Question

Before analyzing the data, use your own experience and intuition to predict what the data will show. Do you think the proportion of babies with low birth weight is higher for women with a history of premature labor? Are the criteria for approximate normality satisfied?

For the 30 random women in the study with a history of premature labor, a proportion of 18/30 = 0.60 (60%) had babies with low birth weight. For the remaining 159 women, a proportion of 41/159 = 0.26 (26%) had babies with low birth weight.

We now investigate the following research question: do the data provide evidence that the proportion of babies born with low birth weight is higher for women with a history of premature labor? This question is answered with a hypothesis test. To conduct the test we use a 1% level of significance.

Answer 1

Premature labor babies with low birth weight:

n1=30

x1=18

The Proportion (p1)= X1/n1= 18/30 =0.60

The remaining women babies with low birth weight:

n2=159

x2=41

The proportion (p2)= x2/n2= 41/159= 0.257862

To test normality of test: the condition must accept np>5 and npq>5

First sample: n1p1= 30*0.6= 18 and n1*p1*(1-p1)= 30*0.6*0.4= 7.2

Second Sample: n2*p2= 159*0.257862= 41 and n2*p2*(1-p2)= 159*0.257892*(1-0.257892)=30.42767

Two samples are meeting the normality requirement.

Two proportion Z test:

Hypothesis Test:

H0: p1 - p2 > 0

Ha: p1  - p2 $\leq$ 0

The significant level= 0.01(One tail)

The significant value: Z(0.01) =2.326

The proportion of success: $\hat{p}=\frac{p_1+p_2}{n_1+n_2}=\frac{18+41}{30+159}=0.312169$

The test statistic:

$Z= \frac{p_1-p_2-0}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1}+\frac{1}{n_2})}}= \frac{0.6-0.257862-0}{\sqrt{0.312169(1-0.312169)(\frac{1}{30}+\frac{1}{159})}}$

$Z= \frac{0.6-0.257862-0}{0.092237514 }=0.342138/0.0922375=3.709319$

P-value: Z Score 3.709319: One tail= 0.000

The test statistic is significant and there is enough evidence to support the proportion of babies with low birth weight is higher for women with a history of premature labor.