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.
