Science Buddies: "Ask an Expert"

white red wine apple cider
1.8 1.8 1.75
1.9 1.7 1.7
1.75 1.6 1.75
2 1.7 1.7
1.8 1.55 1.55

1.850 1.670 1.690 Average
0.100 0.097 0.082 Std Dev
0.010 0.010 0.007 Variance

The averages don't match the normalized percentages you posted, so I'm guessing that the variation in the starting weights requires normalization.

Sorry, lost the Excel spread sheet column spacing.

I didn't post percentages but the actual weight differences for each trial.

lil bit, can you post the starting weights for the trials?

The starting weights for the white were:87.1; 86.7;86.15;86.6;86.6.
The starting weights for red were: 89.0; 89.8; 88.3; 87.7; 88.15.
The stareting weights for apple weer: 87.45; 86.4; 86.45; 86.3; 86.7.

lil bit wrote:The starting weights for the white were:87.1; 86.7;86.15;86.6;86.6.
The starting weights for red were: 89.0; 89.8; 88.3; 87.7; 88.15.
The stareting weights for apple weer: 87.45; 86.4; 86.45; 86.3; 86.7.

I recalculated the standard deviation using the normalization of the starting masses. So, for example, for the white vinegar trial, every thing ideally should weigh 87.1 g. So, the results for the 86.7 trial were 99.5% of what they should be, because the starting weight was only 99.5% of what is should be. (Which is really good- all the standard deviations on these are really fantastic. Obviously you took a lot of care in your measuring.)

The "norm factor" is the percent off the max weight. The "gas" is your experimental value. "norm gas" is the amount of gas you would have measured if you had used a standard weight. Mathematically, I used [(1-norm factor)*gas+ gas). Then I calculated the average and the standard deviation.

white norm factor gas norm gas
87.1 1 1.8 1.8
86.7 0.995407577 1.9 1.908725603
86.15 0.989092997 1.75 1.769087256
86.6 0.994259472 2 2.011481056 ave 1.859925373
86.6 0.994259472 1.8 1.810332951 std 0.099564815

red
89 0.991091314 1.8 1.816035635
89.8 1 1.7 1.7
88.3 0.983296214 1.6 1.626726058
87.7 0.976614699 1.7 1.739755011 ave 1.692199332
88.15 0.981625835 1.55 1.578479955 std 0.093376481

cider
87.45 1 1.75 1.75
86.4 0.987993139 1.7 1.720411664
86.45 0.988564894 1.75 1.770011435
86.3 0.986849628 1.75 1.77301315 ave 1.715345912
86.7 0.991423671 1.55 1.56329331 std 0.087550544

If you look, you see there is overlap between the normalized red and the white vinegar data. That is, 1.8, 1.7, 1.76 could appear in both groups. The standard deviation also says there is some overlap[1]. So, usually, you report the average + and - the standard deviation. So the range of values for the white vinegar based on your trial is 1.86 +/- 0.09, which means that you would believe numbers that were 1.76 to 1.96 (rounded). For red, 1.79 to 1.60 is the range. If you were a scientist, you would do more averages, and more sophisticated statistics to tell if these numbers are different or the same, since it looks "suggestive" that the white vinegar has a higher average value. But, with the data we have, since the ranges do overlap, I'd have to say the results are the same for all vinegars.

Given that this reaction should depend on the amount of acid and baking soda, you might expect that if all are 5% acid, and all are 50 mL, you should get the same result.

Louise

[1] from the wikipedia-
"The standard deviation is the most common measure of statistical dispersion, measuring how widely spread the values in a data set are. If the data points are close to the mean, then the standard deviation is small. Conversely, if many data points are far from the mean, then the standard deviation is large. If all the data values are equal, then the standard deviation is zero."

I'm guessing that the variation in the starting weights requires normalization.

lil bit, Boy was I WRONG! You and your lab assistant did a FANTASTIC job of measuring!

I used Excel and normalized and graphed your data and am attempting to have the science buddies administrators forward it to you.

If you throw out the highest two trials from the white and the bottom two from the red, the rest of the data is in a very tight clump.

I agree. Nice data.

Craig_Bridge wrote:

I'm guessing that the variation in the starting weights requires normalization.

lil bit, Boy was I WRONG! You and your lab assistant did a FANTASTIC job of measuring!

I used Excel and normalized and graphed your data and am attempting to have the science buddies administrators forward it to you.

If you throw out the highest two trials from the white and the bottom two from the red, the rest of the data is in a very tight clump.

Not that you should throw out data... but I totally agree with Craig's main point about data quality. When I calculated the deviations I was very impressed. My data rarely looks this good!


Louise

I'm not sure I follow about the normalization of the begininning weights but I calculated the standard deviation of the results by subtracting the average from each result, squaring the difference, adding the squares, dividing by 5 and getting the square root of that. A friend gave me the calcualtion. I got 1.85 +/- .089; 1.67 +/- .096 and 1.69 +/- .073. So white vinegar: 1.939 to 1.76
red vinegar: 1.766 to 1.574
apple cider: 1.76 to 1.62
We made a mistake on the intial average of the red wine results, it's 1.67 not 1.65. So does this show a difference or not???

lil bit wrote:I'm not sure I follow about the normalization of the begininning weights but I calculated the standard deviation of the results by subtracting the average from each result, squaring the difference, adding the squares, dividing by 5 and getting the square root of that. A friend gave me the calcualtion. I got 1.85 +/- .089; 1.67 +/- .096 and 1.69 +/- .073. So white vinegar: 1.939 to 1.76
red vinegar: 1.766 to 1.574
apple cider: 1.76 to 1.62
We made a mistake on the intial average of the red wine results, it's 1.67 not 1.65. So does this show a difference or not???

I think there is no difference between the vinegars. Your standard deviation calculation is fine. If you have microsoft excel they will do this calculation for you- "stdev(cells)", since this calculation is a pain.

However you look at the data (my corrections or your calculations), your standard deviations are very small. You can see my calculated "normalized" standard deviations are quite small. So, you get 1.85 +/- 0.089 if you don't correct the starting mass, and I get 1.8599 +/- 0.0996 (I put extra digits in, so you can find it in the table I posted... on the right hand side).

The normalization thing... So, if you assume for white vinegar that 87.1 g was the starting mass you wanted, then the second trial of 86.7 g was a bit too light. It is only 99.5% of the mass, instead of the whole mass. So, the result was "incorrect" by 99.5%. So, if I know that your measured 1.75 g is only 99.5% of the value, then I can calculate that 100% of the value should be 1.769 g. So, I did this for all trials. This removes all variations in the starting weight, and gives you a better idea of the "true" variations.

BUT, you measured your masses so accurately, that the few percent error from the starting mass variation hardly has an impact.

Louise

Hi,
Maybe this question has already been answered, but do these weights include the mass of baking soda that was added? In other words, do these values reflect the total mass lost from baking soda and vinegar, or just the mass lost from vinegar?

Chris

Chris, yes the weights do include the baking soda.

OK, thanks. I have nothing to add then.