Question 13.5: Shoplifting Incidents In a large department store, the owner......

Step 1 State the hypotheses and identify the claim.

H_{0} : There is no difference in the number of shoplifting incidents before and after the increase in security.

H_{1} : There is a difference in the number of shoplifting incidents before and after the increase in security (claim).

Step 2 Find the critical value from Table K because n \leq 30. Since n=7 and \alpha=0.05 for this two-tailed test, the critical value is 2 . See Figure 13-2.

Step 3 Find the test value.

a. Make a table as shown.

\begin{array}{|l|c|c|c|c|c|c|} \hline Day & Before, \boldsymbol{X}_{\boldsymbol{B}} & After, \boldsymbol{X}_{\boldsymbol{A}} & \begin{array}{c}\text { Difference } \\ \boldsymbol{D}=\boldsymbol{X}_{\boldsymbol{B}}-\boldsymbol{X}_{\boldsymbol{A}}\end{array} & \begin{array}{c}\text { Absolute } \\ \text { value } \boldsymbol{D} \mid\end{array} & Rank & \begin{array}{c}\text { Signed } \\ \text { rank }\end{array} \\ \hline Mon. & 7 & 5 & & & & \\ Tues. & 2 & 3 & & & & \\ Wed. & 3 & 4 & & & & \\ Thurs. & 6 & 3 & & & & \\ Fri. & 5 & 1 & & & & \\ Sat. & 8 & 6 & & & & \\ Sun. & 12 & 4 & & & & \\ \hline \end{array}

b. Find the differences (before minus after), and place the values in the Difference column.

\begin{aligned}& 7-5=2 \quad 6-3=3 \quad 8-6=2 \\& 2-3=-1 \quad 5-1=4 \quad 12-4=8 \\& 3-4=-1\end{aligned}

c. Find the absolute value of each difference, and place the results in the Absolute value column. (Note: The absolute value of any number except 0 is the positive value of the number. Any differences of 0 should be ignored.)

\begin{array}{rlr}|2|=2 & |3|=3 & |2|=2 \\|-1|=1 & |4|=4 & |8|=8 \\|-1|=1 & &\end{array}

d. Rank each absolute value from lowest to highest, and place the rankings in the Rank column. In the case of a tie, assign the values that rank plus 0.5.

\begin{array}{|l|l|l|l|l|l|l|l|} \hline \text{Value }& 2 & 1 & 1 & 3 & 4 & 2 & 8 \\ \hline \text{Rank }& 3.5 & 1.5 & 1.5 & 5 & 6 & 3.5 & 7 \\ \hline \end{array}

e. Give each rank a plus or minus sign, according to the sign in the Difference column. The completed table is shown here.

\begin{array}{|l|c|c|c|c|c|c|} \hline Day & Before, \boldsymbol{X}_{\boldsymbol{B}} & After, \boldsymbol{X}_{\boldsymbol{A}} & \begin{array}{c}\text { Difference } \\ \boldsymbol{D}=\boldsymbol{X}_{\boldsymbol{B}}-\boldsymbol{X}_{\boldsymbol{A}}\end{array} & \begin{array}{c}\text { Absolute } \\ \text { value }|\boldsymbol{D}|\end{array} & Rank & \begin{array}{c}\text { Signed } \\ \text { rank }\end{array} \\ \hline Mon. & 7 & 5 & 2 & 2 & 3.5 & +3.5 \\ Tues. & 2 & 3 & -1 & 1 & 1.5 & -1.5 \\ Wed. & 3 & 4 & -1 & 1 & 1.5 & -1.5 \\ Thurs. & 6 & 3 & 3 & 3 & 5 & +5 \\ Fri. & 5 & 1 & 4 & 4 & 6 & +6 \\ Sat. & 8 & 6 & 2 & 2 & 3.5 & +3.5 \\ Sun. & 12 & 4 & 8 & 8 & 7 & +7 \\ \hline \end{array}

f. Find the sum of the positive ranks and the sum of the negative ranks separately.

Positive rank sum \quad(+3.5)+(+5)+(+6)+(+3.5)+(+7)=+25 Negative rank sum \quad(-1.5)+(-1.5) \quad=-3

g. Select the smaller of the absolute values of the sums (|-3|), and use this absolute value as the test value w_{s}. In this case, w_{s}=|-3|=3.

Step 4 Make the decision. Reject the null hypothesis if the test value is less than or equal to the critical value. In this case, 3>2; hence, the decision is to not reject the null hypothesis.

Step 5 Summarize the results. There is not enough evidence at \alpha=0.05 to support the claim that there is a difference in the number of shoplifting incidents before and after the increase in security. Hence, the security increase probably made no difference in the number of shoplifting incidents.