Question

32. Find the t critical value for a left tailed test, sample size = 11, significance level = 0.01 A) -2.82 B) 2.82 C) -2.76 D) 2.68 E) None of the Above Use the following table for problems 33-36 Men Women Total RH Negative 6 3 9 RH Positive 78 81 159 Total 84 84 168 33. Find the probability that a randomly selected person is a man. A) 6/9 B) 6/168 C) 6/84 D) 84/168 E) None of the Above 34. Find the probability that a randomly selected person is RH negative given she is a woman. A) 3/9 B) 3/84 C) 3/168 D) 81/84 E) None of the Above 35. Find the probability of randomly selecting a Negative RH individual A) 1 B) 9/84 C) 9/168 D) 159/168E) None of the Above 36. Find the probability of randomly selecting a person that is a woman and is is RH negative A) 6/9 B) 6/84 C) 6/168 D) 3/168 E) None of the Above

Answer 1

132 we have to find t criticail value for a left tailed test sample sire-la df: 11-1:10 and significanu len(2) = 0 oL with web il we have to find t such that PETLA] = 0.0L using t dishibution we get PIL-2,76) on comparing this gut 0.0L ho above me Ita 2.76 imen. 78 te option @ is correct Given faible Rutve) Prive) Toril 6 84 81 4.8% 9 isy 168 here lokal persons are women 3 Total reln(s) = 168

33: Pla man) nman) nes) 84 168 - f A man) 86 168 le ophion option is correct or 34 PERHE-ve) women PERHE-ve) n women] f(women) HERH(-ve) n women his n/women) hes) 31168 84168 3 84 Z 3 PERMtive)Worten}34. option te M cineet

35 [ Selecting to HERH(-ve) h(s) 9 168 tee sellching RH (-ve)] = 7/10 te option © y arrect 36 Il sellching women and RH (Eve) pH eve) nl women and Phl-ve)) nis) 3 168 selecting women and women and entive)]= 3 772 M conect i e option 6