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

Step-by-step explanation:

Answer 2

`z=\frac{0.21 -0.1}{\sqrt{(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<<\alpha` 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%.

Step-by-step 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=(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 \\eq 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{(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{(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<<\alpha` 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%.