Bernoulli problem

First, calculate $\hat{p} $ , the sample proportion = the number of foreigners in the sample divided by the number of students in the sample = 21/225 $\approx$ 0.0933.

The test statistic is $z=\frac{\hat{p}-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}$ where $p_{0}$ is the value of p under the null hypothesis, which in this case is 0.1.

$\frac{\frac{21}{225}-0.1}{\sqrt{\frac{(0.1)(0.9)}{225}}} = -\frac{1}{3}$

The distrubution of Z is asymptotically standard normal.

$H_{0}: p \geqslant 0.1$

$H_{A}: p < 0.1$

Since the alternative hypothesis is to the left, we should look at the tail to the left of the test statistic to get the p-value.

In R: pnorm(-1/3) gives us 0.3694413
In excel: NORMDIST(-1/3, 0, 1, TRUE) gives us the same value
There are also tables where you can look these p-values up if you google 'z table' or 'standard normal cdf table'

Conclusion: The p-value is greater than .05, so we fail to reject the null hypothesis. There is insufficient evidence that the true proportion of foreign students is less than 10%.