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.

Answer

Let's start by focusing on the business school. We can see that, of the 1200 males who applied to the university, 300 (or 25%) were accepted into the business school. Of the 800 females who applied to the university, 200 (or 25%) were accepted into the business school. So, the business school looks to be in good shape, as an equal percentage of males and females, namely 25%, were accepted into it.

Now, for the engineering school. We can see that, of the 1200 males who applied to the university, 240 (or 20%) were accepted into the engineering school. Of the 800 females who applied to the university, 160 (or 20%) were accepted into the engineering school. So, the engineering school also looks to be in good shape, as an equal percentage of males and females, namely 20%, were accepted into it.

We probably don't have to drag this out any further. If we look at each column in the table, we see that the proportion of males accepted into each school is the same as the proportion of females accepted into each school... which therefore happens to equal the proportion of students accepted into each school, regardless of gender. Therefore, we can conclude that males and females are distributed equally among the four schools.

Answer

Let's again start by focusing on the business school. In this case, of the 1200 males who applied to the university, 240 (or 20%) were accepted into the business school. And, of the 800 females who applied to the university, 240 (or 30%) were accepted into the business school. So, the business school appears to have different rates of acceptance for males and females, 20% compared to 30%.

Now, for the engineering school. We can see that, of the 1200 males who applied to the university, 480 (or 40%) were accepted into the engineering school. Of the 800 females who applied to the university, only 80 (or 10%) were accepted into the engineering school. So, the engineering school also appears to have different rates of acceptance for males and females, 40% compared to 10%.

Again, there's no need drag this out any further. If we look at each column in the table, we see that the proportion of males accepted into each school is different than the proportion of females accepted into each school... and therefore the proportion of students accepted into each school, regardless of gender, is different than the proportion of males and females accepted into each school. Therefore, we can conclude that males and females are not distributed equally among the four schools.

Answer

We are interested in testing the null hypothesis:

\(H_0 : p_{RF} =p_{AF} \text{ and } p_{RO} =p_{AO} \text{ and } p_{RR} =p_{AR} \text{ and } p_{RN} =p_{AN}\)

against the alternative hypothesis:

\(H_A : p_{RF} \ne p_{AF} \text{ or } p_{RO} \ne p_{AO} \text{ or } p_{RR} \ne p_{AR} \text{ or } p_{RN} \ne p_{AN}\)

The observed data were given to us in the table above. So, the next thing we need to do is find the expected counts for each cell of the table:

It is in the calculation of the expected values that you can readily see why we have (2−1)(4−1) = 3 degrees of freedom in this case. That's because, we only have to calculate three of the cells directly.

Once we do that, the remaining five cells can be calculated by way of subtraction:

Now that we have the observed and expected counts, calculating the chi-square statistic is a straightforward exercise:

\(Q=\frac{(2-6.942)^2}{6.942}+ ... +\frac{(5-10.65)^2}{10.65} =31.88 \)

The chi-square test tells us to reject the null hypothesis, at the 0.05 level, if Q is greater than a chi-square random variable with 3 degrees of freedom, that is, if \(Q > 7.815\). Because \(Q = 31.88 > 7.815\), we reject the null hypothesis. There is sufficient evidence at the 0.05 level to conclude that the distribution of unnecessary transfusions differs among attending physicians and residents.

Answer

Here's the table of expected counts:

which results in a chi-square statistic of 8.006:

\(Q=\frac{(60-50.886)^2}{50.886}+ ... +\frac{(57-48.132)^2}{48.132}=8.006 \)

The chi-square test tells us to reject the null hypothesis, at the 0.05 level, if Q is greater than a chi-square random variable with 3 degrees of freedom, that is, if Q > 7.815. Because Q = 8.006 > 7.815, we reject the null hypothesis. There is sufficient evidence at the 0.05 level to conclude that the desire to ride a bicycle depends on age.

Answer

Well, without even knowing the formal definition of an order statistic, we probably don't need a rocket scientist to tell us that, in order to find the order statistics, we should probably arrange the data in increasing numerical order. Doing so, the observed order statistics are:

\(y_1=602<y_2=633<y_3=709<y_4=742<y_5=781\)

Answer

The key to finding the desired probability is to recognize that the only way that the fifth order statistic, \(Y_5\), would be less than one is if at least 5 of the random variables \(X_1, X_2, X_3, X_4, X_5, \text{ and }X_6\) are less than one. For the sake of simplicity, let's suppose the first five observed values \(x_1, x_2, x_3, x_4, x_5\) are less than one, but the sixth \(x_6\) is not. In that case, the observed fifth-order statistic, \(y_5\), would be less than one:

The observed fifth order statistic, \(y_5\), would also be less than one if all six of the observed values \(x_1, x_2, x_3, x_4, x_5, x_6\) are less than one:

The observed fifth order statistic, \(y_5\), would not be less than one if the first four observed values \(x_1, x_2, x_3, x_4\) are less than one, but the fifth \(x_5\) and sixth \(x_6\) is not:

Again, the only way that the fifth order statistic, \(Y_5\), would be less than one is if 5 or 6... that is, at least 5... of the random variables \(X_1, X_2, X_3, X_4, X_5, \text{ and }X_6\) are less than one. For the sake of simplicity, we considered just the first five or six random variables, but in reality, any five or six random variables less than one would do. We just have to do some "choosing" to count the number of ways that we can get any five or six of the random variables to be less than one.

