Question - Understanding the Null Hypothesis in Probability of Gender Distribution

The null hypothesis states that each child has an equal chance of being a girl, which is a probability of 0.5. Since there are four children, the probability that all children are girls or all are boys (the two extreme cases) is $0.5^4 = 0.0625$ for each case. Together, the probability for either all girls or all boys is $0.0625 + 0.0625 = 0.125$, hence $H_0: p = 0.125$.

The test statistic for a proportion is $Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$ where $\hat{p}$ is the sample proportion, $p_0$ is the hypothesized population proportion, and $n$ is the sample size.

Here, $\hat{p} = \frac{21}{120}$, $p_0 = 0.125$, and $n = 120$. So:

$Z = \frac{\frac{21}{120} - 0.125}{\sqrt{\frac{0.125(1-0.125)}{120}}}$

Compute the Z-value and then compare it to the critical value for a one-tailed test at the 10% significance level, which is $Z = 1.28$ for $Z > 0$ (looking at a standard Z-table and finding the value corresponding to 0.9 since it is one-tailed).

If the calculated $Z$-value is greater than the critical value, we reject the null hypothesis. Otherwise, we do not reject the null hypothesis.