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Chapter 11: Q. 43 (page 726)

Gastric freezing was once a recommended treatment for ulcers in the upper intestine. The use of gastric freezing stopped after experiments showed it had no effect. One randomized comparative experiment found that 28of the 82gastric-freezing patients improved, while 30of the 78patients in the placebo group improved. We can test the hypothesis of 鈥渘o difference鈥� in the effectiveness of the treatments in two ways: with a two-sample z test or with a chi-square test.

(a) Minitab output for a chi-square test is shown below. State appropriate hypotheses and interpret the P-value in context. What conclusion would you draw?

Chi-Square Test: Gastric freezing, Placebo Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts

(b) Minitab output for a two-sample z test is shown below. Explain how these results are consistent with the test in part (a).

Short Answer

Expert verified

(a) H0: There is no association between the variables

Ha: There is an association between the variables

If there is no association between the variables, then we have a probability of 57.0%of obtaining a similar or more extreme sample.

(b) We can deduce from this P-value that there is no difference in the population proportions, which is the case if the variables are unrelated.

As a result, the two-sample z-test and the chi-square test are equivalent, as seen by the similar P-values.

Step by step solution

01

Part (a) Step 1: Given information

The given data is

02

Part (b) Step 2: Explanation

The null hypothesis states that there is no difference between the two groups:

H0: There is no difference in the improvement rates for the two treatments.

The alternative hypothesis states that there is a difference between the two groups.

Ha: There is a difference in the improvement rates for the two treatments.

The P-value is given in the output as:

P=0.570=57.0%

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, if the null hypothesis is true.

If there is no association between the variables, then we have a probability of 57.0%of obtaining a similar or more extreme sample.

03

Part (b) Step 1: Given information

The data given is

04

Part (b) Explanation

Find the hypothesis

H0:p1-p2=0

H0:p1-p2鈮�0

The P-value is given in the output as:

P=0.570=57.0%

We can deduce from this P-value that there is no difference in the population proportions, which is the case if the variables are unrelated.

As a result, the two-sample z-test and the chi-square test are equivalent, as seen by the similar P-values.

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