Q. 6.1 - The Practice of Statistics for AP

Question

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 $170$ new 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:

$\begin{array}{|cccc|}\hline \text{ Exclusive Territory }& & & \\ \text{ Success }& \text{ Yes }& \text{ No }& \text{ Total }\\ \text{ Yes }& 108& 15& 123\\ \text{ No }& 34& 13& 47\\ \text{ Total }& 142& 28& 170\\ \hline\end{array}$

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

Step-by-Step Solution

Answer

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

Step 1: 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

$\begin{array}{|cccc|}\hline \text{ Exclusive Territory }& & & \\ \text{ Success }& \text{ Yes }& \text{ No }& \text{ Total }\\ \text{ Yes }& 108& 15& 123\\ \text{ No }& 34& 13& 47\\ \text{ Total }& 142& 28& 170\\ \hline\end{array}$

The level of significance $=0.01$

Step 2: 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.

$\text{ Expected Count }=\frac{\text{ row total }{}^{*}\text{ coloumn total }}{\text{ Total }}$

$\text{ Formula used: }{\chi }^2=\sum \left(\frac{(\text{ Observed }-\text{ Expected }{)}^2}{\text{ Expected }}\right)$

$\text{ Degree of freedom }=(\text{ no. of rows }-1)*(\text{ no. of columns }-1)$

The data is collected from random samples of $170$ new 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 $1700$ franchise 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.

Step 3: Calculation

The expected values are calculated as shown below

Exclusive

Territory

Success	Yes	No	Total
Yes	$\frac{123\times 142}{170}=102.74$	$\frac{123\times 28}{170}=20.26$	$123$
No	$\frac{47\times 142}{170}=39.26$	$\frac{47\times 28}{170}=7.74$	$47$
Total	$142$	$28$	$170$

The test statistic is calculated as shown below

${\chi }^2=\sum \left(\frac{(\text{ Observed -Expected }{)}^2}{\text{ Expected }}\right)$

${\chi }^2=\frac{(108-102.74{)}^2}{102.74}+\frac{(15-20.26{)}^2}{20.26}+\frac{(34-39.26{)}^2}{39.26}+\frac{(13-7.74{)}^2}{7.74}$

${\chi }^2=5.91$

Degree of freedom $=(2-1{)}^{*}(2-1)=1$

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

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