed Unequal
(continued)
Forming interval estimates:
§ The population variances are
assumed unequal, so a pooled
variance is not appropriate
§ use a t value with n degrees
of freedom, where
σx
2 and σy
2 known
σx
2 and σy
2 unknown
*
σx
2 and σy
2
assumed equal
σx
2 and σy
2
assumed unequal
1)
/(n
n
s
1)
/(n
n
s
)
n
s
(
)
n
s
(
y
2
y
2
y
x
2
x
2
x
2
y
2
y
x
2
x
-
÷
÷
ø
ö
ç
ç
è
æ
+
-
÷
÷
ø
ö
ç
ç
è
æ
ú
ú
û
ù
ê
ê
ë
é
+
=
v

70. Chap 10-70
*
Test Statistic,
σx
2 and σy
2 Unknown, Unequal
σx
2 and σy
2
assumed equal
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal
1)
/(n
n
s
1)
/(n
n
s
)
n
s
(
)
n
s
(
y
2
y
2
y
x
2
x
2
x
2
y
2
y
x
2
x
-
÷
÷
ø
ö
ç
ç
è
æ
+
-
÷
÷
ø
ö
ç
ç
è
æ
ú
ú
û
ù
ê
ê
ë
é
+
=
v
Where t has n degrees of freedom:
The test statistic for
μx – μy is:
Y
2
y
X
2
x
0
n
σ
n
σ
D
)
y
x
(
t
+
-
-
=

71. Chap 10-71
Lower-tail test:
H0: μx – μy ³ 0
H1: μx – μy < 0
Upper-tail test:
H0: μx – μy ≤ 0
H1: μx – μy > 0
Two-tail test:
H0: μx – μy = 0
H1: μx – μy ≠ 0
Decision Rules
a a/2 a/2
a
-ta -ta/2
ta ta/2
Reject H0 if t < -tn-1, a Reject H0 if t > tn-1, a Reject H0 if t < -tn-1 , a/2
or t > tn-1 , a/2
Where t has n - 1 d.f.
Two Population Means, Independent
Samples, Variances Unknown

72. Chap 10-72
Pooled Variance t Test: Example
You are a financial analyst for a brokerage firm. Is there a
difference in dividend yield between stocks listed on the
NYSE & NASDAQ? You collect the following data:
NYSE NASDAQ
Number 21 25
Sample mean 3.27 2.53
Sample std dev 1.30 1.16
Assuming both populations are
approximately normal with
equal variances, is
there a difference in average
yield (a = 0.05)?

73. Chap 10-73
Calculating the Test Statistic
( ) ( ) ( ) ( ) 1.5021
1)
25
(
1)
-
(21
1.16
1
25
1.30
1
21
1)
n
(
)
1
(n
S
1
n
S
1
n
S
2
2
2
1
2
2
2
2
1
1
2
p =
-
+
-
+
-
=
-
+
-
-
+
-
=
( ) ( ) ( ) 2.040
25
1
21
1
5021
.
1
0
2.53
3.27
n
1
n
1
S
μ
μ
X
X
t
2
1
2
p
2
1
2
1
=
÷
ø
ö
ç
è
æ
+
-
-
=
÷
÷
ø
ö
ç
ç
è
æ
+
-
-
-
=
The test statistic is:

74. Chap 10-74
Solution
H0: μ1 - μ2 = 0 i.e. (μ1 = μ2)
H1: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2)
a = 0.05
df = 21 + 25 - 2 = 44
Critical Values: t = ± 2.0154
Test Statistic: Decision:
Conclusion:
Reject H0 at a = 0.05
There is evidence of a
difference in means.
t
0 2.0154
-2.0154
.025
Reject H0 Reject H0
.025
2.040
2.040
25
1
21
1
5021
.
1
2.53
3.27
t =
÷
ø
ö
ç
è
æ
+
-
=

75. Chap 10-75
Two Population Proportions
Goal: Test hypotheses for the
difference between two population
proportions, Px – Py
Population
proportions
Assumptions:
Both sample sizes are large,
nP(1 – P) > 9

76. Chap 10-76
Two Population Proportions
Population
proportions
(continued)
§ The random variable
is approximately normally distributed
y
y
y
x
x
x
y
x
y
x
n
)
p
(1
p
n
)
p
(1
p
)
p
(p
)
p
p
(
Z
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
-
+
-
-
-
-
=

