Chapter 8 - Comparing Two Population Means
8.1
In this chapter, we will compare the means of two
populations. For example,
a. Does Advil fight pain better than Tylenol?
b. Does Longhorn Steakhouse have a longer wait
than Outback Steakhouse?
c. Are men older than women when they graduate
from college?
d. Are SAT scores higher after taking a preparation
course than before?
e. Are the starting salaries for computer science
majors higher than those of marketing majors?
The method we first consider requires not only that
the samples selected from the two populations be
random but that they be independent samples.
Independent Samples: Two samples are
independent if the values in one sample are not
related to values in the other sample.
THEOREM. If independent and random samples of
sizes n1 and n2 (n1 and n2  30) are drawn from two
populations with means 1 and 2 and with standard
deviations of 1 and 2, respectively, then the
sampling distribution of X 1 - X 2 has the following
properties:
1. E( X 1 - X 2 )
= 1 - 2
2. S.D( X 1 - X 2 ) =  1 / n1   2 / n2
3. X 1 - X 2 is approximately
distributed.
That is,
X 1 - X 2 ~ N(1 - 2,  1 / n1   2 / n 2 )
2
2
2
normally
2
Notes:
1. A CI for 1 - 2 has the form
( X 1 - X 2 )  Z  1 / n1   2 / n2
 /2
2
2
2. If 1 and 2 are unknown, we will use s1 and s2
to estimate 1 and 2, respectively.
Example 1.The same standardized test is given to
students selected at random from two schools. The
following table summarizes the results:
School A: sample size = 50, mean = 68.7,
sd = 6.0
School B: sample size = 40, mean = 65.1,
sd = 6.4
(a) Construct a 95% C.I. estimate for 1 - 2.
(b) Construct an 84% C.I. estimate for 1 - 2.
Solution (a) [1.012, 6.188]
(b) [1.738, 5.462]; z = 1.41
Example 2. A small amount of selenium, from 50 to
200 micrograms per day, is considered essential to
good health. Suppose that two random samples of
30 adults were selected, one from region A and
another from region B of the United States. A day's
intake of selenium from both, liquids and solids, was
recorded for the 30 adults from region A and region
B. The following was observed:
Region A: mean = 167.1, sd = 24.3 micrograms.
Region B: mean = 140.9, sd = 17.6 micrograms.
(a) Construct a 90% C.I. estimate for 1 - 2.
(b) Construct a 68% C.I. estimate for 1 - 2.
Solution (a) [17.16, 35.24]
(b) [20.78, 31.62] z = 0.99
Hypothesis Testing for 1 - 2
Example 3. A sampling of 50 women and 40
men produced a mean age at death for women of
75.5 years and for men of 67.9 years with a sd of
16.2 and 18.3 respectively. At the 0.05 level of
significance, do women live longer than men?
Solution. Let
1 = mean age for women at death.
2 = mean age for men at death.
Step1: State the null and alternative hypotheses.
H : 1 - 2 = 0
H : 1 - 2 > 0
0
a
Step 2: Write the given information using
proper symbols.
x1 = 75.5, n1 = 50, s1 = 16.2
x2 = 67.9, n2 = 40, s2 = 18.3,  = 0.05
S.D( X 1 - X 2 )=  1 / n1   2 / n2 = 3.6907
2
2
Step 3: Find the test statistic.
Z = (VariableSD Mean)
= [(75.5 – 67.9) – 0]/3.6907
= 2.06
Step 4: Find the p-value
p-value = P[Z > 2.06] = 0.0197
Step 5: Conclusion
Since the p-value < , reject the null hypothesis.
Yes, women live longer than men.
Example 4. Last year a sample of 100 new cars
manufactured by a Japanese car-maker averaged
33.8 mpg. This year a sample of 100 new cars
of the same car-maker averaged 35.1 mpg. If
each sample had a sd of 10 mpg, can the carmaker conclude that they have significantly
increased the performance their cars? Use 0.01
as the level of significance.
Solution. Let
1 = average mpg of last year’s cars.
2 = average mpg of this year’s cars.
Step1: State the null and alternative hypotheses.
H : 1 - 2 = 0
H : 1 - 2 < 0
0
a
Step 2: Write the given information using
proper symbols.
x1 = 33.8, n1 = 100, s1 = 10
x2 = 35.1, n2 = 100, s2 = 10,  = 0.01
S.D( X 1 - X 2 )=  1 / n1   2 / n2 = 1.414
2
2
Step 3: Find the test statistic.
Z = (VariableSD Mean)
= [(33.8 – 35.1) – 0]/1.414
= -0.92
Step 4: Find the p-value
p-value = P[Z <-0.92] = 0.1788
Step 5: Conclusion
Since the p-value >= , do not reject the null
hypothesis.
No, they have not significantly increased the
performance of their cars.
Example 5. A reading test is given to an
elementary school class that consists of 42
Anglo-American
children
and
40
Mexican-American children. The means for the
Anglo-American
and
Mexican-American
children were 74 and 70, respectively whereas
the sds for the Anglo-American and the
Mexican-American children were 8 and 10,
respectively. Is the difference between the
means of the two groups significant at the 0.05
level?
Solution. Let
1 = mean score for Anglo-Americans
2 = mean score for Mexican-Americans.
Step1: State the null and alternative hypotheses.
H : 1 - 2 = 0
H : 1 - 2  0
0
a
Step 2: Write the given information using
proper symbols.
x1 = 74, n1 = 42, s1 = 8
x2 = 70, n2 = 40, s2 = 10,  = 0.05
S.D( X 1 - X 2 )=  1 / n1   2 / n2 = 2.006
2
2
Step 3: Find the test statistic.
Z = (VariableSD Mean)
= [(74. – 70) – 0]/2.006
= 1.994
Step 4: Find the p-value
p-value = P[|Z| > 1.994] =2(0.0233) =
0.0466
Step 5: Conclusion
Since the p-value < , reject the null hypothesis.
Yes, there is a significant difference between the
means of the two groups.
HW: 9-19(odd), 29, 30 (pp. 396-400)
8.2 Inferences For Two Populations Means
Using Independent Samples (Standard
Deviations Assumed Equal)
Assumptions:
1. Two independent samples of data,
2. Both populations are normally distributed,
3. Variances of both populations are equal, but
unknown,
4. Both samples are small (n1, n2 < 30).
Because we are assuming that the variances of
two populations are equal, we say that there is a
 =
common variance,  . In other words,
 =  . The way that we estimate  is with a
value called the pooled estimate of common
variance, s .
2
2
2
1
2
2
2
2
p
s
2
p
=
(n1  1) s12  (n2  1) s 2 2
n1  n2  2
Distribution of the Pooled t-Statistic
Under the assumptions above, the variable
t=
( x1  x 2)  ( 1   2)
sp
1
1

