Six bottles of wine of the same variety were analyzed for residual sugar content
ID: 967025 • Letter: S
Question
Six bottles of wine of the same variety were analyzed for residual sugar content with the following results:
Bottle
Percent (w/v) Residual Sugar
1
2
3
4
5
6
0.99, 0.84, 1.02
1.02, 1.13, 1.017, 1.02
1.25, 1.32, 1.13, 1.20, 1.12
0.72, 0.77, 0.61, 0.58
0.90, 0.92, 0.73
0.70, 0.88, 0.72, 0.73
a) Evaluate the standard deviation s for each set of data.
b) Pool the data to obtain an absolute standard deviation for the method.
Bottle
Percent (w/v) Residual Sugar
1
2
3
4
5
6
0.99, 0.84, 1.02
1.02, 1.13, 1.017, 1.02
1.25, 1.32, 1.13, 1.20, 1.12
0.72, 0.77, 0.61, 0.58
0.90, 0.92, 0.73
0.70, 0.88, 0.72, 0.73
Explanation / Answer
First, let's calculate the average of each set of data:
1: (0.99+0.84+1.02)/3 =0.95
SD = [(0.95-0.99)2 + (0.95-0.84)2 + (0.95-1.02)2 / 2 ]1/2 = 0.096
2. (1.02+1.13+1.017+1.02)/4 = 1.0468
SD = [(1.0468-1.02)2 + (1.0468-1.13)2 + (1.0468-1.017)2 + (1.0468-1.02)2 / 4]1/2 = 0.056
Now, doing the same procedure with the remaining samples we have:
3. 1.225; SD = 0.0802
4. 1.12; SD = 0
5. 0.67; SD = 0.0898
6. 0.7971; SD = 0.0974
Finally, all this data:
(0.0802+0+0.0898+0.0974+0.096+0.056/6 = 0.0699 or 0.07
Hope this helps