A couple wants to have 2 boys
A couple wants to have 2 boys and 2 girls. They will continue tohave children until they have reached the goal. Assuming that theprobability that they have 54% chance to have a boy and 46% chanceto have a girl in every pregnancy. No twin or more are possible. Assuming that their goal is possible and has been fulfilledtoday!
Create at least four hypotheses based on the data fileprovided!
1) Use a different test for each hypothesis.
2) Construct the hypothesis statements and their statistic forms(H0 and H1).
Answer:
1.Given that,possible chances (x)=2sample size(n)=4success rate ( p )= x/n = 0.5success probability,( po )=0.54failure probability,( qo) = 0.46null, Ho:p=0.54alternate, H1: p!=0.54level of significance, alpha = 0.05from standard normal table, two tailed z alpha/2 =1.96since our test is two-tailedreject Ho, if zo < -1.96 OR if zo > 1.96we use test statistic z proportion = p-po/sqrt(poqo/n)zo=0.5-0.54/(sqrt(0.2484)/4)zo =-0.161| zo | =0.161critical valuethe value of |z alpha| at los 0.05% is 1.96we got |zo| =0.161 & | z alpha | =1.96make decisionhence value of |zo | < | z alpha | and here we do not rejectHop-value: two tailed ( double the one tail ) – Ha : ( p != -0.16051) = 0.87248hence value of p0.05 < 0.8725,here we do not reject HoANSWERS—————null, Ho:p=0.54alternate, H1: p!=0.54test statistic: -0.161critical value: -1.96 , 1.96decision: do not reject Hop-value: 0.87248we do not have enough evidence to support the claim that theprobability that they have 54% chance to have a boy
2.Given that,possible chances (x)=2sample size(n)=4success rate ( p )= x/n = 0.5success probability,( po )=0.46failure probability,( qo) = 0.54null, Ho:p=0.46alternate, H1: p!=0.46level of significance, alpha = 0.05from standard normal table, two tailed z alpha/2 =1.96since our test is two-tailedreject Ho, if zo < -1.96 OR if zo > 1.96we use test statistic z proportion = p-po/sqrt(poqo/n)zo=0.5-0.46/(sqrt(0.2484)/4)zo =0.161| zo | =0.161critical valuethe value of |z alpha| at los 0.05% is 1.96we got |zo| =0.161 & | z alpha | =1.96make decisionhence value of |zo | < | z alpha | and here we do not rejectHop-value: two tailed ( double the one tail ) – Ha : ( p != 0.16051 )= 0.87248hence value of p0.05 < 0.8725,here we do not reject HoANSWERS—————null, Ho:p=0.46alternate, H1: p!=0.46test statistic: 0.161critical value: -1.96 , 1.96decision: do not reject Hop-value: 0.87248we do not have enough evidence to support the claim thatprobability that 46% chance to have a girl in every pregnancy
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