Comparing Two Population Means (matched pairs and independent samples)
1.
The personnel manager of a large retail clothing store suspects a difference in the mean amount of break time taken by
workers during the weekday shifts compared to that of the weekend shifts. It is suspected that the weekday workers take
X 1 = 53 minutes of break time per
8-hour shift and S1 = 7.3 minutes. A random sample of 40 weekend workers had a mean X 2 = 47 minutes and S 2 =
longer breaks on the average. A random sample of 46 weekday workers had a mean 1
9.1 minutes. Test the manager’s suspicion at the 5% level of significance.
Two-Sample Z-Test and CI
Sample
1
2
N
46
40
Mean
53.00
47.00
StDev
7.30
9.10
SE Mean
1.1
1.4
Difference = mu (1) - mu (2)
Estimate for difference: 6.00
95% lower bound for difference: 3.01
Z-Test of difference = 0 (vs >): Z-Value = 3.34
2.
P-Value = 0.001
Two growth hormones are being considered. A random sample of 10 rats were given the first hormone and their average
X 1 = 2.3 pounds with standard deviation S1 = 0.4 pound. For the second hormone, a random sample
of 15 rats showed their average weight gain to be X 2 = 1.9 pounds with standard deviation S 2 = 0.2 pound. Assume
weight gain was
the weight gains follow a normal distribution. Using a 10% level of significance, can we say there is a difference in
average weight gains for the two growth hormones?
Two-Sample T-Test and CI
Sample
1
2
N
10
15
Mean
2.300
1.900
StDev
0.400
0.200
SE Mean
0.13
0.052
Difference = mu (1) - mu (2)
Estimate for difference: 0.400
90% CI for difference: (0.156, 0.644)
T-Test of difference = 0 (vs not =): T-Value = 2.93
3.
P-Value = 0.013
DF = 23
A local claims that the waiting time for its customers to be served is the lowest in the area. A competitor's bank checks the
waiting times at both banks. The sample statistics are listed below. Test the local bank's claim. Use   0.05.
Local Bank
n1 = 15
Competitor Bank
n2 = 16
X 1 = 5.3 minutes
X 2 = 5.6 minutes
S1 = 1.1 minutes
S2 = 1.0 minutes
Two-Sample T-Test and CI
Sample
1
2
N
15
16
Mean
5.30
5.60
StDev
1.10
1.00
SE Mean
0.28
0.25
Difference = mu (1) - mu (2)
Estimate for difference: -0.300
95% upper bound for difference: 0.344
T-Test of difference = 0 (vs <): T-Value = -0.79
P-Value = 0.217
DF = 29
4. Suppose that simple random samples of college freshman are selected from two universities - 15
students from school A and 20 students from school B. On a standardized test, the sample from school
A has an average score of 1000 with a standard deviation of 100. The sample from school B has an
average score of 950 with a standard deviation of 90. What is the 95% confidence interval for the
difference in test scores at the two schools, assuming that test scores came from normal distributions in
both schools?
Two-Sample T-Test and CI
Difference = mu (1) - mu (2)
Estimate for difference: 50.0
95% CI for difference: (-15.6, 115.6)
T-Test of difference = 0 (vs not =): T-Value = 1.55
Both use Pooled StDev = 94.3719
P-Value = 0.130
DF = 33
Answer: 95% confidence interval is -15.6 to 115.6. That is, we are 95% confident that the true
difference in population means is within: -15.6 to 115.6.
5. The local baseball team conducts a study to find the amount spent on refreshments at the ball park.
Over the course of the season they gather simple random samples of 50 men and 100 women. For men,
the average expenditure was $20, with a standard deviation of $3. For women, it was $15, with a
standard deviation of $2. What is the 99% confidence interval for the spending difference between men
and women? Assume that the two populations are independent and normally distributed.
Two-Sample Z-Test and CI
Difference = mu (1) - mu (2)
Estimate for difference: 5.000
99% CI for difference: (3.759, 6.241)
Z-Test of difference = 0 (vs not =): Z -Value = 10.66
P-Value = 0.000
Answer: The 99% confidence interval is $3.76 to $6.24. That is, we are 99% confident that men
outspend women at the ballpark by at least $3.76 and at most $6.24.
6.
SAT prep courses are often very expensive, but are they worth it? The data below is from nine different randomly
selected students who took the SAT twice (once before taking Kaplan’s test prep course and one after taking the Kaplan
test prep course). At the 5% level of significance, test the claim that the after scores are better:
Student
1
2
3
4
Before Score
480
510
530
540
After Score
460
500
530
520
Difference
20
10
0
20
**The standard deviation of the differences is 25.2212
7.
5
550
580
-30
6
560
580
-20
7
600
560
40
8
620
640
-20
9
660
690
-30
A salesman for a shoe company claimed that runners would record quicker times, on the average, with the company's brand
of sneaker. A track coach decided to test the claim. The coach selected eight runners. Each runner ran two 100-yard dashes
on different days. In one 100-yard dash, the runners wore the sneakers supplied by the school; in the other, they wore the
sneakers supplied by the salesman. Each runner was randomly assigned the sneakers to wear for the first run. Their times,
measured in seconds, were as follows:
Runners
Company Sneaker
School Sneaker
1
10.8
11.4
2
12.3
12.5
3
10.7
10.8
4
12.0
11.7
5
10.6
10.9
6
11.5
11.8
7
12.1
12.2
8
11.2
11.7
X D = -.225 and s D = .276. Assume the population of differences is approximately normal.
