In: Statistics and Probability
Given,
a)the probability of playing soccer p(s)=0.43
b)the probability of playing at least one of basketball and soccer
P(B or S)=P(B)+P(S)-P(B and S)
=0.53+0.43-0.35=0.61
c)the probability of a student playing soccer, given that they play basketball
P(S|B)=P(B and S)/P(B)
P(S|B)=0.35/0.53=0.6604
d)Are playing soccer and basketball independent:
no, playing soccer and basketball are not indepedent.because P(B and S)P(B)*P(S)
e)
Probability of playing neither basketball or Soccer= 1 - probability of playing at least one of basketball and soccer
= 1 - 0.61 = 0.39
Null Hypothesis H0: p = 0.39
Alternative Hypothesis H0: p 0.39
given,
n=29, x=6
p̂=6/29=0.2069
Since it is observed that ∣z∣=2.022>zc=1.96, it is then concluded that the null hypothesis is rejected.
Using the P-value approach: The p-value is p = 0.0432, and since p=0.0432<0.05, it is concluded that the null hypothesis is rejected.
Conclusion
It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is different than p0, at the α=0.05 significance level.