A traffic engineer believes that the number of cars passing through a certain intersection between 2-6 pm on weekdays follows a normal distribution with mean 750. A new highway is opened, and it is hypothesized that the number of cars passing through the intersection will change as a result. A sample of 15 weekdays is taken, and the average number of cars passing through the intersection is computed to be 720, with a sample standard deviation of 42. Decide whether this reduction in traffic is statistically significant, using significance level = 0.05.

Respuesta :

Answer:

[tex]t=\frac{720-750}{\frac{42}{\sqrt{15}}}=-2.766[/tex]    

[tex]df=n-1=15-1=14[/tex]  

[tex]p_v =P(t_{(14)}<-2.766)=0.0076[/tex]  

We see that the p value is lower than the significance level given of 0.05. So then we have enough evidence to reject the null hypothesis and we can say that we have a significantly reduction in the traffic.

Step-by-step explanation:

Data given

[tex]\bar X=720[/tex] represent the sample mean

[tex]s=42[/tex] represent the sample standard deviation

[tex]n=15[/tex] sample size  

[tex]\mu_o =750[/tex] represent the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 750 (reduction), the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 750[/tex]  

Alternative hypothesis:[tex]\mu < 750[/tex]  

The statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

Calculate the statistic

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

[tex]t=\frac{720-750}{\frac{42}{\sqrt{15}}}=-2.766[/tex]    

P-value

The degrees of freedom are given by:

[tex]df=n-1=15-1=14[/tex]  

Since is a one sided lef tailed test the p value would be:  

[tex]p_v =P(t_{(14)}<-2.766)=0.0076[/tex]  

Conclusion  

We see that the p value is lower than the significance level given of 0.05. So then we have enough evidence to reject the null hypothesis and we can say that we have a significantly reduction in the traffic.

RELAXING NOICE
Relax