Chapter 26, Problem 27E

a)

To determine

To explain whether the survey is retrospective, prospective or an experiment.

Answer to Problem 27E

It is an experiment.

Explanation of Solution

Experimental study: Assigning people or things to groups and applying some treatment to one group, while the other group does not receive the treatment.

Therefore, in given survey, treatment was inflicted on the subjects. Hence, this survey is an experiment.

b)

To determine

To find which test is appropriate.

Answer to Problem 27E

Chi-square test of Homogeneity.

Explanation of Solution

When we want to check whether the distribution of one variable is the same in multiple group then we use Chi-square test of Homogeneity.Therefore, here research wants to check the rate of urinary infections is same for all the three groups. Hence,Chi-square test of Homogeneity is appropriate.

c)

To determine

To state null and alternative hypotheses.

Answer to Problem 27E

  $H_0$ : The rate of urinary infections is same for all the three groups.

  $H_a$ : The rate of urinary infections is different among the groups.

Explanation of Solution

The null and alternative hypotheses:

  $H_0$ : The rate of urinary infections is same for all the three groups.

  $H_a$ : The rate of urinary infections is different among the groups.

d)

To determine

To check the conditions.

Answer to Problem 27E

All the conditions satisfied.

Explanation of Solution

There are three main conditions of Chi-square test:

Counts, Independent and expected counts are at least 5.

The data has given counts so the condition of counts satisfied.

Assuming that subjects were randomly assigned to groups, so condition of independent is satisfied.

The smallest expected count is

  $\frac{50\times 46}{150}=15.3333$

Which is larger than 5 so all the conditions satisfied.

e)

To determine

To find the degrees of freedom.

Answer to Problem 27E

The degrees of freedom = 2

Explanation of Solution

Formula:

  $df=(r-1)(c-1)$

The number of rows = r = 3

The number of columns = c = 2

Therefore, the degrees of freedom are:

  $df=(3-1)\times (2-1)=2$

f)

To determine

To find ${\chi }^2$ and P-value.

Answer to Problem 27E

  ${\chi }^2$ = 7.78 and P-value = 0.0205

Explanation of Solution

Using excel,

  

Therefore,

  ${\chi }^2$ = 7.78

P-value = 0.0205

g)

To determine

To state the conclusion.

Answer to Problem 27E

There is sufficient evidence to reject the claim of same proportions in all the three groups.

Explanation of Solution

Given:

  ${\chi }^2$ = 7.78

P-value = 0.0205

Decision: The p-value< 0.05, reject H0.

Conclusion: There is sufficient evidence to reject the claim of same proportions in all the three groups.

h)

To determine

To analyze the residuals.

Explanation of Solution

  

The negative value indicates that those who drank cranberry juice were less likely to develop urinary tract infections. The positive value indicates that those who drank lactobacillus drink were more likely to develop urinary tract infections.