Question

6. A clinical trial was conducted using a new method designed to increase the probability of conceiving a girl. As of this​ writing, 961 babies were born to parents using the new​ method, and 898 of them were girls. 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 girl. 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. H 0 : p > 0.5

  H 1 : p = 0.5

B. H 0 : p = 0.5

    H 1 : p ≠0.5

C. H 0 : p = 0.5

     H 1 : p > 0.5

D. H 0 : p < 0.5

    H 1 : p = 0.5

E. H 0 : p = 0.5

    H 1 : p < 0.5

F. H 0 : p ≠ 0.5

    H 1 = 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 about the null​ hypothesis?

A. Fail to reject the null hypothesis because the​ P-value is less than or equal to the significance​ level, a.

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

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

D. Reject the null hypothesis because the​ P-value is less than or equal to the significance​ level, alpha.

What is the final​ conclusion?

A. 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 girl.

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 girl.

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 girl.

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

Answer 1

Solution :

This is the right tailed test .

The null and alternative hypothesis is

H₀ : p = 0.5

H_a : p >  0.5

n = 961

x = 898

= x / n = 898 / 961 = 0.9344

P₀ = 0.5

1 - P₀ = 1 - 0.5 = 0.5

z = - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.9344 - 0.5/ [(0.5 * 0.5) / 961]

= 26.93

Test statistic = 26.93

This is the right tailed test .

P(z > 26.93) = 1 - P(z < 26.93) = 1 - 1 = 0

P-value = 0

= 0.01

P-value <

D. Reject the null hypothesis because the​ P-value is less than or equal to 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 girl.