Chapter 11.2, Problem 6.1CYU

To determine

To test if there is any convincing evidence of an association between an exclusive territory clause and business survival for Mc Donald’s franchises at significance level 0.01.

Answer to Problem 6.1CYU

Explanation of Solution

Given:

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

  

The level of significance = 0.01

Concept used:

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 independence and the samples can be at most 10% of the population.

Formula used: $\begin{array}{l}ExpectedCount=\frac{rowtotal*coloumntotal}{Total}\\ {\chi }^2={\displaystyle \sum \left(\frac{{(Observed-Expected)}^2}{Expected}\right)}\\ Degreeoffreedom=(no.ofrows-1)*(no.ofcolumns-1)\end{array}$

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 as independent. Hence all conditions are met to carry out 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.

Calculation:

The expected values are calculated as shown below

  $\begin{array}{cccc}& Exclusive& territory& \\ Success& Yes& No& Total\\ Yes& \frac{123*142}{170}=102.74& \frac{123*28}{170}=20.26& 123\\ No& \frac{47*142}{170}=39.26& \frac{47*28}{170}=7.74& 47\\ Total& 142& 28& 170\end{array}$

The test statistic is calculated as shown below

  $\begin{array}{l}{\chi }^2={\displaystyle \sum \left(\frac{{(Observed-Expected)}^2}{Expected}\right)}\\ {\chi }^2=\frac{{\left(108-102.74\right)}^2}{102.74}+\frac{{\left(15-20.26\right)}^2}{20.26}+\frac{{\left(34-39.26\right)}^2}{39.26}+\frac{{\left(13-7.74\right)}^2}{7.74}\\ {\chi }^2=5.91\end{array}$

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.

Conclusion:

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