A research team conducted a study of soft-drink preferences among residents in a test market prior to an advertising campaign for a new cola product. Of the participants, 130 are teenagers and 130 are adults. The researchers secured the following results:
Cola
Noncola
Teenagers 50 80
Adults 90
40 Calculate an appropriate measure of association, and decide how to present the results. How might this information affect the advertising strategy? The first factor to consider would be to determine the null hypothesis (actual age of the consumer and soft drink preference). The null hypothesis: H0= There is no relationship between the soft-drink preference and the age of the consumer.The statistical test:Critical values of Chi-SquareFigure 1. Chi-Square formula. Retrieved from http://www.stat.yale.edu/Courses/1997-
98/101/chisq.htm(Formulas taken from pages 490 and 491, Cooper and Schindler, 2014)a= 0.05d.f. (degrees of freedom) = 1χ2 critical value = 3.841Chi-Square Test: Cola, Non-ColaExpected counts are printed below observed countsChi-Square contributions are printed below expected counts Cola Non-Cola TotalTeens 50 80 130 70.00 60.00 5.714 6.667Adults 90 40 130 70.00 60.00 5.714 6.667Total 140 120 260Chi-Sq = 24.762, DF = 1, P-Value = 0.000Interpreting the testThe calculated chi-square value of 24.762 is larger than the critical value of 3.841 which means the null hypothesis (H) must be rejected. The alternative hypothesis does not specify thetypeof association, so close attention to the data is required to interpret the information provided by the test. The alternative hypothesis (Ha) is accepted, there is a relationship between the soft-drink preference and the age (teenage and adult, exact age not considered) of the consumer. The distribution of the statisticX2ischi-squarewith (r-1)(c-1) degrees of freedom, whererrepresents the number of rows in the two-way table andcrepresents the number of columns. The distribution is denoted(df), where df is the number of degrees of freedom.The chi-square distribution is defined for all positive values. TheP-valuefor the chi-square test isP (