Categorical data of 2 by 2 table I am trying to compare expert and Novice perceptual differences about Apple watch features. The data contained Negative and Positive viewpoints. 
ex. Data for design feature:
Novice : Negative (9) Positive (22).
Expert : Negative (27) Positive (12).
I used chi square to see if there is any significant differences between the two groups.
I need to find out which group likes this feature more? is there is any test i can do it beside chi square.
 A: Since you have a 2 x 2 table, after you conduct the chi-square test of association, there's nothing else you need to do. An association between Novice/Expert and Positive/Negative implies that the proportions for each level of experience are different. If you had a larger table, you would need to unpack the results more in order to come to a similar conclusion. 
Novice is 71% positive (22/(9+22)).
Expert is 31% positive (12/(12+27)).
R code:
Matrix = matrix(c(22,9,12,27), nrow=2, byrow=TRUE, 
         dimnames=list(c("Novice", "Expert"),c("Positive", "Negative")))

Matrix

   ###        Positive Negative
   ### Novice      22        9
   ### Expert      12       27


chisq.test(Matrix)

   ### Pearson's Chi-squared test with Yates' continuity correction
   ### 
   ### X-squared = 9.6215, df = 1, p-value = 0.001923


prop.table(Matrix, margin=1)

   ###          Positive  Negative
   ### Novice 0.7096774 0.2903226
   ### Expert 0.3076923 0.6923077

A: The chi-squared test will answer the question "Are the sentiments positive vs. negative independent of the experience level of the user?" The null hypothesis is that they are statistically independent. I find that p-value is 0.0019, so you may reject the null hypothesis and assert that there is a dependence between these features.
But you specifically ask "is there is any test i can do it beside chi square?" So I'll assume that you are not interested in/not happy with any $\chi^2$ results. So, to answer your question, YES, there is another test you can do.
You may use a T test of means to determine if the mean sentiments of the experts is statistically significantly different from the mean sentiments of novices. If "like" has a value of 1 and don't-like has a value of 0, then your t-test will be between the novices with data (1 1 1 1 . . . 1 0 . . . 0)  22 1's and 9 0's and the experts with data (1..1 0..0) 12 1s and 27 zeros. I get a p value of .00062, or a rejection of the null that the two groups have the same average sentiments about the new Apple watch features.
