Download Confidence Intervals on the Mean of Paired Differences,

Statistics 215 Lab Materials
Confidence Intervals on the Mean of Paired Differences,
µd
We need some notation before we introduce the formula for this confidence interval. µd is the mean of the
differences between matched (or paired) elements of the first and second populations. We will estimate µd
€
with d , the average of the paired differences in the two samples. Since each observation from the first
sample is paired with one observation from the second sample, we only need a single number for the
sample size. We will use n. The standard deviation of these n differences, will be sd.
Confidence interval for the mean of paired difference, µ d (Small Sample)
The following (1-α)*100% CI for the mean difference of paired difference can be used when
n (the number of paired observations) is more than 2 and the original data is approximately Normal (which
implies that the differences will be Normally distributed also)
d ± t(n −1,1−α ) *
2
€
sd
n
where d is the mean of the differences of the paired observations and sd is the standard deviation of the
differences of the paired observations. Since the observations are paired, we treat the differences as a
single population.
sd
Note that s d =
n
Example:
Representatives of Vanguard airlines want to claim that their rates are lower than their competitor
Southwest Airlines. They sample 26 fares for routes that both airlines fly. The values for these routes are
given in the table below. Vanguard fares on those routes have an average cost of 121.89 with a standard
deviation of 38.58. Southwest fares have a mean cost of 126.01 and a standard deviation of 36.35 for those
same routes. For the differences the mean is –4.12 (Vanguard minus Southwest) and the standard deviation
of the differences is 5.29. Assuming that the data comes from a Normal distribution, create a 95%
confidence interval for the difference in the means between Vanguard and Southwest.
Since n>2 and we have assumed a Normal distribution for the data we can use the formula given above.
The population is paired since we are looking at the same routes for two different airlines. First note that
d = -4.12, and sd = 5.29. These numbers were calculated on the values in the “diff” column of the table
below. This example provides additional information about the standard deviations of the values for
Vanguard and Southwest that is irrelevant to the confidence interval we are making.
Route
Boston to Detroit
Boston to Memphis
Bostion to Charlotte
Charlotte to Memphis
Charlotte to Dallas
Vanguard fare
Southwest fare
138.79
109.99
139.64
76.83
76.09
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139.01
110.76
142.51
81.78
83.98
diff
-0.22
-0.77
-2.86
-4.95
-7.89
Statistics 215 Lab Materials
Charlotte to Salt Lake City
Charlotte to Denver
Dallas to Boston
Dallas to Los Angeles
Dallas to Denver
Dallas to Chicago
Dallas to Des Moines
Denver to Boston
Denver to Detroit
Denver to Salt Lake City
Des Moines to Seattle
Memphis to New Orleans
Omaha to Albuquerque
Omaha to Seattle
Philadelphia to Boston
Philadelphia to Denver
Pittsburgh to Omaha
Seattle to Los Angeles
Seattle to Portland
Tulsa to Dallas
Winston-Salem to Omaha
d ± t(n −1,1−α ) *
2
184.62
114.90
64.57
171.12
186.17
124.87
100.42
140.48
144.70
114.81
95.57
150.37
139.06
40.26
193.23
77.56
139.78
90.40
105.77
120.82
128.44
186.08
122.91
78.39
177.80
180.75
127.94
96.54
145.23
146.40
114.41
102.44
148.66
153.58
49.73
198.27
89.79
139.87
100.93
109.90
122.49
126.11
-1.46
-8.01
-13.82
-6.68
5.42
-3.08
3.87
-4.75
-1.70
0.40
-6.87
1.71
-14.52
-9.47
-5.04
-12.23
-0.09
-10.53
-4.13
-1.67
2.34
sd
n
= −4.11 ± t(25,0.975) *
€
= −4.11 ± 2.060 *
€
5.29
26
5.29
26
= −4.11 ± 2.137
€
= (-6.247, -1.973)
€
We are 95% confident that the mean difference between Vanguard and Southwest (Vanguard minus
Southwest) on identical routes is between $ –6.247 and $-1.973. Since zero is not inside this interval, we
can claim with 95% confidence that there is a difference between the mean cost of flights on identical
routes of these two airlines. Consequently, Vanguard can claim that with 95% confidence, that their rates
are lower than Southwest’s.
Confidence interval for the mean of paired difference, µ d (Large Sample)
When n is large (at least 30 paired differences in the sample) we can use the following formula to compute
a confidence interval on µd
d ± z(1−α ) *
2
€
sd
n
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Statistics 215 Lab Materials
Example:
Suppose d = 10.45, sd = 4.19 and n = 56. We want to create a 90% CI for the difference of paired means.
d ± z(1−α ) *
2
€
sd
n
= 10.45 ± z(0.975) *
= 10.45 ± 1.96 *
€
4.19
56
4.19
56
= 10.45 ± 1.097
€
€
= (9.353, 11.547)
Hence, we are 95% confidence that the mean of paired differences between the two populations is between
9.353 and 11.547.
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