In a best-of-seven series, if both teams are equally matched, which test would assess whether the observed distribution of series lengths fits the 50-50 model?

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Multiple Choice

In a best-of-seven series, if both teams are equally matched, which test would assess whether the observed distribution of series lengths fits the 50-50 model?

Explanation:
The key idea is testing whether observed frequencies across categories match a specified distribution. Here, a best-of-seven series can end in four, five, six, or seven games, so the outcome is a four-category categorical variable. If both teams are truly evenly matched, the model gives fixed probabilities for each length based on combinatorics (for example, shorter sweeps are less likely than mid-range lengths, and the total probabilities sum to 1). A chi-square goodness-of-fit test compares the observed counts in each length category to the counts expected under that 50-50 model, telling you if the observed distribution fits the theoretical one. The degrees of freedom are three because there are four categories and the model specifies all four probabilities, so df = 4 − 1 = 3. Other tests don’t fit this goal: a chi-square test of independence would look for association in a contingency table, not a single-category distribution; a paired t-test compares means of paired measurements; and linear regression models a numeric outcome as a function of a predictor.

The key idea is testing whether observed frequencies across categories match a specified distribution. Here, a best-of-seven series can end in four, five, six, or seven games, so the outcome is a four-category categorical variable. If both teams are truly evenly matched, the model gives fixed probabilities for each length based on combinatorics (for example, shorter sweeps are less likely than mid-range lengths, and the total probabilities sum to 1). A chi-square goodness-of-fit test compares the observed counts in each length category to the counts expected under that 50-50 model, telling you if the observed distribution fits the theoretical one.

The degrees of freedom are three because there are four categories and the model specifies all four probabilities, so df = 4 − 1 = 3. Other tests don’t fit this goal: a chi-square test of independence would look for association in a contingency table, not a single-category distribution; a paired t-test compares means of paired measurements; and linear regression models a numeric outcome as a function of a predictor.

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