Q. 43 - The Practice of Statistics for AP

Question

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 $28$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) 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). 

Step-by-Step Solution

Answer

(a) $H_0$: There is no association between the variables

$H_a$: 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.

Part (a) Step 1: Given information

The given data is 

Part (b) Step 2: Explanation

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

$H_0$ : 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.

$H_a$ : 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.

Part (b) Step 1: Given information

The data given is

Part (b) Explanation

Find the hypothesis

$H_0:p_1-p_2=0$

$H_0:p_1-p_2\ne 0$

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.