77. Chap 10-77
Test Statistic for
Two Population Proportions
Population
proportions
The test statistic for
H0: Px – Py = 0
is a z value:
( )
y
0
0
x
0
0
y
x
n
)
p
(1
p
n
)
p
(1
p
p
p
z
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
-
+
-
-
=
y
x
y
y
x
x
0
n
n
p
n
p
n
p
+
+
=
ˆ
ˆ
ˆ
Where

78. Chap 10-78
Decision Rules: Proportions
Population proportions
Lower-tail test:
H0: px – py ³ 0
H1: px – py < 0
Upper-tail test:
H0: px – py ≤ 0
H1: px – py > 0
Two-tail test:
H0: px – py = 0
H1: px – py ≠ 0
a a/2 a/2
a
-za -za/2
za za/2
Reject H0 if z < -za Reject H0 if z > za Reject H0 if z < -za/2
or z > za/2

79. Chap 10-79
Example:
Two Population Proportions
Is there a significant difference between the
proportion of men and the proportion of
women who will vote Yes on Proposition A?
§ In a random sample, 36 of 72 men and 31 of
50 women indicated they would vote Yes
§ Test at the .05 level of significance

80. Chap 10-80
§ The hypothesis test is:
H0: PM – PW = 0 (the two proportions are equal)
H1: PM – PW ≠ 0 (there is a significant difference between
proportions)
§ The sample proportions are:
§ Men: = 36/72 = .50
§ Women: = 31/50 = .62
.549
122
67
50
72
50(31/50)
72(36/72)
n
n
p
n
p
n
p
y
x
y
y
x
x
0 =
=
+
+
=
+
+
=
ˆ
ˆ
ˆ
§ The estimate for the common overall proportion is:
Example:
Two Population Proportions
(continued)
M
p̂
W
p̂

81. Chap 10-81
The test statistic for PM – PW = 0 is:
Example:
Two Population Proportions
(continued)
.025
-1.96 1.96
.025
-1.31
Decision: Do not reject H0
Conclusion: There is not
significant evidence of a
difference between men
and women in proportions
who will vote yes.
( )
( )
1.31
50
.549)
(1
.549
72
.549)
(1
.549
.62
.50
n
)
p
(1
p
n
)
p
(1
p
p
p
z
2
0
0
1
0
0
W
M
-
=
÷
ø
ö
ç
è
æ -
+
-
-
=
-
+
-
-
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
Reject H0 Reject H0
Critical Values = ±1.96
For a = .05

82. Chap 10-82
Population
Variance
2
2
2
1
n
σ
1)s
(n-
=
-
c
follows a chi-square distribution with
(n – 1) degrees of freedom
§ Goal: Test hypotheses about the
population variance, σ2
§ If the population is normally distributed,
Hypothesis Tests of
one Population Variance

83. Chap 10-83
Confidence Intervals for the
Population Variance
Population
Variance
The test statistic for
hypothesis tests about one
population variance is
2
0
2
2
1
n
σ
1)s
(n-
=
-
χ
(continued)

84. Chap 10-84
Decision Rules: Variance
Population variance
Lower-tail test:
H0: σ2 ³ σ0
2
H1: σ2 < σ0
2
Upper-tail test:
H0: σ2 ≤ σ0
2
H1: σ2 > σ0
2
Two-tail test:
H0: σ2 = σ0
2
H1: σ2 ≠ σ0
2
a a/2 a/2
a
Reject H0 if Reject H0 if Reject H0 if
or
2
,
1
n a
-
χ
2
,1
1
n a
-
-
χ 2
,1
1
n 2
/
a
-
-
χ 2
,
1
n 2
/
a
-
χ
2
,1
1
n
2
1
n a
-
-
- < χ
χ
2
,
1
n
2
1
n a
-
- > χ
χ
2
,
1
n
2
1
n 2
/
a
-
- > χ
χ
2
,1
1
n
2
1
n 2
/
a
-
-
- < χ
χ