If you think about it, then, we have a binomial probability calculation here. If the event \(\{X_i<1\}\), \(i=1, 2, \cdots, 5\) is considered a "success," and we let \(Z\) = the number of successes in six mutually independent trials, then \(Z\) is a binomial random variable with \(n=6\) and \(p=0.25\):

\(P(X_i\le1)=\dfrac{1}{2}\int_{0}^{1}x dx=\dfrac{1}{2}\left[\dfrac{x^2}{2}\right]_{x=0}^{x=1}=\dfrac{1}{2}\left(\dfrac{1}{2}-0\right)=\dfrac{1}{4}\)

Finding the probability that the fifth order statistic, \(Y_5\), is less than one reduces to a binomial calculation then. That is:

\(P(Y_5<1)=P(Z=5)+P(Z=6)=\binom{6}{5}\left(\dfrac{1}{4} \right)^5\left(\dfrac{3}{4} \right)^1+\binom{6}{6}\left(\dfrac{1}{4} \right)^6\left(\dfrac{3}{4} \right)^0=0.0046\)

The fact that the calculated probability is so small should make sense in light of the given p.d.f. of \(X\). After all, each individual \(X_i\) has a greater chance of falling above, rather than below, one. For that reason, it would unusual to observe as many as five or six \(X\)'s less than one.

Answer

Recalling the definition of a cumulative distribution function, \(G_5(y)\) is defined to be the probability that the fifth order statistic \(Y_5\) is less than some value \(y\). That is:

\(G_5(y)=P(Y_5 < y)\)

Well, in our above work, we found the probability that the fifth order statistic \(Y_5\) is less than a specific value, namely 1. We just need to generalize our work there to allow for any value \(y\). Well, if the event \(\{X_i<y\}\), \(i=1, 2, \cdots, 5\) is considered a "success," and we let \(Z\) = the number of successes in six mutually independent trials, then \(Z\) is a binomial random variable with \(n=6\) and probability of success:

\(P(X_i\le y)=\dfrac{1}{2}\int_{0}^{y}x dx=\dfrac{1}{2}\left[\dfrac{x^2}{2}\right]_{x=0}^{x=y}=\dfrac{1}{2}\left(\dfrac{y^2}{2}-0\right)=\dfrac{y^2}{4}\)

Therefore, the cumulative distribution function \(G_5(y)\) of the fifth order statistic \(Y_5\) is:

\(G_5(y)=P(Y_5 < y)=P(Z=5)+P(Z=6)=\binom{6}{5}\left(\dfrac{y^2}{4}\right)^5\left(1-\dfrac{y^2}{4}\right)+\left(\dfrac{y^2}{4}\right)^6\)

for \(0<y<2\).

Answer

All we need to do to find the probability density function \(g_{5}(y)\) is to take the derivative of the distribution function \(G_5(y)\) with respect to \(y\). Doing so, we get:

\(g_5(y)=G_{5}^{'}(y)=\binom{6}{5}\left(\dfrac{y^2}{4}\right)^5\left(\dfrac{-2y}{4}\right)+\binom{6}{5}\left(1-\dfrac{y^2}{4}\right)5\left(\dfrac{y^2}{4}\right)^4\left(\dfrac{2y}{4}\right)+6\left(\dfrac{y^2}{4}\right)^5\left(\dfrac{2y}{4}\right)\)

Upon recognizing that:

\(\binom{6}{5}=6\) and \(\binom{6}{5}\times5=\dfrac{6!}{5!1!}\times5=\dfrac{6!}{4!1!}\)

we see that the middle term simplifies somewhat, and the first term is just the negative of the third term, therefore they cancel each other out:

\(g_{5}(y)=\left(\begin{array}{l}
6 \\
5
\end{array}\right)\color{red}\cancel {\color{black}\left(\frac{y^{2}}{4}\right)^{5}\left(\frac{-2 y}{4}\right)}\color{black}+\frac{6 !}{4 ! 1 !}\left(1-\frac{y^{2}}{4}\right)\left(\frac{y^{2}}{4}\right)^{4}\left(\frac{2 y}{4}\right)+\color{red}\cancel {\color{black}6\left(\frac{y^{2}}{4}\right)^{5}\left(\frac{2 y}{4}\right)}\)

Therefore, the probability density function of the fifth order statistic \(Y_5\) is:

\(g_5(y)=\dfrac{6!}{4!1!}\left(1-\dfrac{y^2}{4}\right)\left(\dfrac{y^2}{4}\right)^4\left(\dfrac{1}{2}y\right)\)

for \(0<y<2\). When we go on to generalize our work on the next page, it will benefit us to note that because the density function and distribution function of each \(X\) are:

\(f(x)=\dfrac{1}{2}x\) and \(F(x)=\dfrac{x^2}{4}\)

respectively, when \(0<x<2\), we can alternatively write the probability density function of the fifth order statistic \(Y_5\) as:

\(g_5(y)=\dfrac{6!}{4!1!}\left[F(y)\right]^4\left[1-F(y)\right]f(y)\)

Done!

