Chapter 8.4, Problem 5E

To determine

To test: Whether there is any sufficient evidence to infer that the proportion of motor vehicle accidents that are due to improper driving differs from the proportion of accidents involve improper driving in 2009 $p=0.544$.

Answer to Problem 5E

There is no enough evidence to infer that the proportion of motor vehicle accidents that are due to improper driving differs from the proportion of accidents involve improper driving in 2009 $p=0.544$.

Explanation of Solution

Given info:

In 2009 the percentage of motor vehicle accidents that do not involve improper driving is 45.6%. To examine this 110 motor vehicle accidents are randomly selected among which 60 accidents involved improper driving. The level of significance is $\alpha =0.10$.

Calculation:

Here, given that the proportion of motor vehicle accidents that do not involve improper driving is 0.456.

That is, $\left(1-p\right)=0.456$

From this, the proportion of motor vehicle accidents that involve motor vehicle accidents is:

$\begin{array}{c}p=\mathrm{1.0.456}\\ =0.544\end{array}$

Thus, the population proportion is $p=0.544$.

The testing hypotheses are given below:

Null hypothesis $H_0$:

$H_0:p=0.544$

Alternative hypothesis $H_1$:

$H_1:p\ne 0.544$

Point estimate:

Here, the sample size is $n=110$.

Among a random sample of 110 motor vehicle accidents, 60 accidents involved improper driving.

Hence, the specified number of characteristics is $x=60$.

The proportion of burglaries due to open or unlocked door is obtained as,

$\begin{array}{c}\widehat{p}=\frac{x}{n}\\ =\frac{60}{110}\\ =0.5454\end{array}$

Thus, the proportion of burglaries due to open or unlocked door is $\widehat{p}=0.5454$.

Requirements for large sample-test about proportion:

• The sample must be drawn randomly from the desired population.
• The sample must be large, that is $np\ge 5$ and $nq\ge 5$.

Here, the sample is drawn randomly.

Check for the large sample:

(i)

$\begin{array}{l}n\times \widehat{p}\\ =110\times 0.5454\\ =59.994\\ >5\end{array}$

(ii)

$\begin{array}{l}n\times \left(1-\widehat{p}\right)\\ =110\times \left(1-0.5454\right)\\ =50.006\\ >5\end{array}$

Hence, the sample is large.

Thus, z-test about proportion is valid in this case.

Decision rule based on classical approach:

If $\mathrm{mod}(z)>z_{\frac{\alpha }{2}}$, then reject the null hypothesis $H_0$.

Critical value:

For the level of significance,

$\begin{array}{c}\alpha =0.10\\ \frac{\alpha }{2}=\frac{0.10}{2}\\ =0.05\end{array}$

Hence, the cumulative area to the left is,

$\begin{array}{c}\text{Area}\text{to}\text{the}\text{left}=1-\text{Area}\text{to}\text{the}\text{right}\\ =1-0.05\\ =0.95\end{array}$

From Table E of the standard normal distribution from Appendix A, the critical value is 1.645.

Thus, the two tailed critical value of z for the level of significance $\alpha =0.10$ is $\left(z_{\alpha }\right)=1.645$.

Test statistic:

The test statistic for the large sample single proportion test is,

$z=\frac{\left(\widehat{p}-p\right)}{\sqrt{\frac{p\times \left(1-p\right)}{n}}}$

The test statistic is obtained as follows:

Here, $p=0.544$ $n=110$ and $\widehat{p}=0.5454$

$\begin{array}{c}z=\frac{\left(\widehat{p}-p\right)}{\sqrt{\frac{p\times \left(1-p\right)}{n}}}\\ =\frac{0.5454-0.544}{\sqrt{\frac{0.544\times \left(1-0.544\right)}{110}}}\\ =\frac{0.0014}{0.4745}\\ =0.003\end{array}$

Thus, the test statistic is $z=0.003$.

Conclusion based on classical approach:

The test statistic value is $z=0.003$ and the critical value is $\left(z_{\alpha }\right)=1.645$.

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

That is, $0.003\left(=z\right)<1.645\left(=z_{\alpha }\right)$

Hence, failed to reject the null hypothesis $H_0$.

Thus, it can be concluded that there no enough evidence to infer that the proportion of motor vehicle accidents that are due to improper driving differs from the proportion of accidents involve improper driving in 2009 $p=0.544$.