Commute times in the U. S. are heavily skewed to the right. We select a random sample of 240 people from the 2000 U. S. Census who reported a non-zero commute time.

In this sample, the mean commute time is 28.9 minutes with a standard deviation of 19.0 minutes. Can we conclude from this data that the mean commute time in the U. S. is less than half an hour? Conduct a hypothesis test at the 5% level of significance.

What is the p-value for this hypothesis test?

If we compare the p value with a significance level for given we see that so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean is lower than 30 minutes or half of and hour at 5% of signficance.    

Step-by-step explanation:

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

For this case since the sample size is large enough n = 240>30 we can assume that the distribution for the sample mean is normal.

Data given and notation    

represent the sample mean

represent the sample standard deviation for the sample    

sample size    

represent the value that we want to test  

represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)    

represent the p value for the test (variable of interest)

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to determine if the mean conmute time in U.S is  less than 30 min, the system of hypothesis would be:    

Null hypothesis:    

Alternative hypothesis:    

We don't know the population deviation, so for this case we can use the t test to compare the actual mean to the reference value, and the statistic is given by:    

(1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic    

We can replace in formula (1) the info given like this:    

Calculate the P-value    

The degrees of freedom are given by:

Since is a one-side lower test the p value would be:    

Conclusion    

If we compare the p value with a significance level for given we see that so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean is lower than 30 minutes or half of and hour at 5% of signficance.