Note. For the differences,
Paired T-Test and CI
Diff.
N
8
Mean
-0.2250
StDev
0.2760
SE Mean
0.0976
95% CI for mean difference: (-0.4557, 0.0057)
T-Test of mean difference = 0 (vs < 0): T-Value = -2.31;
8.
P-Value = 0.027
Twenty-four males age 25-29 were selected from the Framingham Heart Study. Twelve were smokers and 12 were
nonsmokers. The subjects were paired, with one being a smoker and the other a nonsmoker. Otherwise, each pair was
similar with regard to age and physical characteristics. Systolic blood pressure readings were as follows:
People
Smokers
Nonsmokers
1
122
114
2
146
134
3
120
114
4
114
116
5
124
138
6
126
110
7
118
112
8
128
116
9
130
132
10
134
126
11
116
108
12
130
116
List the differences A - B and verify that X D = 6 and s D = 8.40. Use a 5% level of significance to determine whether the
data indicate a difference in mean systolic blood pressure levels for the populations from which the two groups were
selected. You may assume that the population of differences is approximately normal.
Paired T-Test and CI
Diff.
N
Mean
StDev
SE Mean
12
6.00
8.40
2.42
95% CI for mean difference: (0.66, 11.34)
T-Test of mean difference = 0 (vs not = 0): T-Value = 2.47
9.
P-Value = 0.031
A local school district is concerned about the number of school days missed by its teachers due to illness. A random sample
of 10 teachers is selected. The number of days absent in one year is listed below. An incentive program is offered in an
attempt to decrease the number of days absent. The number of days absent in one year after the incentive program is listed
below. Test the claim that the incentive program cuts down on the number of days missed by teachers. Use  = 0.05.
Assume that the distribution is normally distributed
Teacher
Days absent before incentive
Days absent after incentive
A
3
1
B
8
7
C
7
7
D
2
0
E
9
8
F
4
2
G
2
0
H
0
1
I
7
5
J
5
5
10. A weight-lifting coach claims that weight-lifters can increase their strength by taking a certain supplement. To test the
theory, the coach randomly selects 9 athletes and gives them a strength test using a bench press. The results are listed
below. Thirty days later, after regular training using the supplement, they are tested again. The new results are listed below.
Test the claim that the supplement is effective in increasing the athletes' strength. Use  = 0.05. Assume that the
distribution is normally distributed.
Athlete
Before
After
A
215
225
B
240
245
C
188
188
D
212
210
E
275
282
F
260
275
G
225
230
H
200
195
I
185
190
Confidence Interval (Two populations)
1. Is there a difference in the total scores for women’s and men’s basketball games? A random sample of
n1 = 55 women’s games had a mean winning score of x1 = 78. Another random sample of n2 = 60 men’s
games had a mean winning score of x 2 = 90. Historical data suggests  1 = 10 and  2 = 16. Find a 95%
confidence interval for the population difference (1   2 ) .
95% CI for difference:
(-16.89, -7.11)
2. Two growth hormones are being considered. A random sample of 10 rats were given the first hormone and
their average weight gain was x1 = 2.3 pounds with standard deviation s1 = 0.4 pound. For the second
hormone, a random sample of 15 rats showed their average weight gain to be x 2 = 1.9 pounds with standard
deviation s2 = 0.2 pound. Assume the weight gains follow a normal distribution. Find a 90% confidence
interval for the difference in average weight gains for the two growth hormones.
Two-Sample T-Test and CI
Sample
1
2
N
10
15
Mean
2.300
1.900
StDev
0.400
0.200
SE Mean
0.13
0.052
Difference = mu (1) - mu (2)
Estimate for difference: 0.400
90% CI for difference: (0.194, 0.606)
T-Test of difference = 0 (vs not =): T-Value = 3.32
Both use Pooled StDev = 0.2949
P-Value = 0.003
DF = 23
3. Stone Tires has developed a new tread which they claim reduces stopping distance on wet pavement. A
random sample of 56 test drives with cars using tires with tread type I (old design) showed that the average
stopping distance on wet pavement was x1 = 183 feet. A random sample of 61 test drives conducted
under similar conditions, but with cars using tires with tread type II (new tread) showed that the average
stopping distance was x 2 = 152 feet. Historical data suggests  1 = 49 feet and  2 = 53 feet.
Find a 90% confidence interval for the population mean difference (1   2 ) of stopping distances for the two
types of tire tread.
a) 15.5 to 46.5
b) 13.5 to 44.5
c) 14.5 to 43.5
d) 16.5 to 45. 5
e) 12.5 to 43.5
4. At a large office supply store, the daily sales of two similar brand-name laser printers are being compared.
A random sample of 16 days showed that Brand I had mean daily sales x1 = $2464 with standard deviation
s1 = $529. A random sample of 19 days showed that Brand II had mean daily sales x 2 = $2285 with sample
standard deviation s2 = $440. Assume sales follow an approximately normal distribution.
Find a 90% confidence interval for the population mean difference in sales (1   2 ) .