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1Neutropenia. Neutropenia is an abnormally low number o neutrophils (a type of white blood cell) in the blood. Chemotherapy often reduces the number of neutrophils to a level that makes the patient susceptible to fever and infections. G. Bucaneve et al. published a study of such cancer patients in the paper "Levofloxacin to Prevent Bacterial Infection in Patients With Cancer and Neutropenia" (New England Journal of Medicine, Vol. 353, No, 10, pp. 977-987). For the study. 375 patients were randomly assigned to receive a daily dose of levofloxacin, and 363 were given a placebo. In the group receiving levofloxacin, fever was present in 243 patients for the duration of neutropenia, whereas fever was experienced by 308 patients in the placebo group.

a. At the1% significance level, do the data provide sufficient evidence to conclude that levofloxacin is effective in reducing the occurrence of fever in such patients?

b. Find a98% confidence level for the difference in the proportions of such cancer patients who would experience fever for the duration of neutropenia

Step by step solution

01

Part(a) Step 1: Given Information

The given values are,

$$\begin{gathered} x_1=243, \\ {n}_1=375, \\ {x}_2=308, \\ {n}_2=363, \\ \mathrm{Î\pm }=0.01 \end{gathered}$$

02

Part(a) Step 2: Explanation

The null hypothesis.

$H_0:p_1=p_2$

The alternative hypothesis.

$H_a:p_1<p_2$

Using MINITAB output.

MINITAB output: Test and Cl two proportions

From the MINITAB, the test statistic is $-6.26$and the $p$- value is $0.000$.

$p$- value is lesser than the level of significance.

$p\text{- value}(=0.000)<\mathrm{Î\pm }(=0.01)$

Using the rejection rule, it can be concluded that there is evidence to reject the null hypothesis $H_0$at $\mathrm{Î\pm }=0.001$. Therefore, the data provide sufficient evidence to conclude that levofloxacin is effective in reducing the occurrence of fever in such patients.

03

Part (b) Step 1: Given Information

The given values are,

$$\begin{gathered} x_1=243, \\ {n}_1=375, \\ {x}_2=308, \\ {n}_2=363, \\ \mathrm{Î\pm }=0.01 \end{gathered}$$

04

Part(b) Step 2: Explanation

Using MINITAB output.

Using the MINITAB output.

MINITAB output: Test and CL for two proportions

From the MINITAB output,$98\%$ confidence interval is $(-0.2727,-0.1283)$

Therefore, there is$98\%$ confidence interval for the difference between the proportion is $(-0.2727,-0.1283)$.

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