Test for Statistical Significance

[Guest]

Would really appreciate a little help with a winter beekeeping experiment I ran, to determine if it is statistically significant. In a population of 84 honey bee colonies, 34 received a winter oxalic acid treatment (to kill mites.) the remaining 50 were not treated. Of the 34 that received the oxalic acid treatment, 19 died (55.9%.) In the group of 50 that were not treated, 13 died (26.0.) Can we reasonably conclude, at a 95% confidence level, that oxalic acid kills bees?

I'm guessing the null hypothesis is that a winter oxalic acid treatment does not kill bees, but after that I'm lost in terms of which set of statistical calculations to use. Was thinking a t calculation, but there's no variance in these data; the bees are either alive or they aren't. And it's not like a problem where one is trying to determine if a sampled set of measurements falls within a certain range. Would appreciate some guidance. Thanks!

[Guest]

13 died naturally of the 50 that were not treated. 13 / 50 = 26%
13 x 34 /50 = 9 would have died naturally out of the 34 even if the 34 had not been treated.
Leaves 10 that have likely died due to acid treatment.
10/19 = 53% of a colony would likely die if colony given acid treatment
9/19 = 47% of a colony dies naturally anyway.

95% confidence = 0.53 +- 1.96 x sqrt(0.53 x 0.47 / 19) = 0.53 +- 0.224 = 0.754 to 0.306 = 14 to 6 treated sample dead, dead due to treatment

OR is it...........

95% confidence = 0.559 +- 1.96 x sqrt(0.559 x 0.441 / 34) = 0.559 +- 0.167 = 0.726 to 0.392 = 25 to 13 treated colony dead, all dead.

Interesting question..I'm not a statisticion....?

[Guest]

Just to clarify some of the beekeeping terminology, a "colony" is the family of bees that lives inside of a hive. During the winter, the colony consists of some 15,000 to 20,000 individual bees. They eat the honey they have stored over the course of the growing season, form a cluster with the outer 1 to 3 inches consisting of a very tightly packed aggregation of bees to create an insulating shell. The interior bees, which take turns serving on the outside, eat honey and shiver to generate heat. Once they begin to raise brood around the winter solstice, they keep the brood nest at 94 degrees F, regardless of the outside temperature. Pretty neat, huh? This dilute oxalic acid treatment - a 3% solution in sugar syrup - is trickled on the bees in late fall when they are broodless and kills varroa mites that have hitched a ride (meals on "wheels") on a bee for the winter. Any bee parasitized by a mite experiences a 20 to 80% reduction in her life span, depending on the number of mites. That's a real problem over the winter when bees must survive a good 5 months to make it to the following spring. So varroa mites unquestionably kill bees, and , although the mite populations must be reduced in summer, this late fall treatment would be great if it did not, as I suspect, also kill bees. The thing is, when a colony dies, ALL the bees die - all 15,000-20,000 of them. So the correct wording is:
10/19=53% of colonies would likely die if treated with oxalic acid. But the next statement, "9/19=47% of colonies die naturally anyway" isn't really true. Isn't it 9/34, the 26% we found from the control group that was left untreated?
And I don't understand these statements at all: "14 to 6 treated sample dead, dead due to treatment" and also the later statement, "25 to 13 treated colony dead, all dead."

[Guest]

Yes 9/34 = 26% is ratio that would have died naturally anyway had the 34 been "not treated"....which is based on the same

ratio as the 13/50 = 26% of the untreated colonies.

But considering 34 colonies as the treated sample the colonies which died is the total of the ones that may have died

naturally plus the extra ones that died due to the treatment. So 19 died in total and based on the proportions 9 expected

to die naturally and 10 died from the treatment.

Out of the 19 of the treated colonies that died 10/19 = 53% of them died from treatment and 9/19 = 47% died naturally.

The 95% confidence level for the population proportion then = 0.53 +- 1.96 x sqrt(0.53 x 0.47 / 19) = 0.53 +- 0.224 =

0.754 to 0.306

Therefore, the 95% confidence interval for the population proportion is between 0.306 and 0.754, i.e. there is a 95%

chance that the proportion of dead colonies due to treatment will fall within this range.

I am not sure this is correct but I was trying to deal with the dead colonies within the treated colonies and splitting up

the ones expected to die naturally and the ones killed by treatment.

The other calculation tries to deal with the 19 dead (9 died naturally plus 10 extra) compared with the whole sample of

34 that were treated. ie. all of the dead colonies no matter how they died.

I would interested to see a solution to this problem done by a mathematician.

[Guest]

So following on from previous last post does this mean:-

If you treat bees the proportion of dead colonies in the population will rise to between 39% and 73%

Also if you treat bees the proportion in the dead colonies that have been killed by the treatment is between 31% and 75%

If you don't treat bees the proportion of dead bees will only be 26%.

So don't treat bees.

[Guest]

and still to follow on, what about 50 the untreated:-

P95% CI = 0.26 +- 1.96 x sqrt( 0.26 x 0.74 / 50) = 0.26 +- 0.122 = 0.382 to 0.0.138

Therefore, the 95% confidence interval for the untreated population, the real proportion is estimated as between 0.382 and

0.0.138, Therefore there is a 95% chance that the proportion of untreated colonies which have died will fall within this

range.

So does that mean if "not treated" the percentage of dead colonies ranges from 14% to 38% rounded.

So don't treat bees.

...need to see a solution done by a mathematician...

[Guest]

Thank you very much. This is enormously helpful!

[Guest]

Hi, I am pretty new to statistical significance testin. I understand each of the two groups of data must be of uniform distribution to carry out the t tests. So what happens if one group is uniform and the other isn't? Do I use non-parametric test or can I still use the t test?

Thank you

[Guest]

I'm not a maths or a stats technician
but the calcs above deals with normal distibution and confidence intervals for proportions.
If P= fraction of sucess and Q= fraction of failed then Q = (1-P)
Let sample size is N.
The "z" value used was for normal standard distributions and for a 95% CI value it is 1.96

The central limit theorem states that when the sample size is large, approximately 95% of the sample data will fall within 1.96 standard errors of the middle of the distribution,
So 95% confident that the P fraction of the sample will be the P fraction of the population + or - 1.96 standard errors.
The formulae used was for CI for proportions, so the 95%CI = P =- 1.96 x sqrt(P x Q / N).
Different forms of this eqn would be used for different type of data whether analysed using means, standard deviation etc etc.

I think "t" values are for small samples??