Rock - Paper - Scissors

Suppose you ask two friends to play rock-paper-scissors, count the times each option is played, and obtain the data shown in the table below. Use a one-way \chi^2-test to evaluate whether these data support the hypothesis that players choose between these three options randomly, or if certain options are favored above others.

Here is the data that I obtained for the Rock, Paper, Scissors data.

H_0: Each option in Rock, Paper, Scissors did not differ between each option.
H_A: One of the options in Rock, Paper, Scissors is more likely to be picked.

R-Studio Code

rps = data.frame(observed = c(43, 21, 35), expected = (33))
dimnames(rps) = list(options = c("Rock", "Paper", "Scissors"), picked = c("Observed", "Expected"))
rps

chisq.test(rps)

                     Pearson's Chi-squared test

          data:  rps
          X-squared = 4.0413, df = 2, p-value = 0.1326

The options picked from the two friends isn’t significant enough to differ between each option. (X^2 = 4.0413 , p=0.1326).

I agree!

H0: Each option in Rock, Paper, Scissors did not differ between each option.
HA: One of the options in Rock, Paper, Scissors is more likely to be picked.

rps = data.frame(observed = c(43, 21, 35), expected = (33))

dimnames(rps) = list(options = c(“Rock”, “Paper”, “Scissors”), picked = c (“Observed”, “Expected”))

rps

                  Observed Expected
 Rock                   43       33
 Paper                  21       33
 Scissors               35       33

(43 + 21+35) = 99

equal probability for expected: 1/3

(43-33)^2 / 33 + (21-33)^2 / 33 + (35-33)^2 / 33 = 7.51515 = 7.52

 chisq.test(rps)

Pearson’s Chi-squared test

data: rps
X-squared = 4.0413, df = 2, p-value = 0.1326

0.1326>0.05

Options picked are not significant enough to differ between each option. We are confident 95% of the time that we would fail to reject the null hypothesis.