Question

a study published in college mathematics journal found that novice players of rock - paper - scissors tend to pick scissors less often than by chance. you want to test the hypotheses $h_0:pi = 1/3$, $h_a:pi < 1/3$ on your friend jaye. jaye plays 6 rounds of rock - paper - scissors, and picks scissors 0 times.
a. the probability that jaye picks scissors 0 times can be written as $p(e_1$ and $e_2$ and... and $e_6)$. where $e_1...e_6$ are events. what is $e_3$?
a. the event that jaye picks scissors 3 times.
b. the event that jaye picks scissors on the 3rd round of the game.
c. the 3rd round of the game.
d. the event that jaye picks something other than scissors on the 3rd round of the game.
b. if the null hypothesis is true, what is $p(e_3)$?
c. compute the theory - based p - value.
d. what assumptions are involved in computing the p - value in part c? select all that apply.
a. jayes choices in different rounds of the game are independent.
b. jayes choices in different rounds of the game are mutually exclusive.
c. jayes probability of picking scissors in each round is 1/3.
d. this sample is representative of jayes choices when playing rock - paper - scissors in general.

Explanation:

Step1: Identify $E_3$

In the context of probability of not picking scissors 0 times in 6 - round game, $E_3$ represents the event in the 3rd round. Since we are looking at the non - occurrence of picking scissors in all rounds, $E_3$ is the event that Jaye picks something other than scissors on the 3rd round of the game.

Step2: Calculate $P(E_3)$ under null hypothesis

Under the null hypothesis $H_0:\pi = 1/3$, where $\pi$ is the probability of picking scissors. The probability of not picking scissors is $P(E_3)=1 - \frac{1}{3}=\frac{2}{3}$.

Step3: Compute the p - value

The number of trials $n = 6$, and the number of successes (picking scissors) $x = 0$. Using the binomial probability formula $P(X=x)=\binom{n}{x}\pi^{x}(1 - \pi)^{n - x}$, with $n = 6$, $x = 0$, and $\pi=\frac{1}{3}$, we have $\binom{6}{0}(\frac{1}{3})^{0}(1-\frac{1}{3})^{6-0}=1\times1\times(\frac{2}{3})^{6}=\frac{64}{729}\approx0.088$.

Step4: Identify assumptions

• Independence: Jaye's choices in different rounds of the game are independent. This is a necessary assumption for using the binomial probability formula.
• Constant probability: Jaye's probability of picking scissors in each round is $\frac{1}{3}$ under the null hypothesis.

Answer:

a. D. The event that Jaye picks something other than scissors on the 3rd round of the game.
b. $\frac{2}{3}$
c. $\frac{64}{729}\approx0.088$
d. A. Jaye's choices in different rounds of the game are independent; C. Jaye's probability of picking scissors in each round is 1/3.