Q6. Noonan syndrome is a genetic condition that can affect the heart, growth blood clotting, and mental and physical development. Noonan et al. examined the stature of men and women with Noonan syndrome. The study contained 29 male and 44 female adults. One of the cut-off values used to assess stature was the third percentile of adult height Eleven of the males fell below the third percentile of adult male height, while 24 of the females fell below the third percentile of female adult height. Does this study provide sufficient evidence for us to conclude that among subjects with Noonan syndrome, proportion of females are more likely than males to fall below the respective third percentile of adult height? Let alpha =5% Assume that the distribution is not normal. a) H_(b)/P_(1)=p_(2)H_(4);p_(1)gt p_(2),z-cest ozloulated value=-1.39,z- table wakbe = Lite possible of the froples H_(3), -1.645 dollot r_(1);p_(2),z- test calluniated value=-1.39,z- cab 1.645reqlectH_(0) H_(4);p_(1)=p_(2),H_(1):p_(1),p^2-test calcalealared value=1.93,x^2- vable val d) 3.841,donotrefectH_(0) H_(4)p_(1)=p_(2),H_(1);p_(2),x^2- test calrealated value=1.93,x^2- table 5.024, do not reject H_(0)

To determine if there is sufficient evidence to conclude that the proportion of females with Noonan syndrome falling below the third percentile of adult height is greater than that of males, we can perform a hypothesis test for the difference in proportions.<br /><br />Let's define the null and alternative hypotheses as follows:<br /><br />$H_{0}: p_{1} = p_{2}$ (The proportion of males and females with Noonan syndrome falling below the third percentile of adult height is equal.)<br />$H_{1}: p_{1} < p_{2}$ (The proportion of females with Noonan syndrome falling below the third percentile of adult height is greater than that of males.)<br /><br />We will use a two-proportion z-test to test this hypothesis. The test statistic is calculated as follows:<br /><br />$z = \frac{\hat{p}_{1} - \hat{p}_{2}}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_{1}} + \frac{\hat{p}(1-\hat{p})}{n_{2}}}}$<br /><br />where $\hat{p}_{1}$ and $\hat{p}_{2}$ are the sample proportions of males and females falling below the third percentile, respectively, and $\hat{p}$ is the pooled sample proportion. The values for $\hat{p}_{1}$, $\hat{p}_{2}$, and $\hat{p}$ can be calculated as follows:<br /><br />$\hat{p}_{1} = \frac{11}{29}$<br />$\hat{p}_{2} = \frac{24}{44}$<br />$\hat{p} = \frac{11 + 24}{29 + 44}$<br /><br />The critical value for a one-tailed test at $\alpha = 0.05$ is $z = -1.645$. If the calculated value of $z$ is less than the critical value, we reject the null hypothesis.<br /><br />Now, let's calculate the test statistic and compare it with the critical value:<br /><br />$z = \frac{\frac{11}{29} - \frac{24}{44}}{\sqrt{\frac{\frac{11 + 24}{29 + 44}(1 - \frac{11 + 24}{29 + 44})}{29} + \frac{\frac{11 + 24}{29 + 44}(1 - \frac{11 + 24}{29 + 44})}{44}}$<br /><br />After calculating the above expression, we get the value of $z$. We then compare this value with the critical value of $-1.645$. If the calculated value of $z$ is less than the critical value, we reject the null hypothesis.<br /><br />Based on the given information, the calculated value of $z$ is $-1.39$, which is greater than the critical value of $-1.645$. Therefore, we do not reject the null hypothesis.<br /><br />Hence, the study does not provide sufficient evidence to conclude that the proportion of females with Noonan syndrome falling below the third percentile of adult height is greater than that of males.