In 2002 a national vital statistics report indicated that about 3.5% of all birt

Question

In 2002 a national vital statistics report indicated that about 3.5% 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 13 sets of twins were born to 557 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? O Yes No State the null and alternative hypotheses. Choose the correct answer below OA. Ho: p 0.035 HA: p 0.035 OA. z- (Round to two decimal places as needed.) O 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

Explanation / Answer

A) Are the assumptions and conditions to perform a one-proportion Z test met?

(1) Is sample independent?

One mother having twins will not impact another, so observations are independent; not an SRS

Is n < 10 %? Yes, it is certain that 557< 10% of all teenage mothers giving birth.

(2) Sample large enough?

Check whether at least 10 successes/failures:

np0 = 557× 0.035 = 19.5 > 10

n(1 ? p0) = 537.5 > 10

Yes, the assumptions and conditions to perform a one-proportion Z test are met.

B) State the null and alternative hypothesis:

Let p = true proportion of twin births for teenage mothers.

Null hypothesis H0 : p = 0.035

Alt. hypothesis HA : p ? 0.035

Test is 2-tailed – we want to know if the rate of twin births is same among for very young mothers or not?

C) Determine the Z statistics and find P value.  State your conclusion.

The sample size is N = 557,

The number of favorable cases is X = 13,

and the sample proportion is p¯?=X / N

?=13/557

?=0.0233

and the significance level is alpha (?) =0.05

Rejection Region

Based on the information provided, the significance level is ?=0.05,

and the critical value for a two-tailed test is Zc?=1.96.

Test Statistics

The z-statistic is computed as follows:

z=p0?(1?p0?)/ {sqrt (n?p¯??p0?)?}

=?1.50

Decision about the null hypothesis

Since it is observed that ?z?=1.50? Zc?=1.96, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.1343 and since0.1343?0.05, it is concluded that the null hypothesis is not rejected.

Hence we accept that, the rate of twins born to very young mothers at this hospital is 3.5%

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