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.

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

And the margin of error is given by:

And replacing we got:

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".

represent the sample mean for the sample  

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:

  (1)

In order to calculate the critical value we need to find first the degrees of freedom, given by:

Since the Confidence is 0.80 or 80%, the value of and , 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

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

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

And the margin of error is given by:

And replacing we got:

i’ll just s take the points thxs