Question

The u.s. department of transportation estimates that 10% of americans carpool. does that imply that 10% of cars will have two or more occupants? a sample of 300 cars traveling southbound on the new jersey turnpike yesterday revealed that 63 had two or more occupants. at the 0.01 significance level, can we conclude that 10% of cars traveling on the new jersey turnpike have two or more occupants?

Answer 1

Answer: I HAVE NO IDEA SORRYYYYYYYY

Explanation:

Answer 2

Answer:

$z=\frac{0.21 -0.1}{\sqrt{\frac{0.1(1-0.1)}{300}}}=6.35$  

$p_v =2*P(z>6.35)=2.15x10^{-10}$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of cars that had two or more occupants is significantly different from 0.1 or 10%.  

Explanation:

1) Data given and notation

n=300 represent the random sample taken

X=63 represent the number of cars that had two or more occupants

$\hat p=\frac{63}{300}=0.21$ estimated proportion of cars that had two or more occupants

$p_o=0.1$ 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 the true proportion of cars that had two or more occupants is 0.1:  

Null hypothesis:$p=0.1$  

Alternative hypothesis:$p \neq 0.1$  

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:  

$z=\frac{0.21 -0.1}{\sqrt{\frac{0.1(1-0.1)}{300}}}=6.35$  

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:  

$p_v =2*P(z>6.35)=2.15x10^{-10}$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of cars that had two or more occupants is significantly different from 0.1 or 10%.