Fission tracks are trails found in uranium-bearing minerals, left by fragments released during fission events. An article reports that fifteen tracks on one rock specimen had an average track length of 12 μm with a standard deviation of 2 μm. Assuming this to be a random sample from an approximately normal population, find a 99% confidence interval for the mean track length for this rock specimen. Round the answers to three decimal places. The 99% confidence interval is ( , ).

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 =12[/tex] represent the sample mean

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

s=2 represent the sample standard deviation

n=15 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=15-1=14[/tex]

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that [tex]t_{\alpha/2}=2.977[/tex]

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

[tex]12-2.977\frac{2}{\sqrt{15}}=10.463[/tex]

[tex]12+2.977\frac{2}{\sqrt{15}}=13.537[/tex]

So on this case the 99% confidence interval would be given by (10.463;13.537)