Chapter 13, Problem 13.6.14RE

a.

To determine

To state: The hypotheses and identify the claim.

Answer to Problem 13.6.14RE

The null hypothesis is, the typical book purchased occur at random.

The alternative hypothesis is, the typical book purchased does not occur at random.

The claim is that, the typical book purchased at random.

Explanation of Solution

Given info:

The data shows Fiction or Nonfiction Books, bookstore owner records the 48 books purchased by customers.

Calculation:

The hypotheses are given below:

Null hypothesis

$H_0:$ The typical book purchased occurs at random.

Alternative hypothesis

$H_1:$ The typical book purchased does not occur at random.

Here, the typical book purchased at random. Hence, the claim is that the typical book purchased occurs at random.

b.

To determine

To find: The critical value.

Answer to Problem 13.6.14RE

The critical value is $\pm 1.64$.

Explanation of Solution

Calculation:

The data represent the value for $\alpha$ is 0.10. The alternative is denoted as the two-tailed test.

It is clear that there are 29 M’s and 19 F’s. That is, $n_1=29,n_2=19$.

From Table E, The Standard Normal Distribution, the critical value for $\alpha =0.10$ and $\pm 1.64$.

Hence, the critical value is $\pm 1.64$.

c.

To determine

To find: The test value.

Answer to Problem 13.6.14RE

The test value is –3.35.

Explanation of Solution

Calculation:

The number of runs from the obtained sequence is,

Run	Letters
1	F, F, F, F, F, F
2	N, N
3	F
4	N, N, N, N
5	F, F, F, F, F, F, F
6	N, N, N
7	F
8	N, N, N, N, N
9	F, F, F
10	N, N, N
11	F, F, F, F, F, F, F, F, F
12	N, N
13	F, F

The number of runs is $G=13$.

The mean number of runs is,

$\begin{array}{c}{\mu }_G=\frac{2n_1{n}_2}{n_1+n_2}+1\\ =\frac{2\times 29\times 19}{29+19}+1\\ =\frac{1,102}{48}+1\\ =22.9583+1\end{array}$

$=23.9583$

The standard deviation of runs is,

$\begin{array}{c}{\sigma }_G=\sqrt{\frac{2n_1{n}_2\left(2n_1{n}_2-n_1-n_2\right)}{{\left(n_1+n_2\right)}^2\left(n_1+n_2-1\right)}}\\ =\sqrt{\frac{\left(2\times 29\times 19\right)\left(\left(2\times 29\times 19\right)-29-19\right)}{{\left(29+19\right)}^2\left(29+19-1\right)}}\\ =\sqrt{\frac{1,102\times 1,054}{2,304\times 47}}\\ =\sqrt{\frac{1,161,508}{108,288}}\end{array}$

     $\begin{array}{c}=\sqrt{10.7261}\\ =3.2751\end{array}$

The test statistic value is,

$\begin{array}{c}z=\frac{G-{\mu }_G}{{\sigma }_G}\\ =\frac{13-23.9583}{3.2751}\\ =\frac{-10.9583}{3.2751}\\ =-3.35\end{array}$

Hence, the test value is $z=-3.35$

d.

To determine

To make: The decision.

Answer to Problem 13.6.14RE

The decision is that, the null hypothesis $H_0$ is rejected.

Explanation of Solution

Decision Rule:

If the negative test value is less than the negative critical value, then reject the null hypothesis $H_0$.

Conclusion:

From the results, the critical value is –1.64, and the test value is –3.35.

Here, the test value is less than the critical values.

Therefore, by the rule, the null hypothesis $H_0$ is rejected.

e.

To determine

To summarize: The results.

Answer to Problem 13.6.14RE

The conclusion is that, there is no evidence to support the claim that the typical book purchased occur at random.

Explanation of Solution

From part (d), the null hypothesis is rejected. Hence, there is no evidence to support the claim that the typical book purchased occur at random.