Statistics Question

Question

The following is based on information from The Wolf in the Southwest: The Making of an Endangered Species, by David E. Brown (University of Arizona Press). Before 1918, the proportion of female wolves in the general population of all southwestern wolves was about 50%. However, after 1918, southwestern cattle ranchers began a widespread effort to destroy wolves. In a recent sample of 37 wolves, there were only 11 females. One theory is that male wolves tend to return sooner than females to their old territories, where their predecessors were exterminated. Do these data indicate that the population proportion of female wolves is now less than 50% in the region? Use α = 0.01.

1. What is the value of the sample test statistic? (Round your answer to two decimal places.)
2. Find the P-value of the test statistic. (Round your answer to four decimal places.)

Answer 1

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.

Answer 2

You need to conduct what is called Z-test: One Population Proportion

This editor is not compatible to showcase the calculations, but essentially you have to test if the hypothesized population proportion of 0.50 is significantly different from the sample proportion of 11/37 = 0.297.

Answer:

In my calculations:

1. the observed z-statistic or test statistic =−2.466

The p-value is p= 0.0068