Question

Suppose that you play the game with three different friends separately with the following results: Friend A chose scissors 100 times out of 400 games, Friend B chose scissors 20 times out of 120 games, and Friend C chose scissors 65 times out of 300 games. Suppose that for each friend you want to test whether the long-run proportion that the friend will pick scissors is less than 1/3.
1) Select the appropriate standardized statistics for each friend from the null distribution produced by applet.

-3.47 (100 out of 400; 25%), -4.17 (20 out of 120; 16.7%), -3.80 (65 out of 300; 21.7%)

-3.80 (100 out of 400; 25%), -3.47 (20 out of 120; 16.7%), -4.17 (65 out of 300; 21.7%)

-3.47 (100 out of 400; 25%), -3.80 (20 out of 120; 16.7%), -4.17 (65 out of 300; 21.7%)

-4.17 (100 out of 400; 25%), -3.80 (20 out of 120; 16.7%), -3.47 (65 out of 300; 21.7%)

Answer 1

Answer:

Friend A

$\hat p_A= \frac{100}{400}=0.25$

$z=\frac{0.25 -0.333}{\sqrt{\frac{0.333(1-0.333)}{400}}}\approx -3.47$  

Friend B

$\hat p_B= \frac{20}{120}=0.167$

$z=\frac{0.167 -0.333}{\sqrt{\frac{0.333(1-0.333)}{120}}}\approx -3.80$  

Friend C

$\hat p_C= \frac{65}{300}=0.217$

$z=\frac{0.217-0.333}{\sqrt{\frac{0.333(1-0.333)}{300}}}\approx -4.17$  

So then the best solution for this case would be:

-3.47 (100 out of 400; 25%), -3.80 (20 out of 120; 16.7%), -4.17 (65 out of 300; 21.7%)

Step-by-step explanation:

Data given and notation

n represent the random sample taken

X represent the number of scissors selected for each friend

$\hat p=\frac{X}{n}$ estimated proportion of  scissors selected for each friend

$p_o=\frac{1}{3}=0.333$ 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)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion that the friend will pick scissors is less than 1/3 or 0.333, the system of hypothesis would be:  

Null hypothesis:$p\geq 0.333$  

Alternative hypothesis:$p < 0.333$  

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 $p_o$.

Calculate the statistic  

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

Friend A

$\hat p_A= \frac{100}{400}=0.25$

$z=\frac{0.25 -0.333}{\sqrt{\frac{0.333(1-0.333)}{400}}}\approx -3.47$  

Friend B

$\hat p_B= \frac{20}{120}=0.167$

$z=\frac{0.167 -0.333}{\sqrt{\frac{0.333(1-0.333)}{120}}}\approx -3.80$  

Friend C

$\hat p_C= \frac{65}{300}=0.217$

$z=\frac{0.217-0.333}{\sqrt{\frac{0.333(1-0.333)}{300}}}\approx -4.17$  

So then the best solution for this case would be:

-3.47 (100 out of 400; 25%), -3.80 (20 out of 120; 16.7%), -4.17 (65 out of 300; 21.7%)