Question

In1998 a national vital statistics report indicated that about 2.2 %of all births produced twins. Is the rate of twin births the same among very young​ mothers? Data from a large city hospital found only 10 sets of twins were born to 516 teenage girls. Test an appropriate hypothesis and state your conclusion. Be sure the appropriate assumptions and conditions are satisfied before you proceed.

Are the assumptions and the conditions to perform a​ one-proportion z-test​ met?

State the null and alternative hypotheses. Choose the correct answer below.

A.H0​:p=0.0220.022

HA​:ps≠0.022

B.H0​:p=0.022

HA​:p<0.022

C.H0​:p=0.022

HA​:p>0.022

D.The assumptions and conditions are not​ met, so the test cannot proceed.

Determine the​ z-test statistic. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

A.z=___(Round to two decimal places as​ needed.)

B.The assumptions and conditions are not​ met, so the test cannot proceed.

Find the​ P-value. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

A.P-value=__

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

B.The assumptions and conditions are not​ met, so the test cannot proceed.

What is your​ conclusion? Choose the correct answer below.

A.Rejec H0. The proportion of twin births for teenage mothers is greater than the proportion of twin births for all mothers.

B.Fail to reject H0. The proportion of twin births for teenage mothers is not different from the proportion of twin births for all mothers.

C.Reject H0. The proportion of twin births for teenage mothers is different from the proportion of twin births for all mothers.

D.The assumptions and conditions are not​ met, so the test cannot proceed.

Answer 1

Answer:

Given,

Null hypothesis Ho : p = 0.022

Alternative hypothesis Ha : p != 0.022

consider,

np = 516*0.022 = 11.352 >= 10

nq = 516*(1-0.022) = 504.648 >= 10

p^ = x/n

= 10/516

= 0.0194

Standard deviation = sqrt(pq/n)

= sqrt(0.022(1-0.022)/516)

= 0.0065

test statistic z = (p^ - p)/SD

substitute values

= (0.0194 - 0.022)/0.0065

z = -0.4

P value = 2*P(z > -0.4)

= 0.6891565 [since from z table]

= 0.6892

Here we observe that, p value > alpha(0.1) , so we fail to reject Ho.