Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the​ study, 357 babies were​ born, and 201 of them were girls. Use the sample data with a 0.01 significance level to test the claim that with this​ method, the probability of a baby being a girl is greater than 0.5. Use this information to answer the following questions.

a. List the null and alternative hypothesis.

b. P-value =

c. What is the conclusion?

d. Does the method appear to be effective?

Answer 1

Solution :

a ) This is the right tailed test .

The null and alternative hypothesis is

H0 : p = 0.50

Ha : p >0.5

n =357

x =201

= x / n = 201 / 357 =0.56

P0 = 0.50

1 - P0 = 1 - 0.50 = 0.50

Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

=0.56 -0.50 / [(0.50*0.50) / 357]

= 2.38

Test statistic = z = 2.38

P(z >2.38 ) = 1 - P(z < 2.38 ) = 1 -0.9913

c ) P-value = 0.0087

= 0.05

P-value <

0.0087 < 0.05

Reject the null hypothesis .

There is sufficient evidence to suggest that

d ) Yes it is the method appear to be effective