1. ## random sample-help!

Hi all,

In a city A, 20% of a random sample of 900 school boys had a certain slight physical defect. In another city B, 18.5% of a random sample of 1600 school boys had the same defect.
Is the difference between the proportions significant ?

2. Originally Posted by okello
Hi all,

In a city A, 20% of a random sample of 900 school boys had a certain slight physical defect. In another city B, 18.5% of a random sample of 1600 school boys had the same defect.
Is the difference between the proportions significant ?
Suppose that $X_1$ ~ Binomial $(n_1, \theta_1)$ and $X_2$ ~ Binomial $(n_2, \theta_2)$ and you want to test $\epsilon = \theta_1 - \theta_2$.

If $n_1$ and $n_2$ are large then $\frac{X_1}{n_1} \approx N \left(\theta_1, \frac{\theta_1 (1 - \theta_1)}{n_1} \right)$ and $\frac{X_2}{n_2} \approx N \left(\theta_2, \frac{\theta_2 (1 - \theta_2)}{n_2} \right)$

and so $\frac{\frac{X_1}{n_1} - \frac{X_2}{n_2} - (\theta_1 - \theta_2)}{\sqrt{\frac{\theta_1 (1 - \theta_1)}{n_1} + \frac{\theta_2 (1 - \theta_2)}{n_2} }} \approx N (0, 1)$.

With $\hat{\epsilon} = \frac{X_1}{n_1} - \frac{X_2}{n_2}$ then $se(\hat{\epsilon}) = \sqrt{\frac{\hat{\theta_1} (1 - \hat{\theta_1})}{n_1} + \frac{\hat{\theta_2} (1 - \hat{\theta_2})}{n_2} }$ and so the test is based on $\frac{\hat{\epsilon} - \epsilon}{se(\hat{\epsilon})} \approx N(0, 1)$.

You are testing the hypothesis H0: $\theta_1 = \theta_2$, that is, $\epsilon = 0$ against (I suppose) a two-sided alternative.

Decide on a level of significance, substitute the data and crunch the numbers.