Question

A couple wants to have 2 boys and 2 girls. They will continue to have children until they have reached the goal.
 Assuming that the probability that they have 54% chance to have a boy and 46% chance to have a girl in every pregnancy. No twin or more are possible. 
Assuming that their goal is possible and has been fulfilled today!

Create at least four hypotheses based on the data file provided!

1) Use a different test for each hypothesis.

2) Construct the hypothesis statements and their statistic forms (H0 and H1).

Answer 1

1.
Given that,
possible chances (x)=2
sample size(n)=4
success rate ( p )= x/n = 0.5
success probability,( po )=0.54
failure probability,( qo) = 0.46
null, Ho:p=0.54
alternate, H1: p!=0.54
level of significance, alpha = 0.05
from standard normal table, two tailed z alpha/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.5-0.54/(sqrt(0.2484)/4)
zo =-0.161
| zo | =0.161
critical value
the value of |z alpha| at los 0.05% is 1.96
we got |zo| =0.161 & | z alpha | =1.96
make decision
hence value of |zo | < | z alpha | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.16051 ) = 0.87248
hence value of p0.05 < 0.8725,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.54
alternate, H1: p!=0.54
test statistic: -0.161
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.87248
we do not have enough evidence to support the claim that the probability that they have 54% chance to have a boy

2.
Given that,
possible chances (x)=2
sample size(n)=4
success rate ( p )= x/n = 0.5
success probability,( po )=0.46
failure probability,( qo) = 0.54
null, Ho:p=0.46
alternate, H1: p!=0.46
level of significance, alpha = 0.05
from standard normal table, two tailed z alpha/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.5-0.46/(sqrt(0.2484)/4)
zo =0.161
| zo | =0.161
critical value
the value of |z alpha| at los 0.05% is 1.96
we got |zo| =0.161 & | z alpha | =1.96
make decision
hence value of |zo | < | z alpha | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.16051 ) = 0.87248
hence value of p0.05 < 0.8725,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.46
alternate, H1: p!=0.46
test statistic: 0.161
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.87248
we do not have enough evidence to support the claim that probability that 46% chance to have a girl in every pregnancy