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Chapter 11: Q 6. (page 716)

Roulette Refer to Exercise 2.

a. Confirm that the expected counts are large enough to use a chi-square distribution to calculate the P-value. What degrees of freedom should you use?

b. Use Table C to find the P-value. Then use your calculator鈥檚 蠂2 cdf command.

c. What conclusion would you draw about whether or not the roulette wheel is operating correctly?

Short Answer

Expert verified

Part (a) The df=2

Part (b) The P-value=0.1108

Part (c) There isn't enough evidence to refute the claim concerning the probability distribution in question.

Step by step solution

01

Part (a) Step 1: Given information

ColorObserved countFrequencyExpected count
Red850.4794.74
Black990.4794.74
Green160.0510.53
Total2001.00200

The chi-square test statistic =4.04

02

Part (a) Step 2: Calculation

The projected counts are large enough to use a chi-square distribution if all predicted counts are at least 5This condition is satisfied.

The degrees of freedom =df=c-1=3-1=2

03

Part (b) Step 1: Given information

The chi-square test statistic =4.04

The degrees of freedom =2

04

Part (b) Step 2: Calculation

The p-value using table for df=2 is,

0.10<P<0.15

The excel formula, =CHIDIST(4.40,2)

P-value =0.1108

05

Part (c) Step 1: Given information

The chi-square test statistic =4.04

The degrees of freedom =2

The p-value=0.1108

06

Part (c) Step 2: Calculation

Here, p-value>0.05fail to reject H0

There isn't enough evidence to refute the claim concerning the probability distribution in question.

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Most popular questions from this chapter

What鈥檚 your sign? The University of Chicago鈥檚 General Social Survey (GSS) is the nation鈥檚 most important social science sample survey. For reasons known only to social scientists, the GSS regularly asks a random sample of people their astrological sign. Here are the counts of responses from a recent GSS of 4344 people:

If births are spread uniformly across the year, we expect all 12 signs to be equally likely. Do these data provide convincing evidence at the 1% significance level that all 12 signs are not equally likely?

All current-carrying wires produce electromagnetic (EM) radiation, including the electrical wiring running into, through, and out of our homes. High-frequency EM radiation is thought to be a cause of cancer. The lower frequencies associated with household current are generally assumed to be harmless. To investigate the relationship between current configuration and type of cancer, researchers visited the addresses of a random sample of children who had died of some form of cancer (leukemia, lymphoma, or some other type) and classified the wiring configuration outside the dwelling as either a high-current configuration (HCC) or a low-current configuration (LCC). Here are the data:

Computer software was used to analyze the data. The output included the value X2=0.435

Which of the following may we conclude, based on the chi-square test results?

a. There is convincing evidence of an association between wiring configuration and the chance that a child will develop some form of cancer.

b. HCC either causes cancer directly or is a major contributing factor to the development of cancer in children.

c. Leukemia is the most common type of cancer among children.

d. There is not convincing evidence of an association between wiring configuration and the type of cancer that caused the deaths of children.

e. There is convincing evidence that HCC does not cause cancer in children.

For these data, 蠂2=69.8with a P-value of approximately 0. Assuming that the researchers used a significance level of 0.05, which of the following is true?

a. A Type I error is possible.

b. A Type II error is possible.

c. Both a Type I and a Type II error are possible.

d. There is no chance of making a Type I or Type II error because the P-value is approximately 0.

e. There is no chance of making a Type I or Type II error because the calculations are correct.

When analyzing survey results from a two-way table, the main distinction between a test for independence and a test for homogeneity is

a. how the degrees of freedom are calculated.

b. how the expected counts are calculated.

c. the number of samples obtained.

d. the number of rows in the two-way table.

e. the number of columns in the two-way table.

In the United States, there is a strong relationship between education and smoking: well-educated people are less likely to smoke. Does a similar relationship hold in France? To find out, researchers recorded the level of education and smoking status of a random sample of 459French men aged 20to60years. The two-way table displays the data.

Is there convincing evidence of an association between smoking status and educational level among French men aged20to60years?
See all solutions

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