Significant figures when measuring to nearest 5-minute of a day

Question

Let's say I'm measuring the weight of a potted plant about once a day. I want to record when I measure it to the nearest $5$-minute (i.e. if I measure at $7:33$, then I would record $7:35$). Let's use these data points for an example:

• Timepoint 1: $1080.9$ grams at $8:50$ on Monday.
• Timepoint 2: $1058.1$ grams at $7:30$ on Tuesday.

If I measure the loss of weight per day I can do it through two methods:

1st method measures in $\frac{\text{g}}{\text{day}}$:

$$\frac{1080.9 \text{ g} - 1058.1 \text{ g}}{0.94 \text{ day}} = 24 \frac{\text{g}}{\text{day}}$$

I use $0.94 \text{ day}$ ($2$ significant figures) because I'm measuring to the nearest $5$ minutes of the day, which is to a hundredth of a day's precision. That gives me $2$ sig figs in my answer of $24 \frac{\text{g}}{\text{day}}$.

2nd method measures in $\frac{\text{g}}{5 \text{ min}}$ (There are $288$ $5$-minutes in a day):

$$\frac{1080.9 \text{ g} - 1058.1 \text{ g}}{272 (5 \text{ min})} = 0.0838 \frac{\text{g}}{5 \text{ min}}$$

There are now $3$ significant figures in my answer.

Now I convert $\frac{\text{g}}{5 \text{ min}}$ to $\frac{\text{g}}{\text{day}}$:

$$\left( 0.0838 \frac{\text{g}}{5 \text{ min}} \right) \left( 288 \frac{5 \text{ min}}{\text{day}} \right) = 24.1 \frac{\text{g}}{\text{day}}$$

I now have the same answer using the same data and measured to the same precision, but my answer is more precise and uses an extra sig fig.

Am I doing this correctly? What is going on?

Answer 1

If you were working with exact values, then you'd say that the plant lost $22.8$ grams in $\frac{17}{18}$ day, and thus has a daily weight loss of $22.8 \div \frac{17}{18} = \frac{2052}{85} \approx 24.141176470588235$ grams per day.

But with weights rounded to the nearest $0.1$ gram, and times rounded to the nearest 5 minutes, then your original data really means:

• Timepoint 1: Between 1080.85 and 1080.95 grams, between 08:47:30 and 08:52:30 on Monday.
• Timepoint 2: Between 1058.05 and 1058.15 grams, between 07:27:30 and 07:32:30 on Tuesday.

The minimum weight loss rate is obtained by dividing the minimum weight loss ($1080.85 - 1058.15 = 22.7$ grams) by the maximum time difference ($1365$ minutes = $\frac{91}{96}$ day), for a rate of $\frac{10896}{455} \approx 23.947252747252747$ grams/day, which differs from the “exact” value by $0.19392372333548735$.

The maximum weight loss rate is obtained by dividing the maximum weight loss ($1080.95 - 1058.05 = 22.9$ grams) by the minimum time difference ($1355$ minutes = $\frac{271}{288}$ day), for a rate of $\frac{32976}{1355} \approx 24.336531365313654$ grams/day, which differs from the “exact” value by $0.19535489472541911$.

So, with reasonable rounding, we can say that the plant loses $\boxed{24.1 \pm 0.2}$ grams per day.

If you rely on the “default” assumption that the uncertainty in a number is half of a unit in the last place, then "24" has an uncertainty of $0.5$, while "24.1" has an uncertainty of $0.05$. The actual uncertainty of about $0.195$ is in between these values, so the choice is not immediately obvious.

---

A problem with “significant digits” analysis, as you have noted, is that it's sensitive to the choice of units. For example, the same person may describe their height as “71 inches” (with 2 sig figs) or “1.80 meters” (with 3 sig figs).

Also, that two numbers with the same absolute uncertainty can have an order-of magnitude difference in relative uncertainty, or vice-versa. For example,

• $100$ (if the zeroes are considered significant) has an implied uncertainty of $0.5$ unit, or $0.5\%$.
• $999$ also has an implied uncertainty of $0.5$ unit, but that's only $0.0\overline{500}\%$ relative error.

And that's what you're dealing with when you convert the value of $1360 (\pm 5)$ minutes to either $0.94$ day ($\pm 0.005$ day = $7.2$ min) or $0.944$ day ($\pm 0.005$ day = $0.72$ min = $43.2$ s). The former has a more accurate expression of the uncertainty, even though it only has 2 “significant digits” instead of 3.

Answer 2

Usually in physics experiments percentage errors of each measurement are added to find the total error. *

If you are measuring to the nearest 5 minutes in a day that’s $\frac{1}{288} = .38$ % error.

EDIT:Correction in weight error. I used the total weight in my original calculation, when I should have used the difference in weight as it is here that the error is most manifest. That is the error is in 24.1g not 1080.9g

The error in your weight measurement is $\frac{.1}{24.1} = .41$% error
Your total error is 0.79%.
Then you calculate your answer using as many significant figures as you can from your measurements and you get 24.1g/day with an error of 0.79% which is 24.1 g/day $\pm$ 0.2g/day.
Though errors are usually rounded i.e. 1% error which still comes out to 24.1g/day $\pm$ 0.2g/day.
You choose the significant figures in your answer according to the total accuracy of your result. Which in this case is one decimal place.

• You might ask, “Why are percentage errors added?” Consider measuring the area of a one meter square with each measurement accurate to 1%. That means your measurement could be $1.01m$ or $.99m$. Then the area calculation is between $1.0201m^2$ and $0.9801m^2$ which is an error of 2%. The .0001 amount is discarded. (linearization - ignore second order terms) A one percent error in each measurement totals to a 2% error in your result.