Question

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.

Answer 1

Answer:

$33.9-1.32\frac{7.2}{\sqrt{25}}=31.999$    

$33.9+1.32\frac{7.2}{\sqrt{25}}=35.801$    

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

And the margin of error is given by:

$ME = t_{\alpha/2}\frac{s}{\sqrt{n}}$

And replacing we got:

$ME = 1.32\frac{7.2}{\sqrt{25}} =1.9008$

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".

$\bar X=33.9$ represent the sample mean for the sample  

$\mu$ 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:

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

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

$df=n-1=25-1=24$

Since the Confidence is 0.80 or 80%, the value of $\alpha=0.2$ and $\alpha/2 =0.1$, 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 $t_{\alpha/2}=1.32$

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

$33.9-1.32\frac{7.2}{\sqrt{25}}=31.999$    

$33.9+1.32\frac{7.2}{\sqrt{25}}=35.801$    

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

And the margin of error is given by:

$ME = t_{\alpha/2}\frac{s}{\sqrt{n}}$

And replacing we got:

$ME = 1.32\frac{7.2}{\sqrt{25}} =1.9008$