A small sample of 10 units has a mean μ and a standard deviation σ of 1. Find a 98% confidence interval of the mean μ
Confidence Interval Formula for μ is as follows: X - tscoreα * s/√n < μ < X + tscoreα * s/√n where: X = sample mean, s = sample standard deviation, tscore = statistic with (n - 1) Degrees of Freedom and α = 1 - confidence Percentage
First find degrees of freedom: Degrees of Freedom = n - 1 Degrees of Freedom = 10 - 1 Degrees of Freedom = 9
Find t-score for α0.01 using 9 degrees of freedom: tscore0.01 = 2.8214 <--- Value can be found on Excel using =TINV(0.02,9) tscore = 2.8214
Locate Value in the t-chart below:
DOFα
α 0.005
α 0.01
α 0.015
α 0.02
α 0.025
α 0.03
α 0.035
α 0.04
α 0.045
α 0.05
1
63.6559
31.821
21.2051
15.8945
12.7062
10.5789
9.0579
7.9158
7.0264
6.3137
2
9.925
6.9645
5.6428
4.8487
4.3027
3.8964
3.5782
3.3198
3.104
2.92
3
5.8408
4.5407
3.8961
3.4819
3.1824
2.9505
2.7626
2.6054
2.4708
2.3534
4
4.6041
3.7469
3.2976
2.9985
2.7765
2.6008
2.4559
2.3329
2.2261
2.1318
5
4.0321
3.3649
3.0029
2.7565
2.5706
2.4216
2.2974
2.191
2.0978
2.015
6
3.7074
3.1427
2.8289
2.6122
2.4469
2.3133
2.2011
2.1043
2.0192
1.9432
7
3.4995
2.9979
2.7146
2.5168
2.3646
2.2409
2.1365
2.046
1.9662
1.8946
8
3.3554
2.8965
2.6338
2.449
2.306
2.1892
2.0902
2.0042
1.928
1.8595
9
3.2498
2.8214
2.5738
2.3984
2.2622
2.1504
2.0554
1.9727
1.8992
1.8331
10
3.1693
2.7638
2.5275
2.3593
2.2281
2.1202
2.0283
1.9481
1.8768
1.8125
11
3.1058
2.7181
2.4907
2.3281
2.201
2.0961
2.0067
1.9284
1.8588
1.7959
12
3.0545
2.681
2.4607
2.3027
2.1788
2.0764
1.9889
1.9123
1.844
1.7823
13
3.0123
2.6503
2.4358
2.2816
2.1604
2.06
1.9742
1.8989
1.8317
1.7709
14
2.9768
2.6245
2.4149
2.2638
2.1448
2.0462
1.9617
1.8875
1.8213
1.7613
15
2.9467
2.6025
2.397
2.2485
2.1315
2.0343
1.9509
1.8777
1.8123
1.7531
16
2.9208
2.5835
2.3815
2.2354
2.1199
2.024
1.9417
1.8693
1.8046
1.7459
17
2.8982
2.5669
2.3681
2.2238
2.1098
2.015
1.9335
1.8619
1.7978
1.7396
18
2.8784
2.5524
2.3562
2.2137
2.1009
2.0071
1.9264
1.8553
1.7918
1.7341
19
2.8609
2.5395
2.3457
2.2047
2.093
2
1.92
1.8495
1.7864
1.7291
20
2.8453
2.528
2.3362
2.1967
2.086
1.9937
1.9143
1.8443
1.7816
1.7247
21
2.8314
2.5176
2.3278
2.1894
2.0796
1.988
1.9092
1.8397
1.7773
1.7207
22
2.8188
2.5083
2.3202
2.1829
2.0739
1.9829
1.9045
1.8354
1.7734
1.7171
23
2.8073
2.4999
2.3132
2.177
2.0687
1.9783
1.9003
1.8316
1.7699
1.7139
24
2.797
2.4922
2.3069
2.1715
2.0639
1.974
1.8965
1.8281
1.7667
1.7109
25
2.7874
2.4851
2.3011
2.1666
2.0595
1.9701
1.8929
1.8248
1.7637
1.7081
26
2.7787
2.4786
2.2958
2.162
2.0555
1.9665
1.8897
1.8219
1.761
1.7056
27
2.7707
2.4727
2.2909
2.1578
2.0518
1.9632
1.8867
1.8191
1.7585
1.7033
28
2.7633
2.4671
2.2864
2.1539
2.0484
1.9601
1.8839
1.8166
1.7561
1.7011
29
2.7564
2.462
2.2822
2.1503
2.0452
1.9573
1.8813
1.8142
1.754
1.6991
Calculate the Standard Error of the Mean:
SEM =
σ
√n
SEM =
1
√10
SEM =
1
3.1622776601684
SEM = 0.3162
Calculate high end confidence interval total: High End = X + tscoreα * s/√n High End = 5 + 2.8214 x 1/√10 High End = 5 + 2.8214 x 0.31622776601684 High End = 5 + 0.89220501903991 High End = 5.8922
Calculate low end confidence interval total: Low End = X - tscoreα * s/√n Low End = 5 - 2.8214 x 1/√10 Low End = 5 - 2.8214 x 0.31622776601684 Low End = 5 - 0.89220501903991 Low End = 4.1078
Now we have everything, display our 98% confidence interval:
4.1078 < μ < 5.8922
You have 2 free calculationss remaining
What this means is if we repeated experiments, the proportion of such intervals that contain μ would be 98%
What is the Answer?
4.1078 < μ < 5.8922
How does the Confidence Interval for the Mean Calculator work?
Free Confidence Interval for the Mean Calculator - Calculates a (90% - 99%) estimation of confidence interval for the mean given a small sample size using the student-t method with (n - 1) degrees of freedom or a large sample size using the normal distribution Z-score (z value) method including Standard Error of the Mean. confidence interval of the mean This calculator has 5 inputs.
What 1 formula is used for the Confidence Interval for the Mean Calculator?