Answer to Question #347487 in Statistics and Probability for mitzykia

Question

Newborn babies are more likely to be boys than girls. A random sample found that 13,173 boys were born among 25,468 newborn children. The sample proportion of boys was 0.5172. Is this sample evidence that the birth of boys is more common than the birth of girls in the entire population? Use 95% confidence level.

Answer 1

The following null and alternative hypotheses for the population proportion needs to be tested:

$H_0:p\le0.5$

$H_a:p>0.5$

This corresponds to a right-tailed test, for which a z-test for one population proportion will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a right-tailed test is $z_c = 1.6449.$

The rejection region for this right-tailed test is $R = \{z: z > 1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{\dfrac{13173}{25468}-0.5}{\sqrt{\dfrac{0.5(1-0.5)}{25468}}}\approx5.5017$$

Since it is observed that $z = 5.5017 > 1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=P(Z>5.5017)=0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$ is greater than $0.5,$ at the $\alpha = 0.05$ significance level.