Solution

When we worked with this example on the previous page, we showed that the cumulative distribution function of \(X\) is:

\(F(x)=\dfrac{x^2}{4} \)

for \(0<x<2\). Therefore, applying the above theorem with \(n=6\) and \(r=1\), the p.d.f. of \(Y_1\) is:

\(g_1(y)=\dfrac{6!}{0!(6-1)!}\left[\dfrac{y^2}{4}\right]^{1-1}\left[1-\dfrac{y^2}{4}\right]^{6-1}\left(\dfrac{1}{2}y\right)\)

for \(0<y<2\), which can be simplified to:

\(g_1(y)=3y\left(1-\dfrac{y^2}{4}\right)^{5}\)

Applying the theorem with \(n=6\) and \(r=4\), the p.d.f. of \(Y_4\) is:

\(g_4(y)=\dfrac{6!}{3!(6-4)!}\left[\dfrac{y^2}{4}\right]^{4-1}\left[1-\dfrac{y^2}{4}\right]^{6-4}\left(\dfrac{1}{2}y\right)\)

for \(0<y<2\), which can be simplified to:

\(g_4(y)=\dfrac{15}{32}y^7\left(1-\dfrac{y^2}{4}\right)^{2}\)

Applying the theorem with \(n=6\) and \(r=6\), the p.d.f. of \(Y_6\) is:

\(g_6(y)=\dfrac{6!}{5!(6-6)!}\left[\dfrac{y^2}{4}\right]^{6-1}\left[1-\dfrac{y^2}{4}\right]^{6-6}\left(\dfrac{1}{2}y\right)\)

for \(0<y<2\), which can be simplified to:

\(g_6(y)=\dfrac{3}{1024}y^{11}\)

Fortunately, when we graph the three functions on one plot:

we see something that makes intuitive sense, namely that as we increase the number of the order statistic, the p.d.f. "moves to the right" on the support interval.

Answer

Because \(r=(13+1)(0.25)=3.5\), the first quartile, alternatively known as the 25th percentile, is:

\(\tilde{q}_1 =y_3+0.5(y_4-y_3) = 0.5y_3+0.5y_4=0.5(49)+0.5(50)=49.5\)

Because \(r=(13+1)(0.4)=5.6\), the 40th percentile is:

\(\tilde{\pi}_{0.40} = y_5 + 0.6(y_6-y_5) =0.4y_5 + 0.6y_6 =0.4(53)+0.6(54)=53.6\)

Because \(r=(13+1)(0.5)=7\), the median is:

\(\tilde{m} =y_7 = 55\)

Answer

Because there are \(n=19\) data points, \(y_{10}=4.80\) serves as a good point estimator for the population median m. Let's go up and down a few spots from there to consider:

\((y_6, y_{14})=(3.15, 5.35)\)

as a possible confidence interval for \(m\). The confidence coefficient associated with the interval \((Y_6, Y_{14})\) is calculated using a binomial table with \(n=19\) and \(p=0.5\):

\(P(Y_6<m<Y_{14})=P(6\le W \le 13)=P(W \le 13) -P(W \le 5)=0.9682-0.0318=0.9364 \)

We can therefore be 93.64% confident that the population median falls in the interval (3.15, 5.35).

Could we do any better? Well, if we were to use the narrower interval \((y_7, y_{13})=(3.90, 5.15)\) instead, its confidence coefficient is not quite as good:

\(P(Y_7<m<Y_{13})=P(7\le W \le 12)=P(W \le 12) -P(W \le 6)=0.9165-0.0835=0.8330 \)

Or, if we were to use the wider interval \((y_5, y_{15})=(3.10, 5.50)\) instead, its confidence coefficient is perhaps a bit too high:

\(P(Y_5<m<Y_{15})=P(5\le W \le 14)=P(W \le 14) -P(W \le 4)=0.9904-0.0096=0.9808 \)

In general, we should aim to get a confidence coefficient at least 90%, but as close to 95% as possible. And, we shouldn't really "shop around" for an interval after we've collected the data. We should decide in advance which confidence interval we are going to use, and commit to use it even after the data have been collected.

Answer

In this case, the mean and variance are:

\(\mu=np=0.5(26)=13\) and \(\sigma^2=np(1-p)=0.5(1-0.5)n=0.25(26)=6.5\)

respectively. Therefore, the approximate confidence coefficient for the interval \((Y_8, Y_{18})\) is:

\(P(Y_8<m<Y_{18})=P(8 \le W \le 17)=P\left(\frac{7.5-13}{\sqrt{6.5}} < Z < \frac{17.5-13}{\sqrt{6.5}} \right)\)

which can be simplified to:

\(P(Y_8<m<Y_{18})=P(-2.16 \le Z \le 1.77)=0.9616 - 0.0154 = 0.9462\)

We can be approximately 94.6% confident that the median time of all escapes is between 359 and 375 seconds.

Answer

Since \((0.75)(26+1)=20.25\), the weighted average of the 20th and 21st order statistic:

\(\tilde{\pi}_{0.75}=y_{20}+0.25(y_{21}-y_{20})=0.75y_{20}+0.25y_{21}=0.75(392)+0.25(393)=392.25\)

serves as a good point estimate of \(\pi_{0.75}\). To find a confidence interval for \(\pi_{0.75}\), let's move up and down a few order statistics from \(y_{20}\) to, say, \(y_{16}\) and \(y_{24}\). In that case, our interval is \((y_{16}, y_{24}=(373, 402)\) with an exact confidence coefficient calculated using a binomial distribution with n = 26 and p = 0.75 as:

\(P(Y_{16}<m<Y_{24})=P(16 \le W \le 23)=P(W \le 23)-P(W \le 15)=0.9742-0.0401=0.9341\)

We can be 93.4% confident that the 75th percentile of all escape times is between 373 and 402 seconds.

