Among 784 checks, 479 had amounts with leading digits of 5, but checks issued in the normal course of honest transactions were expected to have 7.9% of the checks with amounts having leading digits of 5. Is there strong evidence to indicate that the check amounts are significantly different from amounts that are normally expected?
$$n = 784$$ $$x = 478.5$$ $$p = .079$$ $$q = .921$$ $$$$
the mean $$np = 61.94$$ $$nq = 722.06$$
both greater than 5, so it can be approximated
standard deviation $$sd = \sqrt{n*p*q} = 7.5527$$
z-score $$z = \frac{478.5 - 61.94}{7.5527} = 55.21999$$
the z-score corresponds with a probability $$z = 55.22 = 0.9999%$$ $$ 1 - 0.9999 = 0.0001 $$
I'm sure the math is correct but I can't understand what the probability is doing for me. Out of 784 checks, 7.9% are expected to have a leading digit of 5, the normal distribution curve with an expected mean of 61.94 (z score 0, prob 50%), so the question is asking for the probability of getting at least (?) a value of 479? So subtract that probability from one, and that's the probability of getting at least 479? So isnt the probability of getting atleast 479 small, and not strong?
Your understanding is correct. In fact $z=55$ is extreme, and the probability is much smaller than 0.0001.
The interpretaton is that the observation (479) is very improbable given the null hypothesis. We can therefore strongly reject the null hypothesis. So yes, there is strong evidence of funny business.