# Download Confidence Intervals for the Population Mean, Using the t Distribution

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```Confidence Intervals for the Population Mean, Using the t Distribution
A sample of taxicabs had the following consumption of gasoline, in gallons, on a particular day: 32, 16,
25, 22, 27, 19, 20. Assume that the distribution of gasoline usage is normal.
A. Find the appropriate t values for 90% and 95% confidence intervals.
B. Find the mean gasoline usage and the standard deviation of gasoline use.
C. Find the sample standard error.
D. Calculate the 90 and 95% confidence intervals for the mean daily use of gasoline by all taxicabs.
Solutions:
A. Since n = 7, degrees of freedom = 7 – 1 = 6. See below to find t values of 1.943 and 2.447.
B. The most appropriate way to find these numbers is with a statistical calculator. You should find
that ⎯x = 23 and s = 5.416026.
_
_
C. The sample standard error s⎯x = s/√n = 5.416026/√7 = 2.047065
D. The confidence interval is given by ⎯x ± t × s⎯x ⇒ for the 90% confidence interval
23 ± 1.943 × 2.0470650 or 23 ± 3.98. We are 90% confident that the average daily gasoline use in
the population of taxicabs is between 19.020and 26.98 gallons, or between 19 and 27 gallons.
For the 95% confidence interval, we have 23 ± 2.447 × 2.0470650 or 23 ± 5. We are 95%
confident that the mean daily gasoline use for all taxicabs is at least 18 but not more than 28
gallons.
Student's t distribution
Confidence Level
This column
is used to
locate the
degrees of
freedom
0.800 0.900
0.950
0.980
Significance Level for Two-Tailed Test
0.200 0.100
0.050
0.020
Significance Level for One-Tailed Test
0.990
df
0.005
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
0.100
3.078
1.886
1.638
1.533
1.476
1.440
1.415
1.397
1.383
1.372
1.363
1.356
1.350
1.345
1.341
1.337
1.333
1.330
0.050
6.314
2.920
2.353
2.132
2.015
1.943
1.895
1.860
1.833
1.812
1.796
1.782
1.771
1.761
1.753
1.746
1.740
1.734
0.025
12.706
4.303
3.182
2.776
2.571
2.447
2.365
2.306
2.262
2.228
2.201
2.179
2.160
2.145
2.131
2.120
2.110
2.101
0.010
31.821
6.965
4.541
3.747
3.365
3.143
2.998
2.896
2.821
2.764
2.718
2.681
2.650
2.624
2.602
2.583
2.567
2.552
0.010
This row is the
confidence levels
expressed as decimal
fractions.
63.657
9.925
5.841
4.604
4.032
3.707
3.499
3.355
3.250
A. df = n – 1 = 7 – 1 = 6. Find
the confidence level in the top
row of labels: 0.900.
3.169
3.106
3.055
3.012
2.977
2.947
2.921
2.898
2.878
A. For the 95% confidence
interval, use the column
headed 0.950.
A sample of 57 adults passed through a federally funded job training program. After sixteen weeks, the 57
were found and their current weekly income was recorded. In the sample, the mean weekly income was
\$412 with standard deviation \$53. Calculate a 98% confidence interval for the mean wage of all workers
who have participated in this particular training program.
Solution:
_
The confidence _interval is ⎯x ± t × s⎯x where s⎯x = s/√n. In this problem, ⎯x = 412, s = 53 and n = 57.
Thus, s⎯x = 53/√57 = 53/7.549834435 = 7.02. For a 98% confidence interval, the t value = 2.395 (see
below). Thus the confidence interval is 412 ± 2.395 × 7.02 ⇒ 412 ± 16.81. With 98% confidence, we can
say the mean wage of all participants in this program is between \$395.19 and \$428.81.
Confidence Level
0.800 0.900 0.950 0.980 0.990
Significance Level for Two-Tailed Test
0.200 0.100 0.050 0.020 0.010
Significance Level for One-Tailed Test
df
47
48
49
50
51
52
53
54
55
56
57
58
59
60
0.100
1.300
1.299
1.299
1.299
1.298
1.298
1.298
1.297
1.297
1.297
1.297
1.296
1.296
1.296
0.050
1.678
1.677
1.677
1.676
1.675
1.675
1.674
1.674
1.673
1.673
1.672
1.672
1.671
1.671
0.025
2.012
2.011
2.010
2.009
2.008
2.007
2.006
2.005
2.004
2.003
2.002
2.002
2.001
2.000
0.010
2.408
2.407
2.405
2.403
2.402
2.400
2.399
2.397
2.396
2.395
2.394
2.392
2.391
2.390
0.005
2.685
2.682
2.680
2.678
2.676
2.674
2.672
2.670
2.668
2.667
2.665
2.663
2.662
2.660
With n = 57, there are n – 1 = 56
degrees of freedom. Use the
column headed 0.980 for a 98%
confidence interval.
```
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