Question

You have just entered a very competitive Rock-Paper-Scissors tournament sponsored by the World Rock Paper Scissors Society. Since the first-place prize is $10,000, you spend some time researching the various strategies to increase your chances of winning. A claim that you see repeatedly in your research is that Paper is a good strategy since Scissors is chosen less than (1/3) of the time in competitive tournaments. To test this claim you obtain a record of 634 Rock-Paper-Scissors choices in competitive tournaments and observe 190 Scissors. Let p denote the population proportion of choices in competitive tournaments that are Scissors. Answer the following questions about the hypothesis test
H0: p = (1/3) vs. HA: p < (1/3).

Question 1. What is the value of the test statistic z? (use 1 decimal place in your answer)
Question 2. What is the P-value? (use 3 decimal places in your answer)
Question 3. What is an appropriate conclusion for this hypothesis test? (1 submission allowed)

a. Do not reject H0: p = (1/3); there is insufficient evidence to conclude that the population proportion p of Scissors choices in competitive Rock-Paper-Scissors tournaments differs from (1/3).
b. Reject H0: p = (1/3); there is sufficient evidence to conclude that p is greater than (1/3).
c. Reject H0: p = (1/3); there is sufficient evidence to conclude that p is less than (1/3).

Answer 1

Answer:

1. $z=\frac{0.300 -0.333}{\sqrt{\frac{0.333(1-0.333)}{634}}}=-1.8$  

2. $p_v =P(z$  

2. c. Reject H0: p = (1/3); there is sufficient evidence to conclude that p is less than (1/3).

Step-by-step explanation:

1) Data given and notation

n=634 represent the random sample taken

X=190 represent the choices in competitive tournaments that are Scissors

$\hat p=\frac{190}{634}=0.300$ estimated proportion of choices in competitive tournaments that are Scissors

$p_o=0.333$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is less than 0.33.:  

Null hypothesis:$p\geq 0.33$  

Alternative hypothesis:$p < 0.33$  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

$z=\frac{0.300 -0.333}{\sqrt{\frac{0.333(1-0.333)}{634}}}=-1.762$  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided $\alpha$. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

$p_v =P(z$  

If we compare the p value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion it's significantly less than 0.333.  

c. Reject H0: p = (1/3); there is sufficient evidence to conclude that p is less than (1/3).