Answer

Testing whether there is a "good fit" between the observed data and the assumed probability model amounts to testing:

\(H_0 : p_F =0.60\)
\(H_A : p_F \ne 0.60\)

Now, letting \(Y_1\) denote the number of females selected, we know that \(Y_1\) follows a binomial distribution with n trials and probability of success \(p_1\). That is:

\(Y_1 \sim b(n,p_1)\)

Therefore, the expected value and variance of \(Y_1\) are, respectively:

\(E(Y_1)=np_1\) and \(Var (Y_1) =np_1(1-p_1)\)

And, letting \(Y_2\) denote the number of males selected, we know that \(Y_2 = n − Y_1\) follows a binomial distribution with n trials and probability of success \(p_2\). That is:

\(Y_2 = n-Y_1 \sim (b(n,p_2)=b(n,1-p_1)\)

Therefore, the expected value and variance of \(Y_2\) are, respectively:

\(E(Y_2)=n(1-p_1)=np_2\) and \(Var(Y_2)=n(1-p_1)(1-(1-p_1))=np_1(1-p_1)=np_2(1-p_2)\)

Now, for large samples (\(np_1 ≥ 5\) and \(n(1 − p_1) ≥ 5)\), the Central Limit Theorem yields the normal approximation to the binomial distribution. That is:

\(Z=\dfrac{(Y_1-np_1)}{\sqrt{np_1(1-p_1)}}\)

follows, at least approximately, the standard normal N(0,1) distribution. Therefore, upon squaring Z, we get that:

\(Z^2=Q_1=\dfrac{(Y_1-np_1)^2}{np_1(1-p_1)}\)

follows an approximate chi-square distribution with one degree of freedom. Now, we could stop there. But, that's not typically what is done. Instead, we can rewrite \(Q_1\) a bit. Let's start by multiplying \(Q_1\) by 1 in a special way, that is, by multiplying it by \(\left( \left( 1 − p_1 \right) + p_1 \right)\):

\(Q_1=\dfrac{(Y_1-np_1)^2}{np_1(1-p_1)} \times ((1-p_1)+p_1)\)

Then, distributing the "1" across the numerator, we get:

\(Q_1=\dfrac{(Y_1-np_1)^2(1-p_1)}{np_1(1-p_1)}+\dfrac{(Y_1-np_1)^2p_1}{np_1(1-p_1)}\)

which simplies to:

\(Q_1=\dfrac{(Y_1-np_1)^2}{np_1}+\dfrac{(Y_1-np_1)^2}{n(1-p_1)}\)

Now, taking advantage of the fact that \(Y_1 = n − Y_2\) and \(p_1 = 1 − p_2\), we get:

\(Q_1=\dfrac{(Y_1-np_1)^2}{np_1}+\dfrac{(n-Y_2-n(1-p_2))^2}{np_2}\)

which simplifies to:

\(Q_1=\dfrac{(Y_1-np_1)^2}{np_1}+\dfrac{(-(Y_2-np_2))^2}{np_2}\)

Just one more thing to simplify before we're done:

\(Q_1=\dfrac{(Y_1-np_1)^2}{np_1}+\dfrac{(Y_2-np_2)^2}{np_2}\)

In summary, we have rewritten \(Q_1\) as:

\( Q_1=\sum_{i=1}^{2}\dfrac{(Y_i-np_i)^2}{np_i}=\sum_{i=1}^{2}\dfrac{(\text{OBSERVED }-\text{ EXPECTED})^2}{EXPECTED} \)

We'll use this form of \(Q_1\), and the fact that \(Q_1\) follows an approximate chi-square distribution with one degree of freedom, to conduct the desired hypothesis test.

Answer

The value of the test statistic \(Q_1\) is:

\(Q_1=\dfrac{(53-60)^2}{60}+\dfrac{(47-40)^2}{40}=2.04\)

We should reject the null hypothesis if the observed number of counts is very different from the expected number of counts, that is if \(Q_1\) is large. Because \(Q_1\) follows a chi-square distribution with one degree of freedom, we should reject the null hypothesis, at the 0.05 level, if:

\(Q_1 \ge \chi_{0.05, 1}^{2}=3.84\)

Because:

\(Q_1=2.04 < 3.84\)

we do not reject the null hypothesis. There is not enough evidence at the 0.05 level to conclude that the data don't fit the assumed probability model.

