The speeds of vehicles traveling on a highway are normally distributed with an unkown population mean and standard deviation. A random sample of 11 vehicles is taken and results in a sample mean of 65 miles per hour and sample standard deviation of 9 miles per hour. Determine and interpret the 90% confidence interval for the mean speeds of vehicles traveling on the highway.

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".

[tex]\bar X =65[/tex] represent the sample mean for the sample

[tex]\mu[/tex] population mean (variable of interest)

s=9 represent the sample standard deviation

n=11 represent the sample size

2) Confidence interval

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

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

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

[tex]df=n-1=11-1=10[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,10)".And we see that [tex]t_{\alpha/2}=1.81[/tex]

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

[tex]65-1.81\frac{9}{\sqrt{11}}=60.09[/tex]

[tex]65+1.81\frac{9}{\sqrt{11}}=69.91[/tex]

So on this case the 90% confidence interval would be given by (60.09;69.91)

We are 90% confident that the true mean for the speeds of vehicles traveling on a highway is between 60.09 and 69.91 miles per hour.