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.

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