Use the scientic method 5 part A clinical trial was conducted using a new method

Question

Use the scientic method 5 part

A clinical trial was conducted using a new method designed to increase the probability of conceiving a boy. As of this writing,

277 babies were born to parents using the new method, and 245 of them were boys. Use a 0.01 significance level to test the claim that the new method is effective in increasing the likelihood that a baby will be a boy. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution.

Which of the following is the hypothesis test to be conducted?

A.H0: p=0.5 H1: p<0.5

B.H0: p=0.5 H1: p0.5

C.H0: p=0.5 H1: p>0.5

D.H0: p>0.5 H1: p=0.5

E.H0: p0.5 H1: p=0.5

F.H0: p<0.5 H1: p=0.5

What is the test statistic? z=__________(Round to two decimal places as needed.)

What is the P-value? P-value=_________(Round to four decimal places as needed.)

What is the conclusion on the null hypothesis?

A.Reject the null hypothesis because the P-value is less than or equal toless than or equal to the significance level, .

B.Fail to reject the null hypothesis because the P-value is greater thangreater than the significance level, .

C.Reject the null hypothesis because the P-value is greater thangreater than the significance level, .

D.Fail to reject the null hypothesis because the P-value is less than or equal toless than or equal to the significance level, .

What is the final conclusion?

A. There is sufficient evidence to support the claim that the new method is effective in increasing the likelihood that a baby will be a boy.

B.There is not sufficient evidence to support the claim that the new method is effective in increasing the likelihood that a baby will be a boy.

C.There is sufficient evidence to warrant rejection of the claim that the new method is effective in increasing the likelihood that a baby will be a boy.

D.There is not sufficient evidence to warrant rejection of the claim that the new method is effective in increasing the likelihood that a baby will be a boy.

Explanation / Answer

Given that,
possibile chances (x)=245
sample size(n)=277
success rate ( p )= x/n = 0.8845
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5
alternate, H1: p>0.5
level of significance, = 0.01
from standard normal table,right tailed z /2 =2.33
since our test is right-tailed
reject Ho, if zo > 2.33
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.88448-0.5/(sqrt(0.25)/277)
zo =12.7979
| zo | =12.7979
critical value
the value of |z | at los 0.01% is 2.33
we got |zo| =12.798 & | z | =2.33
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 12.79793 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p>0.5
test statistic: 12.7979
critical value: 2.33
decision: reject Ho
p-value: 0

C.Reject the null hypothesis because the P-value is greater thangreater than the significance level, .

C.There is sufficient evidence to warrant rejection of the claim that the new method is effective in increasing the likelihood that a baby will be a boy.

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