1. An experiment was conducted to determine the dissolved oxygen in ground water at one specific location. The experimental uncertainty of the dissolved oxygen was estimated as ±1.5 mg/L. Show all your work. a. The first set of six measurements were: 8.8, 3.1, 4.2, 6.2, 7.6 and 3.6 mg/L. Determine the mean, sample standard deviation, and the standard error of the mean (SEM). b. After examining the statistics of part "a," the experimentalists collect an additional six measurements: 5.2, 8.6, 6.3, 1.8, 6.8 and 3.9 mg/L. Combine all 12 measurements and determine the mean, sample standard deviation, and the standard error of the mean (SEM). c. Complete the following to report the results of part "b": "With 95% confidence, the dissolved oxygen was xxx mg/L ±xxx mg/L. The population mean is probable within ±xxx mg/L. The probable range of the mean of dissolved oxygen is xxx to xxx mg/L (for 95% confidence). d. Comment on the differences in part "a" and "b" regarding the mean, sample standard deviation, and the standard error of the mean (SEM). In particular, mention why these differences were or were not expected. e. Comment on the value of the experimental uncertainty compared to the values of the statistical quantities. f. Comment on why ANOVA should or should not be used to evaluate this experiment.

Respuesta :

Answer:A, I. Mean X=8.8+3.1+4.2+6.2+7.6+3.6/6=5.58.

ii. S.D. for ungrouped data is given as √E[x-X]²/√6

[x-X] ²= 8.8-5.58+3.1-5.58+4.2-5.58+6.2-5.58+7.6-5.58+3.6-5.58=0.02² S.D=√0.02²/√6.

=0.02/2.449=0.00816.

iii. Standard error of the mean =

S.D./√N = 0.00816/6=0.00136.

B. If additional six measurements

were added to the previous one:

I. Mean X= adding total 12 samples and divide by 12 gives 5.51.

ii. Sample S.D. for all the 12 experiment by following same step in A for the 12 experiment gives 0.02/√12 =0.02/3.46=0.0057

iii. Standard error of the mean for the 12 experiment is S.D./√N= 0.0057/12=0.00048.

Explanation: