Question

4 Four bottles of red wine, Merlot, were analyzed for residual sugar content and the results obtained are the following: % w/

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Answer #1

a) Bottle 1

mean = \frac{0.97+0.86+1.20}{3} =1.01

spread = range = upper value -lower value = 1.2-0.86 = 0.34

standard deviation =\sqrt{\frac{1}{n}\sum (x_{i}-\bar{x})^{2}} = \sqrt{\frac{1}{3}\sum (x_{i}-{1.01})^{2}}= 0.141656

coefficient of variation = \frac{standard devaiation}{mean}*100 =\frac{0.141656}{1.01}*100 = 14.025%

Bottle 2

mean = \frac{1.25+1.32+1.13+1.20+1.12}{5} =1.204

spread ,range = 1.32-1.12=0.20

standard deviation= \sqrt{\frac{1}{5}\sum (x_{i}- 1.204)^2} =0.07499

coefficient of variation = (0.07499/1.204) *100 =6.228%

Bottle 3

mean = \frac{0.9+0.92+0.73}{3} =0.85

range = 0.92-0.73 =0.19

standard deviation = \sqrt{\frac{1}{3}\sum (x_{i}-0.85)^2} =0.066

coefficient of variation = (0.066/0.85)*100 = 7.764%

Bottle4

mean = \frac{0.72+0.77+0.61+0.58}{4} =0.67

range = 0.77-0.58 =0.19

standard deviation = \sqrt{\frac{1}{4}\sum (x_{i}-0.67)^2} =0.06957

coefficient of variation = (0.06957/0.67)*100 = 10.38%

bottle 2 has least coefficient of variation. So bottle 2 is more precised.

precision order: bottle 2,bottle 3,bottle4,bottle1

b) pooled standard deviation = \sqrt{\frac{(n_{1}-1)sd1^{2}+(n_{2}-1)sd2^{2}+(n_{3}-1)sd3^{2}+(n_{4}-1)sd4^{2}}{n_{1}+n_{2}+n_{3}+n_{4} -k}}\\= \sqrt{\frac{(3-1)*0.141659^{2}+(5-1)*0.07499^{2}+(3-1)*0.066^{2}+(4-1)*0.06957^{2}}{3+5+3+4 -4}} \\=\sqrt{\frac{0.08585879}{11}}\\=0.088347

  
  
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