Chapter 8.4, Problem 9E

To determine

To test: Whether there is any sufficient evidence to infer that the proportion of burglaries that are due to open or unlocked door differs from the proportion 0.3.

Answer to Problem 9E

There is no enough evidence to infer that the proportion of burglaries that are due to open or unlocked door differs from the proportion 0.3.

Explanation of Solution

Given info:

The statement was that about 30% of the burglaries are due to open or unlocked door. To test this, a random sample of 130 burglaries are tested and 85 were not due to open or unlocked door. The level of significance is $\alpha =0.05$.

Calculation:

The testing hypotheses are given below:

Null hypothesis $H_0$:

$H_0:p=0.3$

Alternative hypothesis $H_1$:

$H_1:p\ne 0.3$

Point estimate:

Here, the sample size is $n=130$.

Among a random sample of 130 burglaries, 85 were not due to open or unlocked door.

And the number of burglaries due to open or unlocked door is $200-85=45$

Hence, the specified number of characteristics is $x=45$.

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

$\begin{array}{c}\widehat{p}=\frac{x}{n}\\ =\frac{45}{130}\\ =0.3462\end{array}$

Thus, the proportion of burglaries due to open or unlocked door is $\widehat{p}=0.3462$.

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)

$\begin{array}{l}n\times \widehat{p}\\ =130\times 0.3462\\ =45.006\\ >5\end{array}$

(ii)

$\begin{array}{l}n\times \left(1-\widehat{p}\right)\\ =130\times \left(1-0.3462\right)\\ =84.994\\ >5\end{array}$

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,

$\begin{array}{c}\alpha =0.05\\ \frac{\alpha }{2}=\frac{0.05}{2}\\ =0.025\end{array}$

Hence, the cumulative area to the left is,

$\begin{array}{c}\text{Area}\text{to}\text{the}\text{left}=1-\text{Area}\text{to}\text{the}\text{right}\\ =1-0.025\\ =0.975\end{array}$

From Table E of the standard normal distribution from Appendix A, the critical value is 1.96.

Thus, the two tailed critical value of z for the level of significance $\alpha =0.05$ is $\left(z_{\alpha }\right)=1.96$.

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, $p=0.3$ $n=130$ and $\widehat{p}=0.3462$

$\begin{array}{c}z=\frac{\left(\widehat{p}-p\right)}{\sqrt{\frac{p\times \left(1-p\right)}{n}}}\\ =\frac{0.3-0.3462}{\sqrt{\frac{0.3\times \left(1-0.3\right)}{130}}}\\ =\frac{-0.0462}{0.0401}\\ =-1.15\end{array}$

Thus, the test statistic is $\mathrm{mod}(z)=1.15$.

Conclusion based on classical approach:

The test statistic value is $z=1.15$ and the critical value is $\left(z_{\alpha }\right)=1.96$.

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

That is, $1.15\left(=z\right)<1.96\left(=z_{\alpha }\right)$

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

Thus, it can be concluded that there is no enough evidence to infer that the proportion of burglaries that are due to open or unlocked door differs from the proportion 0.3.