Question

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 653.5 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 new system. Round the to three decimal places.

Answer 1

`653.5-2.01(311.7)/(√(50))=564.897`

`653.5+2.01(311.7)/(√(50))=742.103`

So on this case the 95% confidence interval would be given by (564.897;742.103)

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=653.5` represent the sample mean

`\mu` population mean (variable of interest)

s=311.7 represent the sample standard deviation

n=50 represent the sample size

Calculate the confidence interval

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

`\bar X \pm t_(\alpha/2)(s)/(√(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(0.025,2.01)".And we see that
`t_(\alpha/2)=2.01`.

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

`653.5-2.01(311.7)/(√(50))=564.897`

`653.5+2.01(311.7)/(√(50))=742.103`

So on this case the 95% confidence interval would be given by (564.897;742.103)