Question

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

290

babies were born to parents using the new​ method, and

246

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.

Upper H 0 : p less than 0.5H0: p<0.5

Upper H 1 : p equals 0.5H1: p=0.5

B.

Upper H 0 : p equals 0.5H0: p=0.5

Upper H 1 : p greater than 0.5H1: p>0.5

C.

Upper H 0 : p equals 0.5H0: p=0.5

Upper H 1 : p not equals 0.5H1: p≠0.5

D.

Upper H 0 : p not equals 0.5H0: p≠0.5

Upper H 1 : p equals 0.5H1: p=0.5

E.

Upper H 0 : p greater than 0.5H0: p>0.5

Upper H 1 : p equals 0.5H1: p=0.5

F.

Upper H 0 : p equals 0.5H0: p=0.5

Upper H 1 : p less than 0.5H1: p<0.5

What is the test​ statistic?

zequals=nothing

​(Round to two decimal places as​ needed.)

What is the​ P-value?

​P-valueequals=nothing

​(Round to four decimal places as​ needed.)

What is the conclusion on the null​ hypothesis?

A.

Fail to rejectFail to reject

the null hypothesis because the​ P-value is

less than or equal toless than or equal to

the significance​ level,

alphaα.

B.

Fail to rejectFail to reject

the null hypothesis because the​ P-value is

greater thangreater than

the significance​ level,

alphaα.

C.

RejectReject

the null hypothesis because the​ P-value is

less than or equal toless than or equal to

the significance​ level,

alphaα.

D.

RejectReject

the null hypothesis because the​ P-value is

greater thangreater than

the significance​ level,

alphaα.

What is the final​ conclusion?

A.There

isis

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 notis 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

isis

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 notis 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.

Click to select your answer(s).

Answer 1

Given that, n = 290 and x = 246

=> sample proportion = 246/290 = 0.848276

We want to test the claim that the new method is effective in increasing the likelihood that a baby will be a boy.

Therefore, the null and alternative hypotheses are,

H0 : p = 0.5

Ha : p > 0.5

Test statistic is,

=> Test statistic = Z = 11.86

p-value = P(Z > 11.86) = 1 - P(Z < 11.86) = 1 - 1 = 0.0000

=> p-value = 0.0000

We reject the null hypothesis because the p-value is less than or equal to the significance level alpha.

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.