Question

Are most student government leaders extroverts? According to Myers-Briggs estimates, about 82% of college student government leaders are extroverts.† Suppose that a Myers-Briggs personality preference test was given to a random sample of 74 student government leaders attending a large national leadership conference and that 57 were found to be extroverts. Does this indicate that the population proportion of extroverts among college student government leaders is different (either way) from 82%? Use α = 0.01.

Answer 1

Answer:

$z=-1.12$  

$p_v =2*P(z<-1.12)=0.263$  

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we don't have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the population proportion of extroverts among college student government leaders is not different from 0.82.  

Step-by-step explanation:

1) Data given and notation

n=74 represent the random sample taken

X=57 represent the number of people found extroverts

$\hat p=\frac{57}{74}=0.770$ estimated proportion of people found  extroverts.

$p_o=0.82$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.959

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 if population proportion of extroverts among college student government leaders is different (either way) from 82%, the system of hypothesis would be on this case:  

Null hypothesis:$p= 0.82$  

Alternative hypothesis:$p \neq 0.82$  

When we conduct a proportion test we need to use the z statisitc, 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:  

$z=\frac{0.770 -0.82}{\sqrt{\frac{0.82(1-0.82)}{74}}}=-1.12$  

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 $\alpha=0.01$. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

$p_v =2*P(z<-1.12)=0.263$  

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we don't have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the population proportion of extroverts among college student government leaders is not different from 0.82.