A research team is working on a project to study the time (in seconds) for high school male runners to finish a 400-meter race. Jimmy, a junior researcher in the team, has randomly selected a sample of 25 male runners from a high school and the time (in seconds) for each of them to complete a 400-meter race was recorded. The sample mean running time was 53 seconds. It is assumed that the running time in a 400-meter race follows a normal distribution with a population standard deviation of 5.5 seconds.(a) Give a point estimate of the population mean running time for a 400-meter race.(b) Calculate the sampling error at 95% confidence level.(c) Construct a 95% confidence interval estimate of the population mean running time for a 400-meter race. The entries in Table I are the probabilities that a random variable having thestandard normal distribution will take on a value between 0 and:. They are givenby the area of the gray region under the curve in the figure.TABLE INORMAL-CURVE AREAS0.000.010.020.030.040.050.060.070.080.090.00.00000.00400.00800.01200.01600.01990.02390.02790.03190.03590.03980.07930.04380.08320.06750.10640.05170.06360.10260.10.04780.05570.05960.07140.07530.20.08710.09100.09480.09870.11030.11410.30.11790.12170.12550.12930.13310.13680.14060.14430.14800.15170.40.15540.15910.16280.16640.17000.17360.17720.18080.18440.18790.50.19150.19500.19850.20190.20540.20880.21230.21570.21900.22240.22570.25800.23240.26420.23570.26730.23890.27040.24220.27340.25170.28230.25490.28520.60.22910.24540.24860.70.26110.27640.27940.80.31590.34130.28810.29100.29390.29670.29950.30230.30510.30780.31060.31330.90.31860.32120.32380.32640.32890.33150.33400.33650.33891.00.34380.34610.34850.35080.35310.35540.35770.35990.36210.36860.38880.40660.37700.39620.41311.10.36430.36650.37080.37290.37490.37900.38100.38300.38490.40320.39070.40820.42360.39970.41620.43060.38690.39250.40990.42511.20.39440.39800.40150.40490.42071.30.41150.41470.41771.40.41920.42220.42650.42790.42920.43191.50.43320.43450.43570.43700.43820.43940.44060.44180.44290.44410.44840.45820.46640.45350.46250.46991.60.44520.44630.44740.44950.45050.45150.45250.45450.45540.46410.45640.46480.45910.46710.46080.46850.46160.46920.46330.47061.70.45730.45991.80.46560.46781.90.47130.47190.47250.47320.47380.47440.47500.47560.47610.47672.00.47720.47780.47830.47880.47930.47980.48030.48080.48120.48170.48210.48610.48340.48710.49010.48380.48750.49040.48460.48810.49090.48570.48902.10.48260.48300.48420.48500.48540.48750.49060.48870.49132.20.48640.48680.48842.30.48930.48960.48980.49110.49162.40.49180.49200.49220.49250.49270.49290.49310.49320.49340.49362.50.49380.49400.49410.49430.49450.49460.49480.49490.49510.49520.49570.49680.49630.49730.49530.49550.49660.49590.49692.60.49560.49600.49610.49620.49640.49742.70.49650.49670.49700.49710.49722.80.49740.49750.49760.49770.49770.49780.49790.49790.49800.49812.90.49810.49820.49820.49830.49840.49840.49850.49850.49860.49863.00.49870.49870.49870.49880.49880.49890.49890.49890.49900.4990Also, for z= 4.0, 5.0 and 6.0, the areas are 0.49997, 0.4999997, and 0.499999999. The entries in Table II are values for which the area to their right under thedistribution with given degrees of freedom (the gray area in the figuure) is equalto a .tTABLE IIVALUE OFtd.f.f0.050f0.025fo.010f0.005d.f16.31412.70631.82163.65712.9204.3036.9659.9252.3533.1824.5415.84142.1322.7763.7474.604452.0152.5713.3654.0321.9432.4473.1433.707671.8952.3652.9983.49971.8602.3062.8963.35591.8332.2622.8213.2509101.8122.2282.7643.16910111.7962.2012.7183.10611121.7822.1792.6813.05512131.7712.1602.6503.01213141.7612.1452.6242.97714151.7532.1312.6022.94715161.7462.1202.5832.92116171.7402.1102.5672.89817181.7342.1012.5522.87818191.7292.0932.5392.86119201.7252.0862.5282.84520211.7212.0802.5182.83121221.7172.0742.5082.81922231.7142.0692.5002.80723241.7112.0642.4922.79724251.7082.06024852.78725261.7062.0562.4792.77926271.7032.0522.4732.77127281.7012.0482.4672.76328291.6992.0452.4622.75629Inf.1.6451.9602.3262.576In.

