Question

A medical journal reported the results of a study in which three groups of 50 women were randomly selected and monitored for urinary tract infections over 6 months. One group drank cranberry juice​ daily, one group drank a lactobacillus​ drink, and the third group drank neither of those​ beverages, serving as a control group. In the control​ group, 18 women developed at least one infection compared with 19 of those who consumed the lactobacillus drink and only 7 of those who drank cranberry juice. Does the study provide supporting evidence for the value of cranberry juice in warding off urinary tract infections in​ women? Complete parts a through f below.

f) If you concluded that the groups are not the​ same, analyze the differences using the standardized residuals of your calculations. Select the correct choice​ below, and if​ necessary, fill in the answer box to complete your choice.

A.

		Beverage Infection No Infection Cranberry Juice Lactobacillus Control
		​(Round to four decimal places as​ needed.)

B. The conclusion does not indicate that the groups are different.

C. Since the assumptions are not​ satisfied, a hypothesis test is not appropriate.

Answer 1

Solution:-

A)

Beverage	Infection	No Infection
Cranberry juice	-	-
Lactobacillus	7	12
Control	1	17

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P₁> P₂
Alternative hypothesis: P_Control < P_{Lactobacillus}

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)
p = 0.21622
SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.1354

z = (p₁ - p₂) / SE

z = - 2.31

where p₁ is the sample proportion in sample 1, where p₂ is the sample proportion in sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.

Since we have a one-tailed test, the P-value is the probability that the z-score is less than -2.31.

Thus, the P-value = 0.01.

Interpret results. Since the P-value (0.01) is less than the significance level (0.05), we have to reject the null hypothesis.