Subscribers to the FAME magazine revealed the following preferences for three categories: Fashion 30, Athletics 24 and Business 15. Following these frequencies of observation, compute the chi-square test statistic. At the 0.05 level of significance, would you conclude they are similar?

CHi-square test statistic is given by the formula: x^2=\sum_{}^{}\frac{\left(O_{ij}-E_{ij}\right)^2}{E_{ij}} O is the observed and E is the expected frequecies find the degrees of freedom df=(r-1)(c-1) r is the number of categories, and c is the number of preference categories both c and r is 3 df=(3-1)(3-1)=4 assuming that under the null hypothesis each category is equally likely, E_{ij}=\frac{total\:\:observation}{number\:\:of\:\:categories}=\frac{30+24+15}{3}=\frac{69}{3}=23 now substitute the values of each x^2=\frac{\left(30-23\right)^2}{23}+\frac{\left(24-23\right)^2}{23}+\frac{\left(15-23\right)^2}{23} x^2=4.957 checking the chi square table at 0.05 significance level the critical chi square is 9.49 since the calculated chi square does not exceed the critical value, we fail to reject the null hypothesis. there is not enough evidence to conclude a difference in preferences