The random variable running time follows normal distribution. We have to construct 95% confidence…

Answered: A research team is working on a project…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

A research team is working on a project to study the time (in seconds) for high school male runners to finish a 400-meter race. Jimmy, a junior researcher in the team, has randomly selected a sample of 25 male runners from a high school and the time (in seconds) for each of them to complete a 400-meter race was recorded. The sample mean running time was 53 seconds. It is assumed that the running time in a 400-meter race follows a normal distribution with a population standard deviation of 5.5 seconds.
(a) Give a point estimate of the population mean running time for a 400-meter race.
(b) Calculate the sampling error at 95% confidence level.
(c) Construct a 95% confidence interval estimate of the population mean running time for a 400-meter race.

The entries in Table I are the probabilities that a random variable having the
standard normal distribution will take on a value between 0 and:. They are given
by the area of the gray region under the curve in the figure.
TABLE I
NORMAL-CURVE AREAS
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.0
0.0000
0.0040
0.0080
0.0120
0.0160
0.0199
0.0239
0.0279
0.0319
0.0359
0.0398
0.0793
0.0438
0.0832
0.0675
0.1064
0.0517
0.0636
0.1026
0.1
0.0478
0.0557
0.0596
0.0714
0.0753
0.2
0.0871
0.0910
0.0948
0.0987
0.1103
0.1141
0.3
0.1179
0.1217
0.1255
0.1293
0.1331
0.1368
0.1406
0.1443
0.1480
0.1517
0.4
0.1554
0.1591
0.1628
0.1664
0.1700
0.1736
0.1772
0.1808
0.1844
0.1879
0.5
0.1915
0.1950
0.1985
0.2019
0.2054
0.2088
0.2123
0.2157
0.2190
0.2224
0.2257
0.2580
0.2324
0.2642
0.2357
0.2673
0.2389
0.2704
0.2422
0.2734
0.2517
0.2823
0.2549
0.2852
0.6
0.2291
0.2454
0.2486
0.7
0.2611
0.2764
0.2794
0.8
0.3159
0.3413
0.2881
0.2910
0.2939
0.2967
0.2995
0.3023
0.3051
0.3078
0.3106
0.3133
0.9
0.3186
0.3212
0.3238
0.3264
0.3289
0.3315
0.3340
0.3365
0.3389
1.0
0.3438
0.3461
0.3485
0.3508
0.3531
0.3554
0.3577
0.3599
0.3621
0.3686
0.3888
0.4066
0.3770
0.3962
0.4131
1.1
0.3643
0.3665
0.3708
0.3729
0.3749
0.3790
0.3810
0.3830
0.3849
0.4032
0.3907
0.4082
0.4236
0.3997
0.4162
0.4306
0.3869
0.3925
0.4099
0.4251
1.2
0.3944
0.3980
0.4015
0.4049
0.4207
1.3
0.4115
0.4147
0.4177
1.4
0.4192
0.4222
0.4265
0.4279
0.4292
0.4319
1.5
0.4332
0.4345
0.4357
0.4370
0.4382
0.4394
0.4406
0.4418
0.4429
0.4441
0.4484
0.4582
0.4664
0.4535
0.4625
0.4699
1.6
0.4452
0.4463
0.4474
0.4495
0.4505
0.4515
0.4525
0.4545
0.4554
0.4641
0.4564
0.4648
0.4591
0.4671
0.4608
0.4685
0.4616
0.4692
0.4633
0.4706
1.7
0.4573
0.4599
1.8
0.4656
0.4678
1.9
0.4713
0.4719
0.4725
0.4732
0.4738
0.4744
0.4750
0.4756
0.4761
0.4767
2.0
0.4772
0.4778
0.4783
0.4788
0.4793
0.4798
0.4803
0.4808
0.4812
0.4817
0.4821
0.4861
0.4834
0.4871
0.4901
0.4838
0.4875
0.4904
0.4846
0.4881
0.4909
0.4857
0.4890
2.1
0.4826
0.4830
0.4842
0.4850
0.4854
0.4875
0.4906
0.4887
0.4913
2.2
0.4864
0.4868
0.4884
2.3
0.4893
0.4896
0.4898
0.4911
0.4916
2.4
0.4918
0.4920
0.4922
0.4925
0.4927
0.4929
0.4931
0.4932
0.4934
0.4936
2.5
0.4938
0.4940
0.4941
0.4943
0.4945
0.4946
0.4948
0.4949
0.4951
0.4952
0.4957
0.4968
0.4963
0.4973
0.4953
0.4955
0.4966
0.4959
0.4969
2.6
0.4956
0.4960
0.4961
0.4962
0.4964
0.4974
2.7
0.4965
0.4967
0.4970
0.4971
0.4972
2.8
0.4974
0.4975
0.4976
0.4977
0.4977
0.4978
0.4979
0.4979
0.4980
0.4981
2.9
0.4981
0.4982
0.4982
0.4983
0.4984
0.4984
0.4985
0.4985
0.4986
0.4986
3.0
0.4987
0.4987
0.4987
0.4988
0.4988
0.4989
0.4989
0.4989
0.4990
0.4990
Also, for z= 4.0, 5.0 and 6.0, the areas are 0.49997, 0.4999997, and 0.499999999.