85. Chap 10-85
Hypothesis Tests for Two Variances
Tests for Two
Population
Variances
F test statistic
H0: σx
2 = σy
2
H1: σx
2 ≠ σy
2 Two-tail test
Lower-tail test
Upper-tail test
H0: σx
2 ³ σy
2
H1: σx
2 < σy
2
H0: σx
2 ≤ σy
2
H1: σx
2 > σy
2
§ Goal: Test hypotheses about two
population variances
The two populations are assumed to be
independent and normally distributed

86. Chap 10-86
Hypothesis Tests for Two Variances
Tests for Two
Population
Variances
F test statistic
2
y
2
y
2
x
2
x
/σ
s
/σ
s
F =
The random variable
Has an F distribution with (nx – 1)
numerator degrees of freedom and
(ny – 1) denominator degrees of
freedom
Denote an F value with n1 numerator and n2
denominator degrees of freedom by
(continued)

87. Chap 10-87
Test Statistic
Tests for Two
Population
Variances
F test statistic 2
y
2
x
s
s
F =
The critical value for a hypothesis test
about two population variances is
where F has (nx – 1) numerator
degrees of freedom and (ny – 1)
denominator degrees of freedom

88. Chap 10-88
Decision Rules: Two Variances
n rejection region for a two-
tail test is:
F
0
a
Reject H0
Do not
reject H0
F
0
a/2
Reject H0
Do not
reject H0
H0: σx
2 = σy
2
H1: σx
2 ≠ σy
2
H0: σx
2 ≤ σy
2
H1: σx
2 > σy
2
Use sx
2 to denote the larger variance.
α
1,
n
1,
n y
x
F -
-
2
/
α
1,
n
1,
n
0 y
x
F
F
if
H
Reject -
-
>
2
/
α
1,
n
1,
n y
x
F -
-
where sx
2 is the larger of
the two sample variances
α
1,
n
1,
n
0 y
x
F
F
if
H
Reject -
-
>

89. Chap 10-89
Example: F Test
You are a financial analyst for a brokerage firm. You
want to compare dividend yields between stocks listed
on the NYSE & NASDAQ. You collect the following data:
NYSE NASDAQ
Number 21 25
Mean 3.27 2.53
Std dev 1.30 1.16
Is there a difference in the
variances between the NYSE
& NASDAQ at the a = 0.10 level?

90. Chap 10-90
F Test: Example Solution
§ Form the hypothesis test:
H0: σx
2 = σy
2 (there is no difference between variances)
H1: σx
2 ≠ σy
2 (there is a difference between variances)
Degrees of Freedom:
§ Numerator
(NYSE has the larger
standard deviation):
§ nx – 1 = 21 – 1 = 20 d.f.
§ Denominator:
§ ny – 1 = 25 – 1 = 24 d.f.
§ Find the F critical values for a = .10/2:
2.03
F
F
0.10/2
,
24
,
20
,
1
n
,
1
n y
x
=
=
-
- 2
/
α

91. Chap 10-91
§ The test statistic is:
1.256
1.16
1.30
s
s
F 2
2
2
y
2
x
=
=
=
a/2 = .05
Reject H0
Do not
reject H0
H0: σx
2 = σy
2
H1: σx
2 ≠ σy
2
F Test: Example Solution
§ F = 1.256 is not in the rejection
region, so we do not reject H0
(continued)
§ Conclusion: There is not sufficient evidence
of a difference in variances at a = .10
F
2.03
F 0.10/2
,
24
,
20 =

92. Chap 10-92
Two-Sample Tests in EXCEL
For paired samples (t test):
§ Tools | data analysis… | t-test: paired two sample for means
For independent samples:
§ Independent sample Z test with variances known:
§ Tools | data analysis | z-test: two sample for means
For variances…
§ F test for two variances:
§ Tools | data analysis | F-test: two sample for variances

93. Chap 10-93
Chapter Summary
§ Compared two dependent samples (paired
samples)
§ Performed paired sample t test for the mean
difference
§ Compared two independent samples
§ Performed z test for the differences in two means
§ Performed pooled variance t test for the differences
in two means
§ Compared two population proportions
§ Performed z-test for two population proportions

94. Chap 10-94
Part 1: Summary
§ Used the chi-square test for a single population
variance
§ Performed F tests for the difference between
two population variances
§ Used the F table to find F critical values
(continued)