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Perform a follow-up analysis of the test in Exercise 40 by finding the individual components of the chi-square statistic. Which cell(s) contributed most to the final result?

Short Answer

Expert verified

It is noted that we obtained the largest chi-square subtotals for the Placebo and Both. This means that Both medicines together lead to better results and thus this treatment should be used (if the side effects are not too significant).

Step by step solution

01

Given information

The given data is

02

Explanation

Observed counts

Expected counts

The expected counts are the row total multiplied by the column total, divided by the sample sizen=6602

The chi-square statistic is the sum of squared deviations (between observed and expected counts) divided by the expected count:

localid="1650645457450" 2=(250-205.8128)2205.8128+(206-205.8128)2205.8128+(211-206.4368)2206.4368+(157-205.9376)2205.9376+(1399-1443.1872)21443.1872+(1443-1443.1872)21443.1872+(1443-1447.5632)21447.5632+(1493-1444.0624)21444.0624=9.487+0+0.101+11.629+1.353+0+0.14+1.658=24.243

It is noted that we obtained the largest chi-square subtotals for the Placebo and Both. This means that Both medicines together lead to better results and thus this treatment should be used (if the side effects are not too significant).

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Most popular questions from this chapter

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).

What is the most important reason that students buy from catalogs? The answer may differ for different groups of students. Here are results for separate random samples of American and Asian students at a large midwestern university

(a) Should we use a chi-square test for homogeneity or a chi-square test of association/independence in this setting? Justify your answer. (b) State appropriate hypotheses for performing the type of test you chose in part (a). Minitab output from a chi-square test is shown below.

(c) Check that the conditions for carrying out the test are met.

(d) Interpret the P-value in context. What conclusion would you draw?

Sorry, no chi-square We would prefer to learn from teachers who know their subject. Perhaps even pre-school children are affected by how knowledgeable they think teachers are. Assign 48three- and four-year-olds at random to be taught the name of a new toy by either an adult who claims to know about the toy or an adult who claims not to know about it. Then ask the children to pick out a picture of the new toy in a set of pictures of other toys and say its name. The response variable is the count of right answers in four tries. Here are the data:

The researchers report that children taught by the teacher who claimed to be knowledgeable did significantly better 2=20.24,P<0.05. Explain why this result isn't valid.

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(a) (18-25)225+(22-25)225+(39-25)225+(21-25)225

(b) (25-18)218+(25-22)222+(25-39)239+(25-21)221

(c) (18-25)25+(22-25)25+(39-25)25+(21-25)25

(d)(18-25)2100+(22-25)2100+(39-25)2100+(21-25)2100

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Which hypotheses would be appropriate for performing a chi-square test?

(a) The null hypothesis is that the closer students get to graduation, the less likely they are to be opposed to tuition increases. The alternative is that how close students are to graduation makes no difference in their opinion.

(b) The null hypothesis is that the mean number of students who are strongly opposed is the same for each of the four years. The alternative is that the mean is different for at least two of the four years.

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