Question

A die was cast \(n=120\) independent times and the following data resulted: \begin{tabular}{c|cccccc} Spots Up & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline Frequency & \(b\) & 20 & 20 & 20 & 20 & \(40-b\) \end{tabular} If we use a chi-square test, for what values of \(b\) would the hypothesis that the die is unbiased be rejected at the \(0.025\) significance level?

Short answer

The null hypothesis that the die is unbiased would be rejected if \(b < 20 - \sqrt{78.33}\) or \(b > 20 + \sqrt{78.33}\), in other words, if the frequency of '1' is too low or too high compared to what is expected for an unbiased die.

Step-by-step solution

Calculate expected frequencies

For an unbiased die, we would expect each outcome to appear \(n/6\) times. So, the expected frequency for each outcome is \(120/6 = 20\).

Calculate the chi-square statistic

The chi-square statistic is the sum of the squared differences between observed and expected frequencies divided by the expected frequency, for each outcome. For this die, the chi-square statistic is given by: \(X^2 = \sum_{i=1}^6 \frac{(O_{i}-E_{i})^2}{E_{i}} = \frac{(b-20)^2}{20} + \sum_{i=2}^5 \frac{(20-20)^2}{20} + \frac{(40-b-20)^2}{20}\). Where, \(O_{i}\) represents the observed frequency and \(E_{i}\) represents the expected frequency for outcome \(i\). Simplifying, \(X^2 = \frac{(b-20)^2}{20} + \frac{(20-b)^2}{20} = \frac{2b^2 -80b + 800}{20} = \frac{b^2 - 40b + 400}{10}\).

Consider the rejection region

The null hypothesis would be rejected if the chi-square statistic is greater than the critical value. The critical value for df=5 and alpha=0.025 is 12.833. Hence, we reject the null hypothesis if \(\frac{b^2 - 40b + 400}{10}>12.833\). Solving this inequality gives the range of values for which the null hypothesis would be rejected.

Solve the inequality

To find the values of \(b\) for which the null hypothesis would be rejected, solve the inequality \(b^2 -40b + 400 > 128.33\). This is a quadratic inequality that can be solved by finding the roots of the quadratic equation \(b^2 -40b + 271.67 = 0\) and testing the intervals defined by the roots. The roots can be found using the quadratic formula: \(b = \frac{40\pm\sqrt{40^2 - 4*1*271.67}}{2*1}\). Solving for \(b\) gives: \(b_1 = 20 - \sqrt{78.33}\), and \(b_2 = 20 + \sqrt{78.33}\). Therefore, the null hypothesis would be rejected if \(b < b_1\) or \(b > b_2\).