Question 11.1: Arrests for Crimes Is there enough evidence to reject the cl......
Arrests for Crimes
Is there enough evidence to reject the claim that the number of arrests for each category of crimes is the same? Use α = 0.05.
Step 1 State the hypotheses and identify the claim.
H_{0} : There is no difference in the number of arrests for each type of crime. (claim)
H_{1} : There is a difference in the number of arrests for each type of crime.
Step 2 Find the critical value. The degrees of freedom are 4-1=3, and at \alpha=0.05 the critical value from Table G in Appendix A is 7.815.
Step 3 Compute the test value. Note the expected values are found by E=n / k=160 / 4=40.
The table looks like this.
\begin{array}{|c|c|c|c|c|} \hline & \begin{gathered}\text { Larceny } \\ \text { thefts }\end{gathered} & \begin{gathered}\text { Property } \\ \text { crimes }\end{gathered} & \begin{gathered}\text { Drug } \\ \text { use }\end{gathered}& \begin{gathered}\text { Driving under } \\ \text { the influence }\end{gathered} \\ \hline \text{Observed }& 38 & 50 & 28 & 44 \\ \text{Expected }& 40 & 40 & 40 & 40 \\ \hline \end{array}
The test value is computed by subtracting the expected value corresponding to the observed value, squaring the result, and dividing by the expected value. Then find the sum of these values.
\begin{aligned}\chi^{2} & =\sum \frac{(0-E)^{2}}{E}=\frac{(38-40)^{2}}{40}+\frac{(50-40)^{2}}{40}+\frac{(28-40)^{2}}{40}+\frac{(44-40)^{2}}{40} \\& =0.1+2.5+3.6+0.4 \\& =6.6\end{aligned}
Step 4 Make the decision. The decision is to not reject the null hypothesis since 6.6<7.815, as shown in Figure 11-1.
Step 5 Summarize the results. There is not enough evidence to reject the claim that there is no difference in the number of arrests for each type of crime.