n1 n2
has the t-distribution with df = n1 + n2 – 2
(See page 403)
Under the assumptions 1-4 above, A CI for 1
- 2 is
( x1  x2 )  t s
 /2
p
1
1

n1 n2
where 1 -  is the
confidence level.
Example 6. A survey of recent graduates with
degrees in Marketing and Accounting revealed
the following starting salaries in thousands of
dollars.
Marketing: 24.6, 27.8, 25.8, 20.8, 23.1, 25.2
24.7, 26.1, 22.6, 23.8
Accounting: 26.7, 24.9, 27.1, 23.1, 27.5, 27.4,
24.9, 26.3, 28.9, 26.4, 28.1, 29.7, 25.5, 24.9,
28.5
Is there an evidence at the 5% significance level
that Accounting majors have higher starting
salaries than Marketing majors?
Solution
1 = mean salary of Marketing majors
2 = mean salary of Accounting majors.
Step1: State the null and alternative hypotheses.
H : 1 - 2 = 0
H : 1 - 2 < 0
0
a
Step 2: Write the given information using
proper symbols.
x1 = 24.45, n1 = 10, s1 = 1.98
x2 = 26.66, n2 = 15, s2 = 1.79,
 = 0.05, df = 10+15 – 2 = 23
The pooled estimate of common variance,
s = (n1  1)ns11 n2(n22 1)s2 = 3.482
2
2
p
2
S.D( X 1 -
X2
)= s
p
1
1

n1 n2
= 0.762
Step 3: Find the test statistic.
t=
( x1  x 2)  ( 1   2)
sp
1
1

n1 n2
= -2.90
Step 4: Find the p-value
P-value = P[t < -2.90] = .004
Step 5: Conclusion
Since the p-value < 0.05, reject the null
hypothesis.
Yes, there is strong evidence that Accounting
majors have higher starting salaries than
Marketing majors.
Example 7. A company assembles large
electronic devices. The assembly method they
use is rather simple and has been used for a long
time. Recently, the company has become aware
of a new assembly technique. The company is
basically satisfied with their current method of
assembly. However, if the company finds that
the new method significantly reduces assembly
time, they would adopt this new method. The
company decides to do a test to compare the two
methods over the course of a day. One
production line will use the current method and
the other production line will try the new
method. The data (in minutes) are shown below:
Current: 37.3, 55.9, 45.4, 52.2, 39.1, 34.4
New : 47.4, 57.3, 39.8, 60.4, 46.7, 36.4
Should the company adopt the new method? Use
a significance level of 10%.
Solution
1 = mean assembly time with the current
method.
2 = mean assembly time with the new method.
Step1: State the null and alternative hypotheses.
H : 1 - 2 = 0
H : 1 - 2 >0
0
a
Step 2: Write the given information using
proper symbols.
x1 = 44.05, n1 = 6, s1 = 8.62
x2 = 48.00, n2 = 6, s2 = 9.42,
 = 0.10, df = 6+6 – 2 = 10
The pooled estimate of the common
variance,
s = (n1  1)ns11 n2(n22 1)s2 = 81.52
2
2
2
p
S.D( X 1 -
X2
)= s
p
1
1

n1 n2
= 5.21
Step 3: Find the test statistic.
t=
( x1  x 2)  ( 1   2)
sp
1
1

n1 n2
= -0.758
Step 4: Find the p-value
p-value = P(t > -0.76) = 0.77
Step 5: Conclusion
Since the p-value >= 0.10, do not reject the null
hypothesis.
The company should not adopt the new method.
Example 8. Find a 90% CI for 1 - 2. Assume
that the variance of both populations to be the
same. You are given the following information:
= 42.13 , n1 = 10, s1 = 0.685
x2 = 43.23, n2 = 10, s2 = 0.750,
df = 10+10 – 2 = 18
s = (n1  1)ns11 n2(n22 1)s2 = 0.516
x1
2
2
2
p
S.D( X 1 -
X2
)= s
p
1
1

n1 n2
= 0.321
A CI for 1 - 2 is
( x1  x2 )  t s
 /2
p
1
1

n1 n2
= (42.13 – 43.23)  1.734(0.321)
= -1.10  0.56
= [-1.66, -0.54]
HW: 11, 17, 18 and 21 (pp. 406-408)