The entries in Table II are values for which the area to their right under the
distribution with given degrees of freedom (the gray area in the figuure) is equal
to a .
t
TABLE II
VALUE OFt
d.f.
f0.050
f0.025
fo.010
f0.005
d.f
1
6.314
12.706
31.821
63.657
1
2.920
4.303
6.965
9.925
2.353
3.182
4.541
5.841
4
2.132
2.776
3.747
4.604
4
5
2.015
2.571
3.365
4.032
1.943
2.447
3.143
3.707
6
7
1.895
2.365
2.998
3.499
7
1.860
2.306
2.896
3.355
9
1.833
2.262
2.821
3.250
9
10
1.812
2.228
2.764
3.169
10
11
1.796
2.201
2.718
3.106
11
12
1.782
2.179
2.681
3.055
12
13
1.771
2.160
2.650
3.012
13
14
1.761
2.145
2.624
2.977
14
15
1.753
2.131
2.602
2.947
15
16
1.746
2.120
2.583
2.921
16
17
1.740
2.110
2.567
2.898
17
18
1.734
2.101
2.552
2.878
18
19
1.729
2.093
2.539
2.861
19
20
1.725
2.086
2.528
2.845
20
21
1.721
2.080
2.518
2.831
21
22
1.717
2.074
2.508
2.819
22
23
1.714
2.069
2.500
2.807
23
24
1.711
2.064
2.492
2.797
24
25
1.708
2.060
2485
2.787
25
26
1.706
2.056
2.479
2.779
26
27
1.703
2.052
2.473
2.771
27
28
1.701
2.048
2.467
2.763
28
29
1.699
2.045
2.462
2.756
29
Inf.
1.645
1.960
2.326
2.576
In.

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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