Because \(n=26\) here, we could have alternatively used the normal approximation to the binomial. In this case, the mean and variance are:

\(\mu=np=0.75(26)=19.5\) and \(\sigma^2 =np(1-p)=26(0.75)(1-0.75)=4.875\)

respectively. Therefore, the approximate confidence coefficient for the interval \((y_{16}, y_{24})\) is:

\(P(Y_{16}<m<Y_{24})=P(16 \le W \le 23)=P\left(\dfrac{15.5-19.5}{\sqrt{4.875}} < Z < \dfrac{23.5-19.5}{\sqrt{4.875}} \right)\)

which can be simplified to:

\(P(Y_{16}<m<Y_{24})=P(-1.81 \le Z \le 1.81)=0.9649-0.0359=0.929\)

As you can see, the normal approximation does quite well, as the approximate probability is 0.929 compared to the exact probability of 0.934.

Answer

We are interested in testing the null hypothesis \(H_0 \colon m = 22\) against the alternative hypothesis \(H_A \colon m < 22\). To do so, we first calculate \(x_i − 22\), for \(i = 1, 2, \dots, 20\). Letting Minitab do the 20 subtractions for us, we get:

We can readily see that \(n+\), the observed number of positive signs, is 5. Therefore, we need to calculate how likely it would be to observe as few as 5 positive signs if the null hypothesis were true. Doing so, we get a P-value of 0.0207:

which is, of course, smaller than \(\alpha = 0.05\). The P-value tells us that it is not likely that we would observe so few positive signs if the null hypothesis were true. Therefore, we reject the null hypothesis in favor of the alternative hypothesis. There is sufficient evidence, at the 0.05 level, to conclude that the median visit length in practices with a large Medicaid load is shorter than 22 minutes.

Incidentally, we can use Minitab to conduct the sign test for us by selecting, under the Stat menu, Nonparametrics and then 1-Sample Sign. Doing so, we get:

Answer

We are interested in testing the null hypothesis \(H_0 \colon m_D = m_{B−A} = 0\) against the alternative hypothesis \(H_A \colon m_D = m_{B−A} > 0\). To do so, we just have to conduct a sign test while treating the differences as the data. If we look at the differences, we see that one of the differences is neither positive or negative, but rather zero. In this case, the standard procedure is to remove the observation that produces the zero, and reduce the number of observations by one. That is, we conduct the sign test using n = 9 rather than n = 10.

Now, n−, the observed number of negative signs, is 1, which yields a P-value of 0.0195:

The P-value is less than 0.05. Therefore, we can reject the null hypothesis. There is sufficient evidence, at the 0.05 level, to conclude that the surgery tends to lower arterial blood pressure.

Again, we can use Minitab to conduct the sign test for us. Doing so, we get:

Answer

We are interested in testing the null hypothesis \(H_0: m = 3.7\) against the alternative hypothesis \(H_A: m ≠ 3.7\). In general, the Wilcoxon signed rank test procedure requires five steps. We'll introduce each of the steps as we apply them to the data in this example.

1. Step 1

   In general, calculate \(X_i − m_0\) for \(i = 1, 2, \dots , n\). In this case, we have to calculate \(X_i − 3.7\) for \(i = 1, 2, \dots , 10\):

   

2. Step 2

   In general, calculate the absolute value of \(X_i − m_0\), that is, \(|X_i − m_0|\) for \(i = 1, 2, \dots , n\). In this case, we have to calculate \(|X_i − 3.7|\) for \(i = 1, 2, \dots , 10\):

   

3. Step 3

   Determine the rank \(R_i, i = 1, 2,\dots , n\) of the absolute values (in ascending order) according to their magnitude. In this case, the value of 0.2 is the smallest, so it gets rank 1. The value of 0.6 is the next smallest, so it gets rank 2. We continue ranking the data in this way until we have assigned a rank to each of the data values:

   

4. Step 4

   Determine the value of W, the Wilcoxon signed-rank test statistic:

   \( W=\sum_{i=1}^{n}Z_i R_i\)

   where \(Z_i\) is an indicator variable with \(Z_i = 0\) if \(X_i − m_0\) is negative and \(Z_i = 1\) if \(X_i − m_0\) is positive. That is, with \(Z_i\) defined as such, W is then the sum of the positive signed ranks. In this case, because the first observation yields a positive \(X_1 − 3.7\), namely 1.3, \(Z_1 = 1\). Because the fifth observation yields a negative \(X_5 − 3.7\), namely −0.9, \(Z_5 = 0\). Determining \(Z_i\) as such for \(i = 1, 2, \dots , 10\), we get:

   

   And, therefore W equals 40:

   \( W=(1)(5)+(1)(1)+ ... +(0)(-8)+(1)(2) =5+1+6+7+9+10+2=40\)

5. Step 5

   Determine if the observed value of W is extreme in light of the assumed value of the median under the null hypothesis. That is, calculate the P-value associated with W, and make a decision about whether to reject or not to reject. Whoa, nellie! We're going to have to take a break from this example before we can finish, as we first have to learn something about the distribution of W.

