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The null hypothesis (denoted as \( \text{H}_0 \)) is a statement we assume to be true before we collect any data. It represents the default or established belief about a situation. In the context of the Chi-Square Test for the California's Daily 4 Lottery digits, the null hypothesis claims that each digit from 0 to 9 is equally likely to appear. This means that, according to the null hypothesis, no digit is more or less likely to be chosen than any other. Our task is to collect data and see if there is enough evidence to refute this assumption. If our calculations and the resulting Chi-Square statistic indicate significant differences from what we'd expect if the digits were equally likely, we can consider rejecting the null hypothesis.

Contrary to the null hypothesis, the alternative hypothesis (denoted as \( \text{H}_1 \)) suggests that there is a difference - that at least one digit in the lottery is more or less likely to occur than the others. This is what we are trying to prove with our test. If our Chi-Square test shows significant results, we have reason to believe that the digits are not equally likely, thus supporting the alternative hypothesis. In this exercise, \( \text{H}_1 \) states that the digits are not all equally likely, meaning probability variations exist among the digits selected.

The expected frequency is a crucial aspect of the Chi-Square test. It is the frequency we'd expect to observe for each category (in this case, each digit) if the null hypothesis were true. For our lottery digits, since we're assuming they are equally likely and there are 10 digits, the expected frequency is calculated as follows: divide the total number of observations by the number of categories. Here, we recorded 240 digits over 60 days. Hence, the expected frequency for each digit is \( \text{Expected Frequency} = \frac{240}{10} = 24 \). This means, under the null hypothesis, we would expect each digit to appear 24 times.

Observed frequency refers to how many times each digit actually appeared in the recorded data. In our exercise, these frequencies are \( 21, 30, 31, 33, 19, 23, 21, 16, 24, \) and \( 22 \). The Chi-Square test then compares these observed frequencies to the expected frequencies to see if there are significant differences. These differences indicate whether some digits appear more or less frequently than what would be expected if they were truly equally likely, as stated by the null hypothesis.

Degrees of freedom in a Chi-Square test refer to the number of categories minus one. This concept is essential because it affects the critical value we use to determine significance. For our Chi-Square test, there are 10 categories (digits 0 to 9). Therefore, the degrees of freedom are calculated as follows: \( \text{Degrees of Freedom} = \text{Number of Categories} - 1 = 10 - 1 = 9 \). When we determine the significance of our Chi-Square statistic, we'll use this degrees of freedom to find the critical value from the Chi-Square distribution table at our chosen significance level (in this case, \( 0.05 \)). If our test statistic exceeds this critical value, we'll reject the null hypothesis, concluding that the digits are not equally likely.