Question 13.6: Hospital Infections A researcher wishes to see if the total ......
Hospital Infections
A researcher wishes to see if the total number of infections that occurred in three groups of randomly selected hospitals is the same. The data are shown in the table. At α = 0.05, is there enough evidence to reject the claim that the number of infections in the three groups of hospitals is the same?
\begin{array}{|c|c|c|} \hline \text{Group }A & \text{Group }B & \text{Group }C \\ \hline 557 & 476 & 105 \\ 315 & 232 & 110 \\ 920 & 80 & 167 \\ 178 & 116 & 155 \\ \hline \end{array}
Source: Pennsylvania Health Care Cost
Containment Council.
Step 1 State the hypotheses and identify the claim.
H_{0} : There is no difference in the number of infections in the three groups of hospitals (claim).
H_{1} : There is a difference in the number of infections in the three groups of hospitals.
Step 2 Find the critical value. Use the chi-square table (Table G) with d.f. =k-1, where k= the number of groups. With \alpha=0.05 and d.f. =3-1=2, the critical value is 5.991.
Step 3 Compute the test value.
a. Arrange all the data from the lowest value to the highest value and rank each value.
\begin{array}{|c|c|c|} \hline \text{Amount }& \text{Group }& \text{Rank }\\ \hline 80 & \mathrm{~B} & 1 \\ 105 & \mathrm{C} & 2 \\ 110 & \mathrm{C} & 3 \\ 116 & \mathrm{~B} & 4 \\ 155 & \mathrm{C} & 5 \\ 167 & \mathrm{C} & 6 \\ 178 & \mathrm{~A} & 7 \\ 232 & \mathrm{~B} & 8 \\ 315 & \mathrm{~A} & 9 \\ 476 & \mathrm{~B} & 10 \\ 557 & \mathrm{~A} & 11 \\ 920 & \mathrm{~A} & 12 \\ \hline \end{array}
b. Find the sum of the ranks for each group.
\begin{array}{ll}\text { Group A } & 7+9+11+12=39 \\\text { Group B } & 1+4+8+10=23 \\\text { Group C } & 2+3+5+6=16\end{array}
c. Substitute in the formula.
H=\frac{12}{N(N+1)}\left(\frac{R_{1}^{2}}{n_{1}}+\frac{R_{2}^{2}}{n_{2}}+\frac{R_{3}^{2}}{n_{3}}\right)-3(N+1)
where
\begin{array}{llll}N & =12 \quad R_{1}=39 & R_{2}=23 & R_{3}=16 \\n_{1} & =n_{2}=n_{3}=4 & &\end{array}
Therefore,
\begin{aligned}H & =\frac{12}{12(12+1)}\left(\frac{39^{2}}{4}+\frac{23^{2}}{4}+\frac{16^{2}}{4}\right)-3(12+1) \\& =5.346\end{aligned}
Step 4 Make the decision. Since the test value of 5.346 is less than the critical value of 5.991, the decision is to not reject the null hypothesis.
Step 5 Summarize the results. There is not enough evidence to reject the claim that there is no difference in the number of infections in the groups of hospitals. Hence, the differences are not significant at \alpha=0.05.