Chapter 13.6, Problem 28E

To determine

To test: Whether the sequence of the people who were tested for drivers wear glasses or contact lenses is random.

Answer to Problem 28E

There is sufficient evidence to infer that the sequence of the people who were tested for drivers wear glasses or contact lenses is random.

Explanation of Solution

Given info:

The data represents the sequence of the people who were tested for drivers who wear glasses (G) or contact lenses(C). The level of significance is $\alpha =0.05$.

Calculation:

The testing hypotheses are given below:

Null hypothesis:

$H_0:$ The sequence of the people who were tested for drivers is random.

Alternative hypothesis:

$H_1:$ The sequence of the people who were tested for drivers is not random.

Critical value:

Here, the test is two tailed test.

For the level of significance $\alpha =0.05$,

From the data it can be seen that there are 31 C’s and 19 G’s.

That is, the sizes are $n_1=31$ and $n_2=19$.

From Table E, The Standard Normal Distribution, the critical value that corresponds to the level of significance $\alpha =0.05$ and $\pm 1.96$.

Thus, the critical value is $\pm 1.96$.

Test statistic:

The number of runs for the given sequence is obtained as follows:

The list of the runs is,

Run	Letters
1	G
2	C,C,C
3	G
4	C,C
5	G,G
6	C,C,C
7	G,G
8	C
9	G
10	C,C
11	G,G
12	C,C,C
13	G,G
14	C,C
15	G
16	C
17	G
18	C,C,C,C
19	G
20	C
21	G,G,G,G
22	C,C,C,C,C,C,C,C,C
23	G

Here, the number of runs is $G=23$.

The mean number of runs is,

$\begin{array}{c}{\mu }_G=\frac{2n_1{n}_2}{n_1+n_2}+1\\ =\frac{2\times 31\times 19}{31+19}+1\\ =\frac{1,178}{50}+1\\ =24.56\end{array}$

The standard deviation of runs is,

$\begin{array}{c}{\sigma }_G=\sqrt{\frac{2n_1{n}_2\left(2n_1{n}_2-n_1-n_2\right)}{{\left(n_1+n_2\right)}^2\left(n_1+n_2-1\right)}}\\ =\sqrt{\frac{\left(2\times 31\times 19\right)\left(\left(2\times 31\times 19\right)-31-19\right)}{{\left(31+19\right)}^2\left(31+19-1\right)}}\\ =\sqrt{\frac{1,178\times 1,128}{2,500\times 49}}\\ =\sqrt{\frac{1,328,784}{122,500}}\end{array}$

     $\begin{array}{c}=\sqrt{10.8472}\\ =3.2935\end{array}$

The test statistic value is,

$\begin{array}{c}z=\frac{G-{\mu }_G}{{\sigma }_G}\\ =\frac{23-24.56}{3.2935}\\ =\frac{-1.56}{3.2935}\\ =-0.4736\end{array}$

Hence, the test value is $z=-0.4736$

Decision rule:

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

Conclusion:

The test value is, $z=-0.4736$ and the critical value is -1.96.

Here, the negative test statistic value is greater than the negative critical value.

That is, $-0.4736\left(=z\right)>-1.96\left(=\text{criticalvalue}\right)$

By the rejection rule “failed to reject the null hypotheses”.

Thus, it can be concluded that there is enough evidence to support the claim that the sequence of the people who were tested for drivers wear glasses or contact lenses is random.