As an aside, it is interesting to note the relationship between using the chi-square goodness of fit statistic \(Q_1\) and the Z-statistic we've previously used for testing a null hypothesis about a proportion. In this case:

\( Z=\dfrac{0.53-0.60}{\sqrt{\frac{(0.60)(0.40)}{100}}}=-1.428 \)

which, you might want to note that, if we square it, we get the same value as we did for \(Q_1\):

\(Q_1=Z^2 =(-1.428)^2=2.04\)

as we should expect. The Z-test for a proportion tells us that we should reject if:

\(|Z| \ge 1.96 \)

Well, again, if we square it, we should see that that's equivalent to rejecting if:

\(Q_1 \ge (1.96)^2 =3.84\)

And not surprisingly, the P-values obtained from the two approaches are identical. The P-value for the chi-square goodness-of-fit test is:

\(P=P(\chi_{(1)}^{2} > 2.04)=0.1532\)

while the P-value for the Z-test is:

\(P=2 \times P(Z>1.428)=2(0.0766)=0.1532\)

Identical, as we should expect!

Answer

We can summarize the observed \((y_i)\) and expected \((np_i)\) counts in a table as follows:

where, for example, the expected number of brown candies is:

\(np_1 = 580(0.40) = 232\)

and the expected number of green candies is:

\(np_4 = 580(0.10) = 58\)

Once we have the observed and expected number of counts, the calculation of the chi-square statistic is straightforward. It is:

\(Q_4=\dfrac{(224-232)^2}{232}+d\frac{(119-116)^2}{116}+\dfrac{(130-116)^2}{116}+\dfrac{(48-58)^2}{58}+\dfrac{(59-58)^2}{58} \)

Simplifying, we get:

\(Q_4=\dfrac{64}{232}+\dfrac{9}{116}+\dfrac{196}{116}+\dfrac{100}{58}+\dfrac{1}{58}=3.784 \)

Because there are k = 5 categories, we have to compare our chi-square statistic \(Q_4\) to a chi-square distribution with k−1 = 5−1 = 4 degrees of freedom:

\(\text{Reject }H_0 \text{ if } Q_4\ge \chi_{4,0.05}^{2}=9.488\)

Because \(Q_4 = 3.784 < 9.488\), we fail to reject the null hypothesis. There is insufficient evidence at the 0.05 level to conclude that the distribution of the color of the candies differs from that specified in the null hypothesis.

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.

Answer

Hmm. So, where do we start? Well, we first have to define some categories. Let's divide up the interval of possible IQs into \(k = 10\) sets of equal probability \(\dfrac{1}{k} = \dfrac{1}{10}\). Perhaps this is best seen pictorially:

So, what's going on in this picture? Well, first the normal density is divided up into 10 intervals of equal probability (0.10). Well, okay, so the picture is not drawn very well to scale. At any rate, we then find the IQs that correspond to the \(k = 10\) cumulative probabilities of 0.1, 0.2, 0.3, etc. This is done in two steps:

1. Step 1

   first by finding the Z-scores associated with the cumulative probabilities 0.1, 0.2, 0.3, etc.

2. Step 2

   then by converting each Z-score into an X-value. It is those X-values (IQs) that will make up the "right-hand side" of each bucket:

   Category	\(X\)	Obs'd	\(p_i = \left(e^{-5.6}5.6^x\right) / x!\)	Exp'd
   1	0,1,2*	5	0.0824	8.24
   2	3	13	0.1082	10.82
   3	4	19	0.1515	15.15
   4	5	16	0.1697	16.97
   5	6	15	0.1584	15.84
   6	7	9	0.1267	12.67
   7	8	12	0.0887	8.87
   8	9	7	0.0552	5.52
   9	10,11,12*	4	0.0539	5.39
   		\(n = 100\)		99.47

   1.  

      Category	Class
      1	(\(-\infty\),79.5)
      2	(79.5, 86.5)
      3	(86.5, 91.6)
      4	(91.6, 95.9)
      5	(95.9, 100.0)
      6	(100.0, 104.1)
      7	(104.1, 108.4)
      8	(108.4, 113.5)
      9	(113.5, 120.5)
      10	(120.5, \(\infty\))

      Now, it's just a matter of counting the number of observations that fall into each bucket to get the observed (Obs'd) column, and calculating the expected number (0.10 × 100 = 10) to get the expected (Exp'd) column:

As illustrated in the table, using the observed and expected numbers, we see that the chi-square statistic is 8.2. We reject if the following is true:

\(Q_9 =8.2 \ge \chi_{10-1, 0.05}^{2} =\chi_{9, 0.05}^{2}=16.92\)

It isn't! We do not reject the null hypothesis at the 0.05 level. There is insufficient evidence to conclude that the data do not follow a normal distribution with a mean of 100 and a standard deviation 16.