Robert Z. answered • 08/15/20

3100+ hours (& counting!) tutoring math -- Prealgebra to Calculus 2

Maya,

You might often see a problem like this analyzed using a normal distribution. The normal distribution is based on having a continuous random variable; however, the number of female wolves is a __discrete__ random variable. Although the normal distribution can be used to approximate the binomial distribution, why do that when the technology of a TI-84 calculator makes it so easy to analyze it with the correct distribution.

The p-value in this case is the probability that the number of female wolves in a sample of 37 is 11 or fewer, given that the underlying proportion is 0.5. This can be found using the binomcdf function with (37, .5, 11) as the input values. This calculation returns a value of 0.0100369259. Since that is greater than α, we **do not reject** the null hypothesis that the proportion is 0.5. *Note that we are not comparing the calculation to α/2, because the wording of the problem indicates that a 1-tail test is appropriate.*

If we go ahead and use the normal approximation, the null hypothesis would be that the true proportion is 0.5, with a standard deviation of √p(1-p)/n = √.5(.5)/37 ≈ .0822

The lower limit of the “do not reject” region is the standard deviation times the z-score for a cumulative distribution graph area of .01. Using the invNorm on the calculator shows that

z = -2.326347877, resulting in 0.5 - 2.326347877(.0822) ≈ **.309. With this method, that is the critical value for the sample proportion.** The sample statistic, however, is 11/37 ≈ .297. It would seem to show that with this approach, our sample proportion is in the “reject” zone.

Not so fast, though! Remember that each integer value “n” in a discrete distribution is represented by the values from (n - .5) to (n + .5) in the continuous approximation. We should take the conservative step of using the sample proportion 11.5/37 ≈ **.311**. This is just slightly above the critical value, putting it back in the “do not reject” zone; using these slightly rounded values give a p-value of normalcdf(0,.311,.0822) ≈ **.0107.**

The different approaches produce slightly different results. The prudent thing to do would be to not reject the null hypothesis at the 1% significance level. If I had to choose a specific answer to your 2 questions, I’d go with:

**1. The sample proportion, rounded to 2 decimal places, is 0.30.**

**2. The p-value of that result is .0107.**