If the player's bet size was independent of their actions, then the expected count for the number of bets would be: Bet Amount Hit Stand Double Split $10 470.59 227.34 162.96 139.11 $500 121.41 58.66 42.04 35.89 However, the actual counts for the actions and bet amounts were: Bet Amount Hit Stand Double Split $10 496 205 158 141 $500 96 81 47 34 We informally showed that there was some differences between these values. Now, with the chisquared distribution, we can do a formal hypothesis test. When using the chisquared distribution to test for independence, the sum R C 2 2 _ (0m EM) 2 Xrnd. Z 2 E, j N X(R1)(o1> i=1 j=1 , where n is the number categories, and OM and Em are the observed and expected counts for the 73th row and jth column, respectively, and where R and C are the number of rows and columns in the crosstable, respectively. a) (1 mark) How many degrees of freedom does the null hypothesis x2 distribution have, in this case? b) (1 mark) Is the hypothesis test onetailed or twotailed, how do you know? c) (4 marks) Find the chisquared test statistic. That is, nd the sum described above. (Hint: The sum has eight terms, and the first term is z 1.3720) (1) (1 mark) Find the pvalue of this data against the null hypothesis of independence. e) (1 mark) For a signicance level of a = 0.05, is there sufcient evidence to reject the null hypothesis that bet amount and game action are independent