Many popular businesses are franchises—think of McDonald’s. The owner of local franchise benefits from brand recognition, national advertising, and detailed guidelines provided by the franchise chain. In return, he or she pays fees to the franchise firm and agrees to follow its policies. The relationship between the local owner and the franchise firm is spelled out in a detailed contract.

One clause that the contract may or may not contain is the entrepreneur’s right to an exclusive territory. This means that the new outlet will be the only representative of the franchise in a specified territory and will not have to compete with other outlets of the same chain. How does the presence of an exclusive-territory clause in the contract relate to the survival of the business?

A study designed to address this question collected data from a random sample of 170new franchise firms. Two categorical variables were measured for each franchisor. First, the franchisor was classified as successful or not based on whether or not it was still offering franchises as of a certain date. Second, the contract each franchisor offered to franchisees was classified according to whether or not there was an exclusive-territory clause. Here are the count data, arranged in a two-way table:

role="math" localid="1650453920307" Exclusive TerritorySuccessYesNoTotalYes10815123No341347Total14228170

Do these data provide convincing evidence of an association between an exclusive territory clause and business survival? Carry out an appropriate test at the α=0.01level.

Short Answer

Expert verified

At 0.01level of significance, there is no association between an exclusive territory clause and business survival.

Step by step solution

01

Given Information

The data of franchisor classified as successful or not and franchises classified as successful or not with exclusive territory clause is given below

Exclusive TerritorySuccessYesNoTotalYes10815123No341347Total14228170

The level of significance=0.01

02

Explanation

The conditions to be met to use a Chi-square test.

- The data should be chosen randomly

- The sample size should be large so that the expected counts are not less than 5.

- There should be independent and the samples can be at most 10%of the population.

Expected Count=row totalcoloumn total*Total

Formula used:χ2=(Observed-Expected)2Expected

Degree of freedom=(no. of rows-1)*(no. of columns-1)

The data is collected from random samples of 170new franchise firms. The sample size is large enough so that the expected counts are greater than 5. We can safely assume that there will be more than 1700franchise firms, so data is considered independent. Hence all conditions are met to carry out the Chi-square test.

Null hypothesis: The exclusive territory clause and business survival are independent.

Alternate hypothesis: The exclusive territory clause and business survival are dependent.

03

Calculation

The expected values are calculated as shown below

Exclusive Territory
SuccessYesNoTotal
Yes123×142170=102.74

123×28170=20.26
123
No47×142170=39.26
47×28170=7.74
47
Total142
28
170

The test statistic is calculated as shown below

χ2=(Observed -Expected)2Expected

χ2=(108-102.74)2102.74+(15-20.26)220.26+(34-39.26)239.26+(13-7.74)27.74

χ2=5.91

Degree of freedom =(2-1)*(2-1)=1

The p-value for 1degree of freedom and test statistic 5.91is 0.015055.

The p-value is more than the level of significance, so we have insufficient evidence at 0.01level of significance to reject the null hypothesis.

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Most popular questions from this chapter

A company claims that each batch of its deluxe mixed nuts contains 52%cashews, 27%almonds, 13%macadamia nuts, and 8%brazil nuts. To test this claim, a quality control inspector takes a random sample of 150nuts from the latest batch. The one-way table below displays the sample data.

(a) State appropriate hypotheses for performing a test of the company’s claim.

(b) Calculate the expected counts for each type of nut. Show your work

The nonprofit group Public Agenda conducted telephone interviews with three randomly selected groups of parents of high school children. There were 202black parents, 202Hispanic parents, and 201white parents. One question asked was “Are the high schools in your state doing an excellent, good, fair, or poor job, or don’t you know enough to say?” Here are the survey results:

(a) Calculate the conditional distribution (in proportions) of responses for each group of parents.

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

(c) Write a few sentences comparing the distributions of responses for the three groups of parents

32. How are schools doing? Refer to Exercises 28 and 30 .

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

(b) Use Table Cto find the P-value. Then use your calculator's χ2cdfcommand.

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

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

The chi-square statistic is

(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

(e)(0.18-0.25)20.25+(0.22-0.25)20.25+(0.39-0.25)20.25+(0.21-0.25)20.25

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 “When asked to create a password, I have used either my mother’s maiden name, or my pet’s name, or my birth date,

or the last four digits of my social security number, or a series of consecutive numbers.” 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.

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