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Rock-paper-scissors is a hand game played by two or more people where players choose to sign either rock, paper, or scissors with their hands. For your statistics class project, you want to evaluate whether players choose between these three options randomly, or if certain options are favored above others. You ask two friends to play rock-paper-scissors and count the times each option is played. The following table summarizes the data: $$ \begin{array}{ccc} \text { Rock } & \text { Paper } & \text { Scissors } \\ \hline 43 & 21 & 35 \end{array} $$ Use these data to evaluate whether players choose between these three options randomly, or if certain options are favored above others. Make sure to clearly outline each step of your analysis, and interpret your results in context of the data and the research question.

Step by step solution

01

Define the Hypotheses

Formulate the null and alternative hypotheses for the test. The null hypothesis \(H_0\) states that the choices of Rock, Paper, and Scissors are made randomly and uniformly. This means each option is equally likely with a probability of \( \frac{1}{3} \) (since there are three options). The alternative hypothesis \(H_a\) states that at least one choice deviates from this uniform probability.

02

Calculate Expected Frequencies

Given the total number of plays, calculate the expected frequency of each choice under the null hypothesis. The total count of plays is \(43 + 21 + 35 = 99\). Assuming randomness, each option should occur \( \frac{1}{3} \) of the time, so the expected frequency for each option is \(99 \times \frac{1}{3} = 33\). This means we expect 33 plays for Rock, 33 for Paper, and 33 for Scissors if choices are random.

03

Perform Chi-Square Test

Using the observed and expected frequencies, calculate the chi-square statistic. The chi-square statistic is given by \( \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \), where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency. Calculate it for Rock, Paper, and Scissors:Rock: \( \frac{(43 - 33)^2}{33} = \frac{100}{33} \approx 3.03\)Paper: \( \frac{(21 - 33)^2}{33} = \frac{144}{33} \approx 4.36\)Scissors: \( \frac{(35 - 33)^2}{33} = \frac{4}{33} \approx 0.12\)Combine these: \( \chi^2 \approx 3.03 + 4.36 + 0.12 = 7.51\).

04

Determine the Degrees of Freedom and Significance Level

Identify the degrees of freedom for the test, which is the number of categories minus one. Here, we have three categories (Rock, Paper, Scissors), so the degrees of freedom is \(3 - 1 = 2\). We'll typically use a significance level (\(\alpha\)) of 0.05.

05

Conclusion from Chi-Square Distribution

Using a chi-square distribution table or calculator, compare the calculated \( \chi^2 \approx 7.51\) with the critical value for 2 degrees of freedom at the 0.05 significance level, which is 5.991. Since \(7.51\) exceeds \(5.991\), we reject the null hypothesis, suggesting that players do not choose their options randomly.

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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, often denoted as \( H_0 \), is a fundamental concept in hypothesis testing in statistics. It assumes that there is no effect or difference in the population and any observed effect is due to sampling error or random chance. In the context of the rock-paper-scissors game, the null hypothesis states that players choose each option with equal probability, indicating randomness.

• This hypothesis serves as a starting point for testing.
• It allows us to quantitatively evaluate assumptions about data.

When performing a chi-square test, the null hypothesis would claim that the frequency of choosing rock, paper, or scissors is uniform. If the evidence suggests a deviation from what is expected under the null, it may be rejected.

Alternative Hypothesis

Contrary to the null hypothesis, the alternative hypothesis is denoted as \( H_a \) and proposes that there is a significant effect or difference. In our scenario, it suggests that players do not choose rock, paper, or scissors with equal likelihood.

• This hypothesis is directional, indicating that at least one choice is preferred.
• Evidence supporting this will allow us to reject \( H_0 \).

In hypothesis testing, identifying the correct alternative hypothesis is crucial as it directly influences the conclusions. If the chi-square test results provide sufficient evidence to reject the null hypothesis, the alternative hypothesis may be accepted, suggesting a preference or inconsistency in player choices.

Degrees of Freedom

Degrees of freedom, a key parameter in many statistical tests, inform us about the number of values free to vary in a calculation. For a chi-square test, it's calculated as the number of categories minus one. In this exercise, we have three categories: rock, paper, and scissors, thus:\[ \text{Degrees of Freedom} = 3 - 1 = 2 \]

• It helps determine the shape of the chi-square distribution.
• Higher degrees of freedom result in a more spread out distribution.

Degrees of freedom are crucial in identifying which chi-square critical value to compare against the test statistic, thereby determining the rejection or acceptance of \( H_0 \). A smaller number offers less flexibility and generally results in higher likelihoods of significant results for larger statistics.

Significance Level

The significance level \( \alpha \) is a threshold chosen before a test to decide if the null hypothesis should be rejected. Commonly set at 0.05, it denotes a 5% risk of concluding that an effect exists when there is none.

• Defines strictness of the evidence required to reject \( H_0 \).
• Lower levels, such as 0.01, indicate more rigorous tests.

In the chi-square test conducted, the observed value of 7.51 is compared to the critical value from a chi-square distribution table based on \( \alpha = 0.05 \) and 2 degrees of freedom. If the calculated value surpasses this threshold, as it does in the exercise, \( H_0 \) is rejected, supporting the alternative hypothesis. This highlights a difference significant enough not to be attributed to random chance alone.