In a random sample of 25 people, the mean commute time to work was 33.9 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a t-distribution to construct a 80% confidence interval for the population mean mu. What is the margin of error of mu? Interpret the results.

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=33.9[/tex] represent the sample mean for the sample

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

s=7.2 represent the sample standard deviation

n=25 represent the sample size

Solution to the problem

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=25-1=24[/tex]

Since the Confidence is 0.80 or 80%, the value of [tex]\alpha=0.2[/tex] and [tex]\alpha/2 =0.1[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,24)".And we see that [tex]t_{\alpha/2}=1.32[/tex]

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

[tex]33.9-1.32\frac{7.2}{\sqrt{25}}=31.999[/tex]

[tex]33.9+1.32\frac{7.2}{\sqrt{25}}=35.801[/tex]

So on this case the 80% confidence interval would be given by (31.999;35.801)

And the margin of error is given by:

[tex] ME = t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]

And replacing we got:

[tex] ME = 1.32\frac{7.2}{\sqrt{25}} =1.9008[/tex]