Question

State the test statistic and P-value. Interpret these values.

For the 30 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

Let $p_1$ be the true proportion of babies born with low birth weight for women with a history of premature labor.

Let $p_2$ be the true proportion of babies born with low birth weight for women with no history of premature labor.

We want to investigate if  the data provide evidence that the proportion of babies born with low birth weight is higher for women with a history of premature labor, that is we want to test if $p_1>p_2$

We want to test the following hypotheses

$\begin{align*} &H_0:p_1=p_2\leftarrow\text{null hypothesis: There is no difference in the proportions}\\ &H_a:p_1>p_2\leftarrow\text{alternative hypothesis: The proportion is higher for women with a history of premature labor}\\ &\alpha=0.01\leftarrow\text{level of significance to test the hypotheses} \end{align*}$

We have the following information from the sample

$\begin{align*} &n_1=30\leftarrow\text{sample size of women with a history of premature labor}\\ &\bar{p}_1=\frac{18}{30}=0.60\leftarrow\text{sample proportion having babies with low birth weight for women with a history of premature labor}\\ &n_2=159\leftarrow\text{sample size of women with no history of premature labor}\\ &\bar{p}_2=\frac{41}{159}=0.26\leftarrow\text{sample proportion having babies with low birth weight for women with no history of premature labor}\\ &(p_1-p_2)_{H_0}=0\leftarrow\text{Hypothesized difference between the proportions} \end{align*}$

The overall proportion of babies with low birth weight is

$\begin{align*} \hat{p}=\frac{18+41}{30+159}=0.3122 \end{align*}$

The estimated standard error of the difference between 2 proportions is

$\begin{align*} \hat{\sigma}_{\bar{p}_1-\bar{p}_2}=\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}=\sqrt{\frac{0.3122(1-0.3122)}{30}+\frac{0.3122(1-0.3122)}{159}}=0.0922 \end{align*}$

We calculate  the test statistics as

$\begin{align*} z=\frac{(\bar{p}_1-\bar{p}_2)-(p_1-p_2)_{H_0}}{\hat{\sigma}_{\bar{p}_1-\bar{p}_2}}=\frac{(0.6-0.26)-0}{0.0922}=3.71 \end{align*}$

this is a one tailed (right tailed ) test. (The alternative hypothesis has ">")

We get the critical value of z for significance level alpha=0.01 using

$\begin{align*} P(Z>z_{\alpha})=0.01\implies P(Z<z_{\alpha})=1-0.01=0.99 \end{align*}$

Using the standard normal tables we can get that for z=2.33 the area under the curve is (0.5+0.4901) = 0.9901

Hence the critical value is 2.33.

We will reject the null hypothesis if the test statistics is greater than the critical value 2.33.

Here the test statistics of 3.71 is greater than the critical value of 2.33. Hence we will reject the null hypothesis.

We can conclude that the data provides sufficient evidence to support the claim the proportion of babies born with low birth weight is higher for women with a history of premature labor.