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Is astrology scientific? The General Social Survey (GSS) asked a random sample of adults their opinion about whether astrology is very scientific, sort of scientific, or not at all scientific. Here is a two-way table of counts for people in the sample who had three levels of higher education:

a. State appropriate hypotheses for performing a chi-square test for independence in this setting.

b. Compute the expected counts assuming that H0is true.

c. Calculate the chi-square test statistic, df, and P-value.

d. What conclusion would you draw?

Short Answer

Expert verified

(a) The appropriate hypothesis is the null hypothesis.

(b) The expected count is,


ASSOCIATE'SBACHELOR'SMASTER'S
YES183.59
251.85103.56
NO50.41
69.15
28.44

c) The test statistics is χ2=10.5819,df=2,P-value=0.005037

(d) There are many convincing pieces of evidence for the association between degree held and the opinion about the astrology.

Step by step solution

01

Part (a) Step 1: Given information 

We need to find out the appropriate hypothesis for performing a chi-square test for independence in this setting.

02

Part (a) Step 2: Explanation 

We know that

The null hypothesis asserts that the variables are unrelated, whereas the alternative hypothesis asserts that they are.

H0is there is no association between degree held and the opinion about the astrology.

Hαis there is an association between degree held and the opinion about the astrology.

03

Part (b) Step 1: Given information 

We need to find the expected counts.

04

Part (b) Step 2: Explanation 

From part (a)

We know that

Expected frequencies are a product of row and column total divided by table total. So,

ROW AND COLUMN NUMBER
EXPECTED FREQUENCY
E11183.59
E12
251.85
E13
103.56
E21
50.41
E22
69.15
E23
28.44
05

Part (c) Step 1: Given information 

We need to find the test statistics, df, P-value.

06

Part (c) Step 2: Explanation 

From parts (a) and (b)

We know that

The squared differences between the actual and predicted frequencies, divided by the expected frequency, make up the chi-square subtotals.

Therefore, test-statistics is, χ2=10.5819

And,

df=2P-value=0.005<P<0.01=0.005037

07

Part (d) Step 1: Given information 

We need to find out the conclusion drawn.

08

Part (d) Step 2: Explanation 

From the parts (a) ,(b) and (c)

We know that

There are many convincing pieces of evidence for the association between degree held and the opinion about astrology.

And this is the conclusion we drew.

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Most popular questions from this chapter

Recent revenue shortfalls in a midwestern state led to a reduction in the state budget for higher education. To offset the reduction, the largest state university proposed a 25%tuition increase. It was determined that such an increase was needed simply to compensate for the lost support from the state. Separate random samples of 50freshmen, 50sophomores, 50juniors, and 50seniors from the university were asked whether they were strongly opposed to the increase, given that it was the minimum increase necessary to maintain the university's budget at current levels. Here are the results:

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

a. The closer students get to graduation, the less likely they are to be opposed to tuition increases.

b. The mean number of students who are strongly opposed is the same for each of the 4years.

c. The distribution of student opinion about the proposed tuition increase is the same for each of the 4years at this university.

d. Year in school and student opinion about the tuition increase are independent in the sample.

e. There is an association between year in school and opinion about the tuition increase at this university.

Last python Refer to Exercises 28 and 30.

a. Verify that the conditions for inference are met.

b. Use Table C to find the P-value. Then use your calculator’s χ2 cdf command.

c. Interpret the P-value from the calculator.

d. What conclusion would you draw using α=0.10?

The two-way table shows the results of the experiment described in

Exercise 28.

Hatching Statuswater temperature:coldwater temperature:neutral
water temperature:hot
Total
Yes16
38
75
129
No11
18
29
58
Total
27
56
104
187

a. State the appropriate null and alternative hypotheses.

b. Show the calculation for the expected count in the Cold/Yes cell. Then provide a

complete table of expected counts.

c. Calculate the value of the chi-square test statistic.

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 that28of the 82gastric-freezing patients improved, while 30of the 78patients 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).

Finger length Is your index finger longer than your ring finger? Or is it the other way around? It isn't the same for everyone. To investigate if there is a relationship between gender and relative finger length, we selected a random sample of 460U.S. high school students who completed a survey. The two-way table shows the results.


Do these data provide convincing evidence at the α=0.10level of an association between gender and relative finger length in the population of students who completed the survey?

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