## Question 388:

1

1. Q388_1_q388_2Proportions.xls

We need to compare two proportions here. The samples are large, but the small number of accidents will likely make the Normal Approximation to the Binomial inaccurate.  To answer the difference between two proportions. You want to know if the observed difference in proportions 20/153,348 - 4/135,035  = 0.0001008 is greater than chance. Notice that there is not much difference at all!

1. The NULL hypothesis is that there is no difference between yellow and red fire truck accidents.
2. The decision rule is if the probability is less than .01 that the observed difference was due to chance we'd reject the NULL hypothesis and conclude there is a difference between accident rates
3. The two sample proportions are 0.000130 and 0.000030
4. To find the z-statistic (which we need to answer where the difference is due to chance).
(See the attached Excel File for the computations)

Find the z-score
Take the observed difference between proportion 0.0001008 and divide by the square-root of a denominator which accounts for the chance which will provide you with a z-score. The denominator is the square root of:

1         1
--   +  --     * PQ
n1       n2

Where P = (x1 + x2)/(n1+n2)  Q is 1-P. The x's are just the total number of accidents and the n's are the sample size of the yellow and red fire-trucks.

P = (20+4)/(153,348+135,035) = 0.0000832
Q = 1-.0.00008 = .999916777
PQ = 0.0000832 *.999916777 = 0.000083216

1/n1 + 1/n2 = 1/153,348 + 1/135,035 = .00001393

So multiply  0.000083216 *.00001393  and take the square root. = 0.0000340428

So the equation is the observed difference 0.0001008 /0.0000340428 = 2.96, which is the z-score we're looking for. If you do a little bit of rounding above the z-score will be off by a bit, but should be within a few hundredths of a point.

That result is the z-score which is the test-statistic. You now look this value up using the z-score to percentile calculator using the 2-sided area. You should get a p-value of .003066, meaning there is about a  .3% chance the difference is due to chance. Since our cut off was 1% or .01, we'd conclude this is greater than chance alone and say that there really is a difference in accident rates. Finally, the observed difference of 0.0001008 is very small, however, given the cost of red and yellow paint are probably the same, if I were a fireman I'd want the truck painted yellow considering it reduces the chance of getting in accident (albeit by a small amount).

Part g.

For the normality assumption, that is our use of the z-score, we need np >5 and nq > 5. We get that for the 1st sample, however with our second sample, nq = 4, which is just below the 5. This means there is a little risk in using the normal approximation to the binomial.