91Ó°ÊÓ

The accompanying table lists results of overtime football games before and after the overtime rule was changed in the National Football League in 2011 . Use a 0.05 significance level to test the claim of independence between winning an overtime game and whether playing under the old rule or the new rule. What do the results suggest about the effectiveness of the rule change? $$\begin{array}{l|c|c} \hline & \text { Before Rule Change } & \text { After Rule Change } \\ \hline \text { Overtime Coin Toss Winner } & & \\ \text { Won the Game } & 252 & 24 \\ \hline \begin{array}{l} \text { Overtime Coin Toss Winner } \\ \text { Lost the Game } \end{array} & 208 & 23 \\ \hline \end{array}$$

Step by step solution

01

- State the Hypotheses

Define the null hypothesis (H_0) and the alternative hypothesis (H_1). Here, H_0: The outcome of the game (win/loss) is independent of the rule change (before/after). H_1: The outcome of the game (win/loss) is dependent on the rule change (before/after).

02

- Construct the Contingency Table

Form a contingency table based on the given data: \[ \begin{array}{|l|c|c|c|} \hline \text{ } & \text{Before Rule Change} & \text{After Rule Change} & \text{Total} \ \hline \text{Won the Game} & 252 & 24 & 276 \ \hline \text{Lost the Game} & 208 & 23 & 231 \ \hline \text{Total} & 460 & 47 & 507 \ \hline \end{array} \]

03

- Calculate Expected Frequencies

Use the formula for expected frequency: \[ E = \frac{(\text{row total}) (\text{column total})}{\text{grand total}} \]. Calculate the expected frequencies for each cell: Expected frequency for Won the Game before rule change: \[ E_{11} = \frac{276 \times 460}{507} \approx 250.4 \], Expected frequency for Won the Game after rule change: \[ E_{12} = \frac{276 \times 47}{507} \approx 25.6 \], Expected frequency for Lost the Game before rule change: \[ E_{21} = \frac{231 \times 460}{507} \approx 209.6 \], Expected frequency for Lost the Game after rule change: \[ E_{22} = \frac{231 \times 47}{507} \approx 21.4 \].

04

- Compute the Chi-Square Statistic

Use the formula for the chi-square statistic: \[ \chi^2 = \sum \frac{(O - E)^2}{E} \], where \(O\) is the observed frequency, and \(E\) is the expected frequency: \[ \chi^2 = \frac{(252 - 250.4)^2}{250.4} + \frac{(24 - 25.6)^2}{25.6} + \frac{(208-209.6)^2}{209.6} + \frac{(23-21.4)^2}{21.4} \approx 0.071 + 0.100 + 0.012 + 0.117 = 0.3 \].

05

- Determine the Degrees of Freedom

Calculate the degrees of freedom using: \[ df = (r-1) (c-1) \], where \(r\) is the number of rows and \(c\) is the number of columns. Here, degrees of freedom: \( df = (2-1)(2-1) = 1 \).

06

- Find the Critical Value and Compare

Using a 0.05 significance level and 1 degree of freedom, the critical value from the Chi-square distribution table is \(3.841\). Compare the chi-square statistic to the critical value: \(0.3 < 3.841\).

07

- Decision and Interpretation

Since the chi-square statistic is less than the critical value, fail to reject the null hypothesis. There is no significant evidence that winning an overtime game is dependent on the rule change. The results suggest that the rule change did not significantly affect the effectiveness of the games.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

null hypothesis

The null hypothesis (denoted as H₀) is a statement that suggests there is no effect or no difference. In the context of a Chi-Square Test for Independence, it means that two categorical variables are independent of each other.
For example, in our exercise, the null hypothesis is that the outcome of winning an overtime game is independent of whether the game was played before or after the rule change in the NFL.
We write it mathematically as: H₀: The outcome of the game (win/loss) is independent of the rule change (before/after).

alternative hypothesis

The alternative hypothesis (denoted as H₁) is the statement that contradicts the null hypothesis. It suggests that there is an effect or a difference between the groups.
In our exercise, the alternative hypothesis asserts that the outcome of winning an overtime game is dependent on whether the game was played before or after the rule change.
This is written as: H₁: The outcome of the game (win/loss) is dependent on the rule change (before/after).

contingency table

A contingency table is a tabular representation that categorizes the data into rows and columns to show the frequency distribution of the variables. It's used to analyze the relationship between categorical variables.
For example, in our exercise, the contingency table shows the frequency of games won and lost by the overtime coin toss winner both before and after the rule change:
Won the Game: - Before Rule Change: 252, After Rule Change: 24
Lost the Game: - Before Rule Change: 208, After Rule Change: 23
By adding totals, we complete the table:
Total Games Before Rule Change: 460
Total Games After Rule Change: 47

expected frequency

Expected frequencies are calculated to determine what we would expect to observe in each category if the null hypothesis is true. They are found using the formula:
\[ E = \frac{(\text{row total}) (\text{column total})}{\text{grand total}} \]
For example, in our contingency table, the expected frequency of games won before the rule change is:
\[ E_{11} = \frac{276 \times 460}{507} \approx 250.4 \]
Similarly, we calculate expected frequencies for other cells.

chi-square statistic

The chi-square statistic measures how much the observed frequencies deviate from the expected frequencies. It's calculated using the formula:
\[ \chi^2 = \sum \frac{(O - E)^2}{E} \]
where O is the observed frequency, and E is the expected frequency.
In our exercise, the chi-square statistic is:
\[ \chi^2 = \sum \frac{(252 - 250.4)^2}{250.4} + \sum \frac{(24 - 25.6)^2}{25.6} + \sum \frac{(208-209.6)^2}{209.6} + \sum \frac{(23-21.4)^2}{21.4} \approx 0.3 \]

degrees of freedom

Degrees of freedom (df) refer to the number of values that are free to vary when calculating a statistic. They are calculated in the context of a contingency table using the formula:
\[ df = (r-1) \times (c-1) \]
where r is the number of rows and c is the number of columns.
For our exercise, there are 2 rows and 2 columns, so the degrees of freedom are:
\[ df = (2-1)(2-1) = 1 \]

significance level

The significance level (often denoted as α) is the threshold at which we reject the null hypothesis. Common values include 0.05 or 0.01.
For our exercise, we use a significance level of 0.05. This means we are willing to accept a 5% chance of incorrectly rejecting the null hypothesis.
Using our degrees of freedom (1), we find the critical value from the chi-square distribution table for α = 0.05 is 3.841.
Since our calculated chi-square statistic (0.3) is less than 3.841, we fail to reject the null hypothesis. Thus, there's no significant evidence that winning an overtime game is dependent on the rule change.