Question

Required information An article presents a new method for timing traffic signals in heavily traveled intersections. The effectiveness of the new method was evaluated in a simulation study. In 50 simulations, the mean improvement in traffic flow in a particular intersection was 652.6 vehicles per hour, with a standard deviation of 311.7 vehicles per hour. Find a 95% confidence interval for the improvement in traffic flow due to the new system. Round the answers to three decimal places. The 95% confidence interval is ( 564.015 , 741.185 ).

Answer 1

Answer:

$652.6-2.01\frac{311.7}{\sqrt{50}}=563.997$    

$652.6+2.01\frac{311.7}{\sqrt{50}}=741.203$    

So on this case the 95% confidence interval would be given by (563.997;741.203)    

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$\bar X=652.6$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s=311.7 represent the sample standard deviation

n=50 represent the sample size  

Soltuion to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=50-1=49$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,49)".And we see that $t_{\alpha/2}=2.01$

Now we have everything in order to replace into formula (1):

$652.6-2.01\frac{311.7}{\sqrt{50}}=563.997$    

$652.6+2.01\frac{311.7}{\sqrt{50}}=741.203$    

So on this case the 95% confidence interval would be given by (563.997;741.203)