Chapter 11.4, Problem 11P

(a)

To determine

State the null and alternative hypothesis.

Explanation of Solution

Calculation:

From the given information the value of $\alpha$ is 0.05, and to test the sequence for randomness.

The null and alternative hypothesis is,

Null hypothesis:

$H_0:$ The sequence of democratic and republican presidential terms is randomly mixed.

Alternative hypothesis:

$H_1:$ The sequence of democratic and republican presidential terms not randomly mixed.

(b)

To determine

Find the number of runs R.

Find the value of $n_1$.

Find the value of $n_2$.

Answer to Problem 11P

The number of runs R is 18.

The value of $n_1$ is 21.

The value of $n_2$ is 17.

Explanation of Solution

Calculation:

The number of runs (blocks) exist in sequence is represented using the random variable R.

There are 21 R’s and 17 D’s in the sequence. That is, $n_1=21,n_2=17$

The block for the sequence is,

$\boxed{DD}\boxed{RR}\boxed{D}\boxed{RRRR}\boxed{D}\boxed{R}\boxed{D}\boxed{RRRR}\boxed{DD}\boxed{RR}\boxed{DDDD}\boxed{RR}\boxed{DD}\boxed{RR}\boxed{D}\boxed{RRR}\boxed{DD}\boxed{R}$

There are 18 blocks in the given sequence.

Hence, the number of runs R is 18, the value of $n_1$ is 21, the value of $n_2$ is 17.

(c)

To determine

Find the sample test statistic R to z.

Answer to Problem 11P

The sample test statistic R to z is –0.60.

Explanation of Solution

Calculation:

The formula of z is,

$z=\frac{R-{\mu }_R}{{\sigma }_R}$

Where

$\begin{array}{c}{\mu }_R=\frac{2n_1{n}_2}{n_1+n_2}+1\\ {\sigma }_R=\sqrt{\frac{\left(2n_1{n}_2\right)\left(2n_1{n}_2-n_1-n_2\right)}{{\left(n_1+n_2\right)}^2\left(n_1+n_2-1\right)}}\end{array}$

The mean of R is,

$\begin{array}{c}{\mu }_R=\frac{2\left(21\right)\left(17\right)}{21+17}+1\\ =\frac{714}{38}+1\\ =\frac{752}{38}\\ =19.79\end{array}$

The standard deviation of R is,

$\begin{array}{c}{\sigma }_R=\sqrt{\frac{\left(2\times 21\times 17\right)\left(\left(2\times 21\times 17\right)-21-17\right)}{{\left(21+17\right)}^2\left(21+17-1\right)}}\\ =\sqrt{\frac{714\times 676}{1,444\times 37}}\\ =\sqrt{9.0339}\\ =3.01\end{array}$

The sample test statistic R to z is,

$\begin{array}{c}z=\frac{18-19.80}{3.01}\\ =\frac{-1.80}{3.01}\\ =-0.60\end{array}$

Hence, the sample test statistic R to z is –0.60.

(d)

To determine

Check whether the null hypothesis is rejected or failed to be rejected.

Find the P-value.

Check the conclusion by using P-value method and critical value method.

Answer to Problem 11P

The null hypothesis is failed to be rejected.

The P-value is 0.5486.

Explanation of Solution

Calculation:

Conclusion:

The sample test statistic R to z is –0.60.

The sample test statistic R to z is between –1.96 and 1.96.

That is, $-1.96\le -0.60\left(=z\right)\le 1.96$.

By the rejection rule, the null hypothesis is failed to be rejected.

Hence, the data is not statistically significant at level 0.05.

P-value:

Step by step procedure to obtain P-value using MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Both Tail, for the region of the curve to shade.
• Enter the X value as –0.60.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.2743 which is one sided value.

The two-tailed P-value is,

$\begin{array}{c}P\text{-value}=2\times 0.2743\\ =0.5486\end{array}$

Hence, the P-value is 0.5486.

Rejection rule:

• If the P-value is less than or equal to $\alpha$, then reject the null hypothesis and the test is statistically significant. That is, $P\text{-value}\le \alpha$.

Conclusion:

The P-value is 0.5486 and the level of significance is 0.05.

The P-value is greater than the level of significance.

That is, $0.5486\left(=P\text{-value}\right)>0.05\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is failed to be rejected.

Hence, the both the conclusions are same.

(e)

To determine

Interpret the conclusion in the context of the application.

Explanation of Solution

Calculation:

From part (d), the null hypothesis is failed to be rejected. This shows that, there is no sufficient evidence that the sequence of democratic and republican presidential terms not randomly mixed at level of significance 0.05.