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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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Most popular questions from this chapter

A study of identity theft looked at how well consumers protect themselves from this increasingly prevalent crime. The behaviors of 61randomly selected college students were compared with the behaviors of 59randomly selected non students.39One of the questions was 鈥淲hen asked to create a password, I have used either my mother鈥檚 maiden name, or my pet鈥檚 name, or my birth date,

or the last four digits of my social security number, or a series of consecutive numbers.鈥� For the students, 22agreed with this statement while 30of the nonstudents agreed.

a) Display the data in a two-way table and perform

the appropriate chi-square test. Summarize the results.

(b) Reanalyze the data using the methods for comparing two proportions that we studied in Chapter10. Compare the results and verify that the chi-square

statistic is the square of the z statistic.

Do men and women participate in sports for the same reasons? One goal for sports participants is social comparison鈥攖he desire to win or to do better than other people. Another is mastery鈥攖he desire to improve one鈥檚 skills or to try one鈥檚 best. A study on why students participate in sports collected data from independent random samples of 67male and 67female undergraduates at a large university. Each student was classified into one of four categories based on his or her responses to a questionnaire about sports goals. The four categories were high social comparison鈥� high mastery (HSC-HM), high social comparison鈥� low mastery (HSC-LM), low social comparison鈥揾igh mastery (LSC-HM), and low social comparison鈥搇ow mastery (LSC-LM). One purpose of the study was to compare the goals of male and female students. Here are the data displayed in a two-way table:

GoalFemaleMaleHSC-HM1431HSC-LM2118LSC-HM215LSC-LM2513

(a) Calculate the conditional distribution (in proportions) of the reported sports goals for each gender.

(b) Make an appropriate graph for comparing the conditional distributions in part (a).

(c) Write a few sentences comparing the distributions of sports goals for male and female undergraduates.

Average ratings (1.3,10.2)The students decided to compare the average ratings of the cafeteria food on the two scales.

(a) Find the mean and standard deviation of the ratings for the students who were given the 1-to-5scale.

(b) For the students who were given the 1-to-5scale, the ratings have a mean of 3.21and a standard deviation of 0.568. Since the scales differ by one point, the group decided to add 1to each of these ratings. What are the mean and standard deviation of the adjusted ratings?

(c) Would it be appropriate to compare the means from parts (a) and (b) using a two-sample t-test? Justify your answer

No chi-squareThe principal in Exercise 9also asked the random sample of students to record whether they did all of the homework that was assigned on each of the 铿乿e school days that week. Here are the data:

Explain carefully why it would not be appropriate to perform a chi-square goodness-of-铿乼 test using these data.

Why men and women play sports Do men and women participate in sports for the same reasons? One goal for sports participants is social comparision鈥攖he desire to win or to do better than other people. Another is mastery鈥攖he desire to improve one鈥檚 skills or to try one鈥檚 best. A study on why students participate in sports collected data from independent random samples of 67 male and 67 female under-graduates at a large university 15 Each student was classified into one of four categories based on his or her responses to a questionnaire about sports goals. The four categories were high social comparison鈥� high mastery (HSC-HM), high social comparison鈥� low mastery (HSC-LM), low social comparison鈥揾igh mastery (LSC-HM), and low social comparison鈥搇ow mastery (LSC-LM). One purpose of the study was to compare the goals of male and female students. Here are the data displayed in a two-way table: Observed Counts for Sports Goals

(a) Check that the conditions for performing the chi-square test are met.

(b) Use Table C to find the P-value. Then use your calculator鈥檚 C2cdf command.

(c) Interpret the P-value from the calculator in context.

(d) What conclusion would you draw? Justify your answer.
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