Question

The Department of Highway Improvements, responsible for repairing a 25-mile stretch of interstate highway, wants to design a surface that will be structurally efficient. One important consideration is the volume of heavy freight traffic on the interstate. State weigh stations report that the average number of heavy-duty trailers traveling on a 25-mile segment of the interstate is 72 per hour. However, the section of highway to be repaired is located in an urban area and the department engineers believe that the volume of heavy freight traffic for this particular section is greater than the average reported for the entire interstate.

To validate this theory, the department monitors the highway for 50 1-hour periods randomly selected throughout the month. Suppose the sample mean and standard deviation of the heavy freight traffic for the 50 sampled hours are:
Sample mean = 74.1 standard deviation = 13.3
a) Do the data support the department's theory? Use α = 0.10 to come to a conclusion based on a large sample test.

Answer 1

`t=(74.1-72)/((13.3)/(√(50)))=1.116`

The degrees of freedom are given by:

`df=n-1=50-1=49`

Now we can calculate the p value with the following probability:

`p_v =P(t_((49))>1.116)=0.135`

Since the p value is higher than the significance level provided of 0.1 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 72 per hour.

Explanation:

Information provided

`\bar X=74.1` represent the sample mean

`s=13.1` represent the sample standard deviation

`n=50` sample size

`\mu_o =72` represent the value that we want to test

`\alpha=0.1` represent the significance level

t would represent the statistic

`p_v` represent the p value

Hypothesis to test

We want to analyze if the true mean for this case is higher than 72, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 72`

Alternative hypothesis:
`\mu > 72`

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the info we got:

`t=(74.1-72)/((13.3)/(√(50)))=1.116`

The degrees of freedom are given by:

`df=n-1=50-1=49`

Now we can calculate the p value with the following probability:

`p_v =P(t_((49))>1.116)=0.135`

Since the p value is higher than the significance level provided of 0.1 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 72 per hour.