Question 11.5: Hospitals and Infections A researcher wishes to see if there......

Step 1 State the hypothesis and identify the claim.

H_{0} : The number of infections is independent of the hospital.

H_{1} : The number of infections is dependent on the hospital (claim).

Step 2 Find the critical value. The critical value using Table G at \alpha=0.05 with (3-1)(3-1)=(2)(2)=4 degrees of freedom is 9.488.

Step 3 Compute the test value. First find the expected values.

\begin{array}{lll}E_{1,1}=\frac{(119)(246)}{582}=50.30 & E_{1,2}=\frac{(119)(136)}{582}=27.81 & E_{1,3}=\frac{(119)(200)}{582}=40.89 \\E_{2,1}=\frac{(79)(246)}{582}=33.39 & E_{2,2}=\frac{(79)(136)}{582}=18.46 & E_{2,3}=\frac{(79)(200)}{582}=27.15 \\E_{3,1}=\frac{(384)(246)}{582}=162.31 & E_{3,2}=\frac{(384)(136)}{582}=89.73 & E_{3,3}=\frac{(384)(200)}{582}=131.96\end{array}

The completed table is shown.

Then substitute in the formula and evaluate to find the test statistic value.

\begin{aligned}\chi^{2}= & \sum \frac{(O-E)^{2}}{E} \\= & \frac{(41-50.30)^{2}}{50.30}+\frac{(27-27.81)^{2}}{27.81}+\frac{(51-40.89)^{2}}{40.89} \\& +\frac{(36-33.39)^{2}}{33.39}+\frac{(3-18.46)^{2}}{18.46}+\frac{(40-27.15)^{2}}{27.15} \\& +\frac{(169-162.31)^{2}}{162.31}+\frac{(106-89.73)^{2}}{89.73}+\frac{(109-131.96)^{2}}{131.96} \\= & 1.719+0.024+2.500+0.204+12.948+6.082 \\& +0.276+2.950+3.995 \\= & 30.698\end{aligned}

Step 4 Make the decision. The decision is to reject the null hypothesis since 30.698>9.488. That is, the test value lies in the critical region, as shown in Figure 11-7.

Step 5 Summarize the results. There is enough evidence to support the claim that the number of infections is related to the hospital where they occurred.