Statistical test for percentages values with different denominators !!!

Question

Hello,

What statistical test can you do with percentage values that do not have the same sample size (do not have the same denominator).

For example : 43/53, 21/30, 11/24 would be 81% , 70% , 46%

these percentages are in reference to a different denominator therefore not equal so how would I do a statistical test on these?

A detailed explanation would be very appreciated, I have been trying really hard to wrap my head around this but can't seem to do so. Thank you

Answer 1

Hi Aren,

There are a couple ways you can do this and the choice depends what exactly you are testing for. I'll list both options that I know and if other tutors want to add more possible tests, they certainly can.

1. Combine all sample sizes and test against some population proportion p. In your example, you would add up all the numerators and all the denominators. Thus: p^= (43 + 21 + 11)/ (53 + 30 + 24)

p^= 0.701. (p^ implies an estimate i.e. a sample proportion). Now, you can test a hypothesis about your true population proportion (p). For example, if you want to know if over 65% of voters favor a particular candidate, you can test:

H0: p= 0.65

HA: p > 0.65

You also need some significance level, alpha. 0.05 (5%) is most common.

To do this, you would use the formula:

z=(p^-p₀)/sqrt(p₀q₀/n)

p^= sample proportion computed above = 0.701

p₀= hypothesized population proportion = 0.65 in example above

q₀= 1 - p₀

n= total sample size = 107

Once you compute this, you would go to a z-table and take 1 - P(Z<z); z= Value computed above. You would compare that with your alpha. If smaller, result is significant and you can reject H0 and conclude true proportion exceeds 0.65. If equal to or larger, you cannot make that conclusion and fail to reject H0.

#2. Treat samples as if they are coming from two different populations and compare the population proportions. I am familiar with a test to compare two populations like this, but I don't know a direct way to compare 3. You could do multiple comparisons for 3, though. That will mean running the test three times. In this case, you would still compute p*= total "successes"/total sample size as above, but only for two samples. Again, for this test, you would need to cut down to 2 samples, not 3, but you could do multiple cross-comparisons--sample 1 vs. sample 2, sample 1 vs. sample 3, sample 2 vs. sample 3. In this case:

H0: p1 = p2 (The population proportions are equal.)

HA: p1 not equal to p2 (Population proportions are not equal.)

Test Statistic for this:

z=(p^1-p^2)/sqrt(p*(1-p*)(1/n₁ + 1/n₂)

Again, you would need to pick an alpha, and again, 5% or 0.05 is most common/

p^1= First sample proportion

p^2=Second sample proportion

p*= total successes/total sample size; two samples combined

n₁= First sample size

n₂= Second sample size

You would then do the computation and check the z-table, but this time, you have a two-sided test, at least in this example, so you would do p= 2*(1-P(Z<z); z=value computed above and check that value against your alpha. Repeat the procedure for all desired comparisons.

I hope this helps.