Download 9-2B: Using Confidence Intervals to Compare the Difference in 2

Section 9 – 3B:
Using Confidence Intervals to Estimate the Difference (µ1 − µ2 )
in Two Population Means using Two Independent Samples
Requirements
1.A random sample of each population is taken. The sample mean for the sample of one population
is x 1 and the sample mean for the sample of a second population is x 2
2. The two samples are independent of each other.
3. Both populations are normal or both sample sizes are greater than 30.
Notation for the Samples of Two Population Means
Population 1
µ1 = population mean
Population 2
µ2 = population mean
n1 = sample size
n2 = sample size
x 1 = Sample Mean from Population 1
x 2 = Sample Mean from Population 2
s1 = Sample Standard Deviation
from Population 1
s2 = Sample Standard Deviation
from Population 2
Creating a Confidence Interval to Estimate the value of
the Difference in 2 Population Means µ1 − µ2
(x1 − x 2 ) − E < (µ1 − µ2 ) < (x1 − x 2 ) + E
if
x1 > x2
if
x2 > x1
or
(x 2 − x1 ) − E < (µ2 − µ1 ) < (x 2 − x1 ) + E
Where E = tα 2 •
( s1 )2
n1
+
(s 2 )2
n2
Use the smallest Degrees of Freedom from Sample 1 and Sample 2.
Note: It does not matter which population is population 1 and population 2. It is important
that the difference between them is positive. The first formula is based on x 1 being larger
than x 2 . The second formula is based on x 2 being larger than x 1 .
You are free to chose which population is selected to be population . If you always chose the
population with the largest x as population 1 then you can always use the first formula.
Section 9– 2B Lecture
Page 1 of 1
©2013 Eitel
Example 1
Is Ortho Grow fertilizer more effective in lawn growth then a generic fertilizer?
Ortho Grow brand fertilizer selected 50 lawns and measured how much the grass had grown in 30
days. The average growth was 2.18 inches with a standard deviation of .64 inches. A generic fertilizer
was also tested on 35 randomly selected lawn. The average growth for the generic brand was 2.37
inches with a standard deviation of .85 inches. Construct a 90% confidence interval for the difference
between the two means. Does it appear there is a difference in the two population means? How
can you tell?
Sample 1
The sample with the larger mean
mean is labeled Sample 1:
x 1 = 2.37
s1 = .85
n1 = 35
Sample 2
The sample with the smaller mean
is labeled Sample 2:
x 1 = 2.18
s1 = .64
n1 = 50
Use the smallest Degrees of Freedom from Sample 1 and Sample 2.
Find the critical value +tα
α = .10 so α / 2 = .05
DF = 34
negative
critical
value
E = tα 2 •
right tail
area = .05
left tail
area = .05
−tα 2 = – 1.691
Find the Maximum Error
2
+tα
E = 1.691•
t
2
= 1.691
( s1 )
(s )
+ 2
2
n1
2
n2
(.85)2 (.64 )2
35
+
50
E = .29
positive
critical
value
Confidence Interval
(x 1 − x 2 ) − E < µ 1 − µ 2 < ( x 1 − x 2 ) + E
(2.37 − 2.18) − .29 < µ1 − µ2 < (2.37 − 2.18) + .29
−.10 < µ1 − µ2 < .48
Conclusion based on the problem: The confidence interval does contain zero.
I am 90% confident that here is no difference in the average growth of the lawn using Ortho
or the generic fertilizer.
Section 9– 2B Lecture
Page 2 of 2
©2013 Eitel
t Distribution: Critical t Values
Degrees of
Freedom
34
Area In One Tail (Right Tail)
0.100
0.050
0.025
0.010
0.005
1.307
1.691
2.032
2.441
2.728
Section 9– 2B Lecture
Page 3 of 3
©2013 Eitel
Example 2
Does completing the homework in Statistics help improve your test score?
Mr. Jensen selected 51 Statistics students at random from his classes who had completed all the
Chapter 9 homework. The average score on the Chapter 9 test for these students was 92.5 points
with a standard deviation of 2 points. Mr. Jensen selected 54 Statistics students at random from his
classes who had not completed all the Chapter 9 homework. The average score on the Chapter 9
test for these students was 83 points with a standard deviation of 4 points. Construct a 95%
confidence interval for the difference between the two means. Dos it appear there is a difference in
the two population means? How can you tell?
Sample 1
The sample with the larger mean
mean is labeled Sample 1:
x 1 = 92.5
s1 = 2
n1 = 51
Sample 2
The sample with the smaller mean
is labeled Sample 2:
x 1 = 83
s1 = 4
n1 = 54
Use the smallest Degrees of Freedom from Sample 1 and Sample 2.
Find the critical value +tα
α = .05 so α / 2 = .025
DF = 50
left tail
area = .025
Find the Maximum Error
2
E = tα
right tail
area = .025
2
E = 2.009
t
t = – 2.009
negative
critical
value
t = 2.009
(s1 )
2
n1
+
( s2 )
2
n2
(2)2 (4 )2
51
+
54
E = 1.23
positive
critical
value
Confidence Interval
( x 1 − x 2 ) − E < µ1 − µ2 < ( x 1 − x 2 ) + E
(92.5 − 83) −1.23 < µ1 − µ2 < (92.5 − 83) − 1.23
8.27 < µ1 − µ2 < 10.73
Conclusion based on the problem: The confidence does not contain zero.
I am 95% confident that there is difference at the .05 significance level in the test scores for
students that complete the chapter homework. They score from 8.3 to 10. 7 points higher on
the test
Section 9– 2B Lecture
Page 4 of 4
©2013 Eitel
t Distribution: Critical t Values
Degrees of
Freedom
50
Area In One Tail (Right Tail)
0.100
0.050
0.025
0.010
0.005
1.299
1.676
2.009
2.403
2.678
Section 9– 2B Lecture
Page 5 of 5
©2013 Eitel