Question

4 Four bottles of red wine, Merlot, were analyzed for residual sugar content and the results obtained are the following: % w/v Residual Sugar Bottles 0.97. 0.86, 1.02 2 1.25, 1.32, 1.13, 1.20. 1.12 0.9, 0.92, 0.73 3 0.72, 0.77, 0.61, 0.58 4 a For each set, determine the mean, spread, standard deviation s and coefficient of variation. Comment on the precision of each data set relative to other sets. (b) Calculate the pooled standard deviation Spooled 16

Answer 1

a) Bottle 1

mean = =1.01

0.97 +0.86 + 1.20 3

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 =
standarddevaiation * 100 mean
$=\frac{0.141656}{1.01}*100$ = 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 = $\sqrt{\frac{(n_{1}-1)sd1^{2}+(n_{2}-1)sd2^{2}+(n_{3}-1)sd3^{2}+(n_{4}-1)sd4^{2}}{n_{1}+n_{2}+n_{3}+n_{4} -k}}\\= \sqrt{\frac{(3-1)*0.141659^{2}+(5-1)*0.07499^{2}+(3-1)*0.066^{2}+(4-1)*0.06957^{2}}{3+5+3+4 -4}} \\=\sqrt{\frac{0.08585879}{11}}\\=0.088347$