Arandom survey of 100 students asked each student to select the most preferred form of recreational activity from five choices. following are the results of the survey: recreational choice baseball jogginggender basketball softball swimming running tennis totalsmale 21 5 9 12 13 60female 9 3 1 15 12 40totals 30 8 10 27 25 1001. test whether the choice is independent of the gender of the respondent. approximate the p-value of the test.2. would we reject the null hypothesis at α = 0.05?

No , we would not reject the null hypothesis at α = 0.05

Step-by-step explanation:

Given that a  random survey of 100 students asked each student to select the most preferred form of recreational activity from five choices. Following are the results of the survey.

(Two tailed chi square test at 5% level for testing independence between gender and sports activities)

We assume H0 to be true and prepare expected values and then calculate chi square as (O-E)^2/E where O = observed value, E = expected value

BasketballSoftballSwimmingrunningTennis  

Males21  (18.00)  [0.50]5  (4.80)  [0.01]9  (6.00)  [1.50]12  (16.20)  [1.09]13  (15.00)  [0.27]60

Females9  (12.00)  [0.75]3  (3.20)  [0.01]1  (4.00)  [2.25]15  (10.80)  [1.63]12  (10.00)  [0.40]40

chisquare8.4097      

p value0.077      

df4      

Since p > 5% we accept null hypothesis.

No , we would not reject the null hypothesis at α = 0.05

answer: x=60

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)