Answer

Recall that we are interested in testing the null hypothesis \(H_0 \colon m = 3.7\) against the alternative hypothesis \(H_A \colon m ≠ 3.7\). The last time we worked on this example, we got as far as determining that W = 40 for the given data set. Now, we just have to use what we know about the distribution of W to complete our hypothesis test. Well, in this case, with n = 10, our sample size is fairly small so we can use the exact distribution of W. The upper and lower percentiles of the Wilcoxon signed rank statistic when n = 10 are:

Therefore, our P-value is 2 × 0.116 = 0.232. Because our P-value is large, we cannot reject the null hypothesis. There is insufficient evidence at the 0.05 level to conclude that the median length of pygmy sunfish differs significantly from 3.7 centimeters.

Answer

We are interested in testing the null hypothesis \(H_0 \colon m = 45\) against the alternative hypothesis \(H_A \colon m ≠ 45\). We can use Minitab's calculator and statistical functions to do the dirty work for us:

Then, summing the last column, we get:

Because we have a large sample (n = 30), we can use the normal approximation to the distribution of W. In this case, our P-value is defined as two times the probability that W ≤ 200. Therefore, using a half-unit correction for continuity, our transformed signed rank statistic is:

\(W'=\dfrac{200.5 - \left(\frac{30(31)}{4}\right)}{\sqrt{\frac{30(31)(61)}{24}}}=-0.6581 \)

Therefore, upon using a normal probability calculator (or table), we get that our P-value is:

\(P \approx 2 \times P(W' < -0.66)=2(0.2546) \approx 0.51 \)

Because our P-value is large, we cannot reject the null hypothesis. There is insufficient evidence at the 0.05 level to conclude that the median age of the onset of diabetes differs significantly from 45 years.

By the way, we can even be lazier and let Minitab do all of the calculation work for us. Under the Stat menu, if we select Nonparametrics, and then 1-Sample Wilcoxon, we get:

Answer

Let X denote the number of cavities in the uncoated teeth, and Y denote the number of cavities in the coated teeth. Then, we're interested in analyzing the differences D = X−Y. Specifically, if m is the median of the differences D = X−Y, then we're interested in testing the null hypothesis \(H_0 \colon m = 0\) against the alternative hypothesis \(H_A \colon m > 0\). Analyzing the differences, we see that two of the differences are 0, and so we eliminate them from the analysis:

Assigning ranks to the remaining 10 data points, we see that we have two sets of tied observations. Because there are four 1s, we assign them the average of the ranks 1, 2, 3, and 4, that is, 2.5. Having used up the ranks 1, 2, 3, and 4, because 2 is the next set of tied observations, the three 2s are assigned the average of the ranks 5, 6, and 7, that is, 6. Having used up the ranks from 1 to 7, the remaining three unique observations get assigned the ranks 8, 9, and 10. When all is said and done, W, the sum of the positive ranks is:

W = 6 + 6 + 2.5 + 2.5 + 9 + 10 + 8 = 44

The statistical table for W:

tells us that the P-value is P(W ≥ 44) = 0.053. There is just barely not enough evidence, at the 0.05 level, to reject the null hypothesis. (I do imagine that a larger sample may have lead to a significant result.)

Alternatively, we could have completely ignored the condition of the data, that is, whether or not there were ties, and put the data into the black box of Minitab, getting:

We see that our analysis is consistent with the results from Minitab, namely 10 data points yield a Wilcoxon statistic of 44.0 and a P-value of 0.051 (round-off error).

Answer

Put your seatbelt on. If not your seatbelt, then at least your bookkeeping hat. This is going to get a bit messy at times. So... let's start with the easy part first... the denominator! If we combine the two samples, we have a total of 3+3 = 6 observations to arrange. Therefore, there are:

\(\binom{3+3}{3}=\binom{6}{3}=\dfrac{6!}{3! 3!}=20\)

possible ways of arranging the x's and y's. Now, that's a small enough number that we can actually enumerate all 20 of the possible arrangements. Here they are:

Now, as the above table suggests, of the 20 possible arrangements, there are two ways of getting R = 2 runs. Therefore:

\(P(R=2)=\frac{2}{20}\)

And, of the 20 possible arrangements, there are four ways of getting R = 3 runs. Therefore:

\(P(R=3)=\frac{4}{20}\)

Piece of cake! What was all this talk about messy bookkeeping? With the snap of a finger, we can determine the entire p.m.f. of R:

Note that the probabilities add to 1. That's a good sign! But finding the specific p.m.f. of R was not the point of this example. We were going to use this example to try to learn something about finding the p.m.f. of R in general. The only reason why we were able to find this p.m.f. of R with such ease was because we could enumerate all 20 of the possible outcomes. What if we can't do that? That is, what if there were so many possible arrangements that it would be crazy to try to enumerate them all? Well, we would go way back to what we did in Stat 414... we would derive a general counting formula! That's what we're working towards here.

