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The Chi-square statistic is a useful tool in testing how well observed data fit a particular expected distribution, such as the binomial distribution. The Chi-square test compares the observed frequencies of outcomes with the expected frequencies if a specified theoretical model—like a binomial distribution—were correct.

In our exercise, the observed frequencies provided are for the values 0 through 4 of the random variable \(X\). We use these in conjunction with expected frequencies, derived under the assumption that \(X\) follows a binomial distribution with parameters \(n=6\) and \(p=0.25\). To calculate the Chi-square statistic, we utilize the formula:

\[ \chi^2 = \sum \frac{(O - E)^2}{E} \]

Here, \(O\) represents observed frequencies and \(E\) represents expected frequencies. This formula essentially measures the sum of the squared differences between observed and expected frequencies, each divided by the expected frequency itself. A larger Chi-square statistic indicates a larger discrepancy between what was observed and what was expected, suggesting a poor fit of the data to the theoretical model.

The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success, \(p\). It is an essential component in assessing the fit of our model. In our example, we are examining whether a binomial distribution with parameters \(n=6\) and \(p=0.25\) is suitable.

The expected probability for each outcome \(X=x\) in a binomial distribution is calculated with the formula:

\[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \]

Here, \(\binom{n}{x}\) is the binomial coefficient representing the number of ways \(x\) successes can occur in \(n\) trials.

The expected frequencies are derived by multiplying these probabilities by the total number of trials, or observations—in our case, this is 50. This forms the basis for comparing against observed frequencies to see if our binomial model satisfactorily describes the data.

The P-value is a crucial part of hypothesis testing. It measures the evidence against a null hypothesis, quantified as the probability that our statistic could be as extreme or more extreme than what we observe, based on chance alone. In simpler terms, it's the probability of getting a Chi-square statistic as large or larger if the null hypothesis were true.

After calculating the Chi-square statistic of 11.008, we determine the P-value using either statistical software or tables that show the Chi-square distribution. The degrees of freedom, in this case, are calculated as \(3\), which is the number of categories minus one parameter estimated.

For our calculated \( \chi^2 = 11.008 \) on 3 degrees of freedom, we find the P-value to be approximately 0.012. Since this value is less than our significance level \( \alpha = 0.05 \), we reject the null hypothesis, concluding that the binomial distribution is not an adequate fit for our observed data. A small P-value indicates the result is statistically significant.