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Chapter 11: Q.4 (page 692)

Roulette Calculate the chi-square statistic for the data in Exercise 2. Show your work.

Short Answer

Expert verified

From the given information, the test statistics are4.0389

Step by step solution

01

Given Information

It is given in the question that,

02

Explanation

Here,

p1=1838=0.4737

p2=1838=0.4737

p3=238=0.0526

The null and alternative hypotheses are:

H0:p1

H0:p2

H0:p3

H伪 : At least one of pi is different

The calculation for test statistics is done as:

03

Step 3: Explanation

The test statistic is

蠂2=鈭�(O鈭�E)2/E=4.0389

Therefore, the test statistic is 4.0389

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Most popular questions from this chapter

Benford鈥檚 lawFaked numbers in tax returns, invoices, or expense account claims often display patterns that aren鈥檛 present in legitimate records. Some patterns are obvious and easily avoided by a clever crook. Others are more subtle. It is a striking fact that the 铿乺st digits of numbers in legitimate records often follow a model known as Benford鈥檚 law.3 Call the 铿乺st digit of a randomly chosen record X for short. Benford鈥檚 law gives this probability model for X (note that a 铿乺st digit can鈥檛 be 0):

A forensic accountant who is familiar with Benford鈥檚 law inspects a random sample of invoices from a company that is accused of committing fraud. The table below displays the sample data.

(a) Are these data inconsistent with Benford鈥檚 law? Carry out an appropriate test at the 伪=0.05level to support your answer. If you 铿乶d a signi铿乧ant result, perform follow-up analysis.

(b) Describe a Type I error and a Type II error in this setting, and give a possible consequence of each. Which do you think is more serious?

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.

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.

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?

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
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