Let's take a look at a case in which r is even, \(r = 4\), say. If you think about it, the only way we can get four runs is if the \(n_1 = 3\) values of X are broken up into 2 runs, which can be done in either of two ways:

and the \(n_2 = 3\) values of Y are broken up into 2 runs, which can be done in either of two ways:

Now, it's just a matter of putting the 2 runs of X and the 2 runs of Y together to make a total of 4 runs in the sequence. Well, there are two ways of getting 2 runs of X and two ways of getting 2 runs of Y. Therefore, the multiplication rule tells us that we should expect there to be 2×2 = 4 ways of getting four runs when the samples are combined. There's just one problem with that calculation, namely that there are 2 ways of starting the sequence off. That is, we could start with an X run:

or we could start with a Y run:

So, the multiplication rule actually tells us that there are \(2 \time 2 \times 2 = 8\) ways of getting four runs when we have \(n_1 = 3\) values of X and \(n_2 = 3\) values of Y. And, we've just enumerated all eight of them!

Now, let's try to generalize the above process. We started by considering a case in which r was even, specifically, r = 4. Note that r is even implies that \(r = 2k\), for a positive integer k. (If \(r = 4\), for example, then \(k = 2\).) Now, the only way we can get \(r = 4\) runs is if the \(n_1 = 3\) values of X are broken up into \(k = 2\) runs and the \(n_2 = 3\) values of Y are broken up into k = 2 runs. We can form the k = 2 runs of the \(n_1 = 3\) values of X by inserting \(k−1 = 1\) divider into the \(n_1 − 1 = 2\) spaces between the X values:

When the divider is here:

we create these two runs:

And, when the divider is here:

we create these two runs:

Note that, in general, there are:

\(\binom{n_1-1}{k-1}\)

ways of inserting k−1 dividers into the \(n_1 − 1\) spaces between the X values, with no more than one divider per space.

Now, we can go through the exact same process for the Y values. It shouldn't be too hard to see that, in general, there are:

\(\binom{n_2-1}{k-1}\)

ways of inserting k−1 dividers into the \(n_2 − 1\) spaces between the Y values, with no more than one divider per space.

Now, if you go back and look, once we broke up the X values into k = 2 runs, and the Y values into k = 2 runs, the next thing we did was to put the two sets of two runs of X and two sets of two runs of Y together. The multiplication rule tells us that there are:

\(\binom{n_1-1}{k-1}\binom{n_2-1}{k-1}\)

ways of putting the two sets of runs together to form \(r = 2k\) runs beginning with a run of x's. And, the multiplication rule tells us that there are:

\(\binom{n_2-1}{k-1}\binom{n_1-1}{k-1}\)

ways of putting the two sets of runs together to form \(r = 2k\) runs beginning with a run of y's. Therefore, the total number of ways of getting \(r = 2k\) runs is:

\(\binom{n_1-1}{k-1}\binom{n_2-1}{k-1}+\binom{n_2-1}{k-1}\binom{n_1-1}{k-1}=2\binom{n_1-1}{k-1}\binom{n_2-1}{k-1}\)

And, therefore putting the numerator and the denominator together, we get:

\(P(R=2k)=\frac{2\binom{n_1-1}{k-1}\binom{n_2-1}{k-1}}{\binom{n_1+n_2}{n_1}}\)

when k is a positive integer.

Whew! Now, we've found the probability of getting r runs when r is even. What happens if r is odd? That is, what happens if \(r = 2k+1\) for a positive integer k? Well, let's consider the case in which \(r = 3\), and therefore \(k = 1\). If you think about it, there are two ways we can get three runs... either we need \(k+1 = 2\) runs of x's and k = 1 run of y's, such as:

and

or we need \(k = 1\) run of x's and \(k+1 = 2\) runs of y's:

and

Perhaps you can see where this is going? In general, we can form \(k+1\) runs of the \(n_1\) values of X by inserting k dividers into the \(n_1−1\) spaces between the X values, with no more than one divider per space, in:

\(\binom{n_1-1}{k}\)

ways. Similarly, we can form k runs of the \(n_2\) values of Y in:

\(\binom{n_2-1}{k-1}\)

ways. The two sets of runs can be placed together to form \(2k+1\) runs in:

\(\binom{n_1-1}{k}\binom{n_2-1}{k-1}\)

ways. Likewise, \(k+1\) runs of the \(n_2\) values of Y, and k runs of the \(n_1\) values of X can be placed together to form:

\(\binom{n_2-1}{k}\binom{n_1-1}{k-1}\)

sets of \(2k+1\) runs. Therefore, the total number of ways of getting \(r = 2k+1\) runs is:

\(\binom{n_1-1}{k}\binom{n_2-1}{k-1}+\binom{n_2-1}{k}\binom{n_1-1}{k-1}\)

And, therefore putting the numerator and the denominator together, we get:

\(P(R=2k+1)=\frac{\binom{n_1-1}{k}\binom{n_2-1}{k-1}+\binom{n_2-1}{k}\binom{n_1-1}{k-1}}{\binom{n_1+n_2}{n_1}} \)

Yikes! Let's put this example to rest!

