Answer Happy

needed data:
8.7361
8.7056
11.0155
4.7527
7.9908
6.4588
8.2348
10.2974
10.6515
6.9701
8.0579
9.2740
8.4163
9.6036
9.7999
7.1401
8.6463
4.7358
11.2907
7.7418
6.5923
8.4921
6.1427
8.9583
7.8787
13.3556
11.2769
4.0062
9.8827
6.2037
8.4899
9.3288
10.4955
8.4539
12.1526
8.0381
9.6550
10.3295
9.1704
10.7619
9.6464
7.4317
5.3893
12.7172
7.7909
9.2067
10.1263
9.2272
7.1905
8.0646
8.7502
11.9579
7.2784
10.5693
9.6172
8.5323
6.8861
8.4317
8.8266
6.0612
9.3844
7.3554
8.8115
9.6724
7.1907
8.4235
9.7001
5.3283
11.0720
13.8489
10.9188
8.3685
9.8572
6.9280
12.7557
10.8814
10.5747
7.2483
9.6399
7.8834
8.3771
7.8600
6.9485
7.1825
8.5802
5.6023
10.2152
8.7644
10.3983
9.5393
9.9886
6.0338
6.9595
8.1060
9.2193
11.2575
8.4201
11.5231
9.9508
11.3482
9.2539
7.6864
6.0372
9.3110
10.6371
8.4148
7.9184
8.3827
6.8068
8.0140
14.2000
14.6000
15.0000
15.4000
9.0000
2.6000
16.0000
10.6000
7.0000
17.2000
2. This problem concerns applications to dissolved oxygen concentration levels in a water science scenario. It is essential to have a sufficient amount of dissolved oxygen in rivers for healty and sustainable aquatic life (for fish and other organisms). In particular, if its level drops below 5 mg/L especially in warm water, aquatic life is endangered Consider a random sample of 120 measurements of oxygen concentration in a few locations of a river in California. The corresponding data set is in OxygenCon worksheet of the same Excel file. It is also in the Matlab file DataOxygen.mat. (a) Obtain both a histogram and a normality plot of the data. Does this data set appear to come from an approximate normal population? Why or why not? (b) Obtain the mean and standard deviation of the sample data to estimate the corresponding population mean and population standard deviation. Moreover, determine the standard error of sample mean (X) estimate.