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

a. State appropriate hypotheses for a chi-square test.

b. Here is Minitab output for a chi-square test. Interpret the P-value. What conclusion would you draw?

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

Step by step solution

01

Part (a) Step 1 : Given information

We have to determine the state the null and alternative hypotheses.

02

Part (a) Step 2 : Simplification

The following are the null and alternate hypotheses:
$H_0$: The rates of improvement in the two therapy groups are identical.

$H_a$:The pace of progress differs between the two therapy groups.

03

Part (b) Step 1 : Given information

We have to state the conclusion.

04

Part (b) Step 2 : Simplification

The p-value for this study is $0.570$. As a result, we may claim that there is a $57$percent chance of getting the sample outcomes, or even more severe, when the improvement rates of the two treatment groups are identical.

Decision: If the P-value is greater than $0.05$, $H_0$is not rejected.

Conclusion: There is no persuasive evidence that the two therapy groups have different improvement rates.

05

Part (c) Step 1 : Given information

We have to compare chi-square test and z-test.

06

Part (c) Step 2 : Simplification

The p-value for this study is $0.570$. The chi-square test and the z-test both have the same p-value. As a result, the conclusions based on the two tests are the same.

Decision: If the P-value is greater than $0.05$, $H_o$is not rejected.

In addition, the chi-square test statistic is $0.322$, and the z-test statistic is $-0.57$.

The property that the square of z becomes the chi-square is well-known.

As a result,

$z^2={(-0.57)}^2=0.322={\chi }^2$

Thispropertyhasbeenmet.

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