Answer

The combined ordered sample, with the x values in blue and the y values in red, looks like this:

Counting, we see that there are 9 runs. We should reject the null hypothesis if the number of runs is smaller than expected. Therefore, the critical region should be of the form r ≤ c. In order to determine what we should set the value of c to be, we'll need to know something about the p.m.f. of R. We can use the formulas that we derived above to determine various probabilities. Well, with \(n_1 = 8 , n_2 = 8 , \text{ and } k = 1\), we have:

\(P(R=2)=\dfrac{2\binom{7}{0}\binom{7}{0}}{\binom{16}{8}}=\dfrac{2}{12,870}=0.00016\)

and:

\(P(R=3)=\dfrac{\binom{7}{1}\binom{7}{0}+\binom{7}{1}\binom{7}{0}}{\binom{16}{8}}=\dfrac{14}{12,870}=0.00109\)

(Note that Table I in the back of the textbook can be helpful in evaluating the value of the "binomial coefficients.") Now, with \(n_1 = 8 , n_2 = 8 , \text{ and } k = 2\), we have:

\(P(R=4)=\dfrac{2\binom{7}{1}\binom{7}{1}}{\binom{16}{8}}=\dfrac{98}{12,870}=0.00761\)

and:

\(P(R=5)=\dfrac{\binom{7}{2}\binom{7}{1}+\binom{7}{2}\binom{7}{1}}{\binom{16}{8}}=\dfrac{294}{12,870}=0.02284\)

And, with \(n_1 = 8 , n_2 = 8 , \text{ and } k = 3\), we have:

\(P(R=6)=\dfrac{2\binom{7}{2}\binom{7}{2}}{\binom{16}{8}}=\dfrac{882}{12,870}=0.06853\)

Let's stop to see what we have going for us so far. Well, so far we've learned that:

\(P(R \le 6)=0.00016+0.00109+0.00761+0.02284+0.06853=0.1002\)

Hmmm. That tells us that if we set c = 6, we'd have a 0.1002 probability of committing a Type I error. That seems reasonable! That is, let's decide to reject the null hypothesis of the equality of the two distribution functions if the number of observed runs \(r ≤ 6\). It's not... we observed 9 runs. Therefore, we fail to reject the null hypothesis at the 0.10 level. There is insufficient evidence at the 0.10 level to conclude that the distribution functions are not equal.

Answer

The combined ordered sample, with the x values in blue and the y values in red, looks like this:

Counting, we see that there are 9 runs. Using the normal approximation to R, with \(n_1 = 10\) and \(n_2 = 11\), we have:

\(\mu_R=\dfrac{2n_1n_2}{n_1+n_2}+1=\dfrac{2(10)(11)}{10+11}+1=11.476\)

and:

\( Var(R)=\dfrac{2n_1n_2(2n_1n_2-n_1-n_2)}{(n_1+n_2)^2(n_1+n_2-1)} =\dfrac{2(10)(11)[2(10)(11)-10-11]}{(10+11)^2(10+11-1)}=4.9637\)

Now, we observed r = 9 runs. Therefore, the approximate P-value, using a half-unit correction for continuity, is:

\(P \approx P\left[Z \le \dfrac{9.5-11.476}{\sqrt{4.9637}} \right]=P[Z \le -0.89]=0.187\)

We fail to reject the null hypothesis at the 0.05 level, because the P-value is greater than 0.05. There is insufficient evidence at the 0.05 level to conclude that the two distribution functions are unequal.

Answer

With just a bit of ordering work (or using Minitab or some other statistical software), we can readily determine that the median weight of the sampled paint cans is 67.9. Labeling each observed weight with either a U if the observed weight is greater than the median (67.9), or an L if the observed weight is less than the median (67.9), we get:

Counting, we see that we observe 7 runs. Because there is no specific direction given in the possible deviation from randomness, we should test for both trend and cyclic effects. That is, we should reject the null hypothesis if we have either too few or too many runs. That is, our rejection region should be of the form:

\({r : r ≤ c \text{ or } r ≥ c}\)

Because we have a sample size of N = 16, we'll use the p.m.f. of R with \(n_1 = 8\) and \(n_2 = 8\). Now, using the formulas with k = 1 and k = 8, we get:

\[P(R=2)=\dfrac{2\binom{7}{0}\binom{7}{0}}{\binom{16}{8}}=\dfrac{2\binom{7}{7}\binom{7}{7}}{\binom{16}{8}} =P(R=16)=\dfrac{2}{12,870} \]

Using the formulas with k = 1 and k = 7, we get:

\[P(R=3)=\dfrac{\binom{7}{1}\binom{7}{0}+\binom{7}{1}\binom{7}{0}}{\binom{16}{8}}=\dfrac{\binom{7}{7}\binom{7}{6}+\binom{7}{7}\binom{7}{6}}{\binom{16}{8}}=P(R=15)=\dfrac{14}{12,870} \]

And, using the formulas with k = 2 and k = 7, we get:

\[P(R=4)=\dfrac{2\binom{7}{1}\binom{7}{1}}{\binom{16}{8}}=\dfrac{2\binom{7}{6}\binom{7}{6}}{\binom{16}{8}} =P(R=14)=\dfrac{98}{12,870} \]

Therefore, the critical region \({r : r ≤ 4 \text{ or } r ≥ 14}\) would yield a test at a significance level of:

\[\alpha = P(R \le 4)+P(R \ge 14)=\dfrac{2+14+98}{12,870}+\dfrac{2+14+98}{12,870}=\dfrac{228}{12,870}=0.018\]

The 7 runs we observed does not fall in our proposed rejection region. Therefore, we fail to reject at the 0.018 significance level. There is insufficient evidence at the 0.018 level to conclude that the process has deviated from randomness.

