Question

In 2000 a national vital statistics report indicated that about 2.5% of all births produced twins. Is the rate of twin births same among very young mothers. Data from a large city hospital found only 10 sets of twins were born to 497 teenage girls. Determine the z statistic and find the p value.

Answer 1

Here we have given that,

n=number of teenage girls = 497

x: number of teenage girls born sets of twins = 10

Now, we estimate the sample proportion as

=sample proportion =

Claim: To check whether the rate of twin births same among very young mothers.

The null and alternative hypothesis are as follows,

v/s

where p is the population proportion young mother produces of twin births.

This is the two-tailed test.

Now, we can find the test statistic is as follows,

Z-statistics=

  =

=

= -0.70

The test statistics is -0.70.

Now we find the P-value,

p-value=2*( P(Z < z-statistics) as this is two-tailed test

               =2* (P( Z < -0.70)

                =2* (0.24196 ) Using standard normal z table see the value corresponding to the z=-0.71

                =0.4839

The p-value is 0.4839

Decision:

= level of significance= 0.05

Here p-value (0.4839) greater than (>) 0.05

Conclusion:

We fail to reject the Ho (Null Hypothesis)

There is not sufficient evidence to support the claim that rate of twin births same among very young mothers.