Rock-paper-scissors. 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 AP 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:
Rock: 43
Paper: 21
Scissors: 35
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.
Independence is satisfied assuming that all selections are random.
Expected Value is equal to \(np\), where the proportion for each is 1/3 assuming an equal chance to pick each one, and the sample size is 99 (43+21+35). All the expected values will be 33, which is greater than 5. Therefore, we have a large enough sample.
The null and alternative hypothesis can be seen below:
\(H_{0}: \text{All proportions are equal}\)
\(H_{a}: \text{At least one of the proportions is different}\)
To tests our hypotheses, we will use a Chi-Square Goodness of Fit test. The equation to calculate Chi-Squared, \(\chi^{2}\), is shown below:
\(\chi^{2} = \sum\frac{\left(O-E\right)^{2}}{E}\)
where \(O\) is the number of observations and \(E\) is the expected value. The degrees of freedom, \(df\), for this test is the number of categories minus one. In this case, it would be 2 degrees of freedom.
After we calculate \(\chi^{2}\), we can calculate the \(p\)-value, which will ultimately tell us if we can reject the null or not. Let us assume that the significance level is 0.05 (\(\alpha=0.05\)).
The following R-code will calculate the \(p\)-value:
n <- c(43,21,35)
samp_size <- sum(n)
p <- c(1/3,1/3,1/3)
E <- samp_size*p
chi2 <- 0
for (i in 1:length(n)) {
chi2 <- chi2 + ((n[i]-E[i])^2)/E[i]
}
num_cat <- length(n)
df <- num_cat-1
pchisq(chi2,df,lower.tail=FALSE)[1] 0.02334025From the previous section, we can see the \(p\)-value is 0.023, which is less than \(\alpha\) (0.05). This means that we have sufficient evidence to reject the null hypothesis in favor of the alternative. Therefore, the proportions of selections are not equal.