16.3 - Unspecified Probabilities

Answer

In order to use the chi-square statistic to test the data, we need to be able to determine the observed and expected number of trials in which we'd get 0, 1, 2, 3, and 4 heads. The observed part is easy... we know those:

It's the expected numbers that are a problem. If the probability p of getting ahead were specified, we'd be able to calculate the expected numbers. Suppose, for example, that p = 1/2. Then, the probability of getting zero heads in four dimes is:

\(P(X=0)=\binom{4}{0}\left(\dfrac{1}{2}\right)^0\left(\dfrac{1}{2}\right)^4=0.0625 \)

and therefore the expected number of trials resulting in 0 heads is 100 × 0.0625 = 6.25. We could make similar calculations for the case of 1, 2, 3, and 4 heads, and we would be well on our way to using the chi-square statistic:

\(Q_4=\sum_{i=0}^{4}\dfrac{(Obs_i-Exp_i)^2}{Exp_i} \)

and comparing it to a chi-square distribution with 5−1 = 4 degrees of freedom. But, we don't know p, as it is unspecified! What do you think the logical thing would be to do in this case? Sure... we'd probably want to estimate p. But then that begs the question... what should we use as an estimate of p?

Answer

Given that four dimes are tossed 100 times, we have 400 coin tosses resulting in 205 heads for an estimated probability of success of 0.5125:

\(\hat{p}=\dfrac{0(8)+1(17)+2(41)+3(30)+4(4)}{400}=\dfrac{205}{400}=0.5125 \)

Using 0.5125 as the estimate of p, we can use the binomial p.m.f. (or Minitab!) to calculate the probability that X = 0, 1, ..., 4:

and then, using the probabilities, the expected number of trials resulting in 0, 1, 2, 3, and 4 heads:

Calculating the chi-square statistic, we get:

\(Q_4=\dfrac{(8-5.65)^2}{5.65}+\dfrac{(17-23.75)^2}{23.75}+ ... + \dfrac{(4-6.90)^2}{6.90} =4.99\)

We estimated the d = 1 parameter in calculating the chi-square statistic. Therefore, we compare the statistic to a chi-square distribution with (5−1)−1 = 3 degrees of freedom. Doing so:

\(Q_4= 4.99 < \chi_{3,0.05}^{2}=7.815\)

we fail to reject the null hypothesis. There is insufficient evidence at the 0.05 level to conclude that the data don't fit a binomial probability model.

Answer

Note that very few observations resulted in 0, 1, or 2 alpha particles being emitted in 0.1 second. And, very few observations resulted in 10, 11, or 12 alpha particles being emitted in 0.1 second. Therefore, let's "collapse" the data at the two ends, yielding us nine "not-so-sparse" categories:

Because \(\lambda\), the mean of X, is not specified, we can estimate it with its maximum likelihood estimator, namely, the sample mean. Using the data, we get:

\(\bar{x}=\dfrac{1(1)+2(4)+3(13)+ ... + 12(1)}{100}=\dfrac{559}{100}=5.6\)

We can now estimate the probability that an observation will fall into each of the categories. The probability of falling into category 1, for example, is:

\(P(\{1\})=P(X=0)+P(X=1)+P(X=2) =\dfrac{e^{-5.6}5.6^0}{0!}+\dfrac{e^{-5.6}5.6^1}{1!}+\dfrac{e^{-5.6}5.6^2}{2!}=0.0824 \)

Here's what our table looks like now, after adding a column containing the estimated probabilities:

Now, we just have to add a column containing the expected number falling into each category. The expected number falling into category 1, for example, is 0.0824 × 100 = 8.24. Doing a similar calculation for each of the categories, we can add our column of expected numbers:

Now, we can use the observed numbers and the expected numbers to calculate our chi-square test statistic. Doing so, we get:

\(Q_{9-1}=\dfrac{(5-8.24)^2}{8.24}+\dfrac{(13-10.82)^2}{10.82}+ ... +\dfrac{(4-5.39)^2}{5.39}=5.7157 \)

Because we estimated d = 1 parameter, we need to compare our chi-square statistic to a chi-square distribution with (9−1)−1 = 7 degrees of freedom. That is, our critical region is defined as:

\(\text{Reject } H_0 \text{ if } Q_8 \ge \chi_{8-1, 0.05}^{2}=\chi_{7, 0.05}^{2}=14.07 \)

Because our test statistic doesn't fall in the rejection region, that is:

\(Q_8=5.77157 < \chi_{7, 0.05}^{2}=14.07\)

we fail to reject the null hypothesis. There is insufficient evidence at the 0.05 level to conclude that the data don't fit a Poisson probability model.