Answer

As reported, the data are ordered, therefore the order statistics are \(y_1 = 0, y_2 = 1, y_3 = 2, y_4 = 2, y_5 = 4, y_6 = 6, y_7 = 6\) and \(y_8 = 7\). Therefore, using the definition of the empirical distribution function, we have:

\(F_n(x)=0 \text{ for } x < 0\)

and:

\(F_n(x)=\frac{1}{8} \text{ for } 0 \le x < 1\) and \(F_n(x)=\frac{2}{8} \text{ for } 1 \le x < 2\)

Now, noting that there are two 2s, we need to jump 2/8 at x = 2:

\(F_n(x)=\frac{2}{8}+\frac{2}{8}=\frac{4}{8} \text{ for } 2 \le x < 4\)

Then:

\(F_n(x)=\frac{5}{8} \text{ for } 4 \le x < 6\)

Again, noting that there are two 6s, we need to jump 2/8 at x = 6:

\(F_n(x)=\frac{5}{8}+\frac{2}{8}=\frac{7}{8} \text{ for } 6 \le x < 7\)

And, finally:

\(F_n(x)=\frac{7}{8}+\frac{1}{8}=\frac{8}{8}=1 \text{ for } x \ge 7\)

Plotting the function, it should look something like this then:

Answer

The probability density function of a Uniform(0, 2) random variable X, say, is:

\( f(x)=\frac{1}{2}\)

for 0 < x < 2. Therefore, the probability that X is less than or equal to x is:

\(P(X \le x) =\int_{0}^{x}\frac{1}{2}dt=\frac{1}{2}x\)

for 0 < x < 2, and we are interested in testing:

• the null hypothesis \(H_0 : F(x)=F_0(x)\) against
• the alternative hypothesis \(H_A : F(x) \ne F_0(x)\)

where F(x) is the (unknown) cdf from which our data were sampled, and

Now, in working towards calculating \(d_n\), we first need to order the eight data points so that \(y_1\le \cdots\le y_8\). The table below provides all the necessary values for finding the KS test statistic. Note that the empirical cdf satisfies \(F_n(y_k)=k/8\), for \(k=0,1,\ldots,8\).

The last two columns represent all the differences we need to check. The largest of these is \(d_8=0.145\). From the table below with \(\alpha=0.05\), the critical value is \(0.46\). So, we can not reject the claim that the data were sampled from Uniform(0,2).

Answer

We are interested in testing the null hypothesis, \(H_0:\) \(X\) is normally distributed with mean 120 and standard deviation 10, against the alternative hypothesis, \(H_A\): \(X\) is not normally distributed with mean 120 and standard deviation 10. Now, in working towards calculating \(d_n\), we again first need to order the ten data points so that \(y_1=105\), \(y_2=108\), etc. Then, we need to calculate the value of the hypothesized distribution function \(F_0(y_k)\) at each of the values of \(y_k\). The standard normal table can help us do this. The probability that \(X\) is less than or equal to 105, for example, equals the probability that \(Z\) is less than or equal to \(-1.5\):

\(F_0(y_1)=P(X\le105)=P\left(Z\le{105-120\over10}\right)=P(Z\le-1.5)=.0668.\)

The table below summarizes the relevant quantities for finding the KS test statistic. Note that the empirical cdf satisfies \(F_n(y_k)=k/10\), except at \(k=5\) because of the tie: \(y_5=y_6=113\).

The last two columns represent all the differences we need to check. The largest of these is 0.3580. From the table below with \(\alpha=0.10\), the critical value is 0.37. Because \(d_{10} = 0.358\), which is just barely less than 0.37, we do not reject the null hypothesis at the 0.10 level. There is not enough evidence at the 0.10 level to reject the assumption that the data were randomly sampled from a normal distribution with mean 120 and standard deviation 10.

Answer

As before, we start by ordering the x values. The formulas for the lower and upper confidence limits tell us that we need to know d and \(F_n (x)\) for each of the 10 data points. Because the \(\alpha\)-level is 0.05 and the sample size n is 10, the table of Kolmogorv-Smirnov Acceptance Limits in the back of our text book, that is, Table VIII, tells us that \(d = 0.41\). We already calculated \(F_n (x)\) in the previous example.

Now, in calculating the lower limit, \(F_L(x) = F_n (x) − d = F_n (x)−0.41\), we see that the lower limit would be negative for the first four data points:

0.1−0.41 = −0.31 and 0.2−0.41 = −0.21 and 0.3−0.41 = −0.11 and 0.4−0.41 = −0.01

Therefore, we assign the first four data points a lower limit of 0, and then just calculate the remaining lower limit values. Similarly, in calculating the upper limit, \(F_U (x) = F_n (x) + d = F_n (x) + 0.41\), we see that the upper limit would be greater than 1 for the last six data points:

0.6+0.41 = 1.01 and 0.7+0.41 = 1.11 and 0.8+0.41 = 1.21 and 0.9+0.41 = 1.31 and 1.0+0.41 = 1.41

Therefore, we assign the last six data points an upper limit of 1, and then just calculate the remaining upper limit values. To summarize,

The last two columns together give us the 95% confidence band for the unknown cumulative distribution function \(F(x)\).