# Why should changing the units of the variable change the required sample size?

I am trying to calculate the survey sample size necessary to estimate average rice yield.

If the standard deviation is 10 tons, then the variance is 100. Plugging this into the sample size formula (at 95% confidence, 5% MoE, 100,000 population) then I need a sample of 16.

But, if I change the unit of measurement to kg, the same standard deviation is 10000kg. The variance then becomes 100,000,000. The required sample size skyrockets to 99,354.

Why should changing the units of the variable change the required sample size? Does this mean that I can get around any sample size problems by simply changing the units of measurement? Apologies if this is too simple a question for this forum, but I've hunted far and wide on the internet for the intuition behind this...

• Welcome to CV! Please give the formula you're using to calculate required sample size. – Scortchi - Reinstate Monica Mar 9 '15 at 12:40
• 5% error in kg is the not the same as 5% error in tons. In the scenario using kg, you're practically asking for a sample that can give a precision 1,000 times higher than that from tons. – Penguin_Knight Mar 9 '15 at 12:42
• @Penguin_Knight A 5% error as a fraction of the mean is unitless, so - at least if it's organized correctly, it shouldn't make a difference what units were used. panglon -- please give the formula you used, showing your calculations. – Glen_b -Reinstate Monica Mar 9 '15 at 12:50
• @Glen_b, yes it is unitless but contextually the total errors allowed are different across two different formulas that use two different units. – Penguin_Knight Mar 9 '15 at 12:51

The problem lies in the $d^2$ in the denominator has not been adjusted. A 5% error in kg is the not the same as 5% error in tons. Using this basic formula that does not assume finite population (aka results may different from your formula's):

$N = \frac{1.96^2 \times SD^2}{d^2}$

> # SD = 10 tons:
> (1.96^2 * (10^2))/(5^2)
[1] 15.3664
>
> # SD = 10000 kg:
> (1.96^2 * (10000^2))/(5^2)
[1] 15366400
>
> # While in fact, the margin of error equivalent to 5% of a ton is 5000% of a kg:
> (1.96^2 * (10000^2))/(5000^2)
[1] 15.3664
>
> # Or, you can get the same thing using ton, making the margin 1000 times smaller:
> (1.96^2 * (10^2))/(.005^2)
[1] 15366400


Then, apply the finite population adjustment:

$final.n = n \times \frac{N}{N + n}$

where $N$ is the population size and $n$ is the sample size based on infinite population, we will recover 16 and 99354.

In the scenario using kg, you're practically asking for a sample that can give a precision 1,000 times higher than that from tons, making your required sample size so much higher.

• This is good, but the kg result doesn't match the numbers in the question. – Glen_b -Reinstate Monica Mar 9 '15 at 12:53
• @Glen_b. Thanks! I realize that, the OP probably uses finite population adjustment. We'll have to wait for the OP to explain what formula was used. – Penguin_Knight Mar 9 '15 at 12:54
• I think that would be best, yes. – Glen_b -Reinstate Monica Mar 9 '15 at 12:55
• @Glen_b, found it! All set. – Penguin_Knight Mar 9 '15 at 13:01
• Wow, thanks for the quick and comprehensive response Penguin_Knight. Indeed, I had just plugged 5 into the formula, and not adjusted for the units. Feeling a little sheepish now... much like those guys who crashed the satellite because they confused lb with newtons. – user70735 Mar 9 '15 at 13:33