Question

Select your answer. A random survey of 212 students asked each student to select the most preferred form of recreational activity from five choices. Here are the results of the survey: Basketball (1) Softball (2) Swimming (3) Jogging (4) Tennis (5) Male 31 16 19 22 23 Female 21 16 13 27 24 Is there a relationship between gender and preferred sport? The appropriate test decision is O reject H, at 0.05 significance level fail to reject H, at 0.05 significance level O accept H, at 0.05 significance level O none of the answers provided here O reject Hį at 0.05 significance level fail to reject H at 0.05 significance level fail to accept H, at 0.05 significance level

Answer 1

Solution: We have to test is there a delationship Spoet between gendee & peeffered lale know that, the chi-square test of independence is wed to determine if there is a significant Helationship beth two catego ei cod vaejables, Categorical vaεlable & we have to test relationship ben them. Therefore, here we we chi-squa de test of independence Here both are Hypothesis Ho: The&e is no celationship ben gendere & peeffered sport ie. Ho: Gender & poeffeced spott ače independent. He: These is delationship bejn genette f pdeffeerd Spoet. lp. Hur The gendes f pdeffeaed spoet aée dependent:

Unded Ho, the test statistic is, g² Ž (0;-E;)2 where, frequency o; is obseeved Ei is expected frequency Expected frequencies are calleated as Ei= Row total x Colunm total observation table y * Basketbol (1) Softball (2) sw Softball (2) swimmingo) Jogging (4) Tennisſa) Total 19 31 22 23 male Female - Total 16 16. 21 27 lol 52 32 32 49 47 212 Obse&vationtable I Oi- Ei (0;- Ei)? SENO O; EL (oi-Ei) Ei 31 27.22 3.78 14.2884 - 0.5249 0,0335 2 16 16.75 -0.75 0.5625 3 19 16.75 2.25 0.3022 5.0625 13.3255 4 22 25.65 -3.65 o-5195 5 23 24.60 2.56 6 21 -3.77 14.212g 0.0576 7 16 24.77 15.24 15.24 23.34 8 13 0.2G -2.24 3.66 5.0176 0.1040 0.5737 0.0037 0.3292 0.5739 0.1157 3.0803 13.3956 27 24 10 22.39 1.61 2.5921 Total

The expected feequencies 1) Eği = 111852 are calculated as follows: 6) Ex= 101452 24.77 - 27.22 212 212 2) E 16 - 111x32 212 16.75 7) Eig=101832 -15.24 212 3) Elg = 111x32 - 16.75 8) E 12 = 101 X32 - 15:24 212 212 4) E22=111449=25.65 g) E27 =101449 = 23.34 212 212 5) E23 = 111x4) 24.60 10) E24=101447 = 22.39 212 2 12 Linded the Ho, the felt statistic is, 2 (0;-Ei) 2 Ei x² = 3.0803 degree of freedom (& 1) (C-1) = (2-1) (5-1) = 1+4=4 i reitical value of mx? at a = 0.05 & dif=4 is 9.4877 Ceitical Value = 9:4877 here, X² < Xceitical hence we accept Ho at 0.05 significance level. & we conclude that, there is no gelationship between gender Spoct. poteffesed