How long does it take a snail to crawl 72 inches?

Assuming this complete problem: "12. An archaeologist uses an accelerator mass spectrometer to find the age of a buried branch. At the 68% confidence level, the spectrometer estimates that the branch was 10,000 years old with a following could the spectrometer estimate as the age of the branch at the 95% confidence level?"

The possible options are:

A. 9500 years old, with a margin of error of 500 years

B. 10000 years old, with a margin of error of 400 years

C. 9500 years olds, with a margin of error of 50 years

D. 10000 years old, with a margin of error of 40 year

Solution to the problem

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".

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

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

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

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

The margin of error for this case is given by this formula:

$ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

For the first case the confidence was 68% so then $\alpha =1-0.68 =0.32$ and $\alpha/2 =0.16$ we can find a critical value on the normal standard distribution that accumulates 0.16 of the area on each tail and this value is $z=\pm 0.994$ , because P(z<-0.944) = 0.16 and P(Z>.944) = 0.16

The margin of error can be expressed like this:

$ME = z_{\alpha/2} SE$

We can solve for the standard error on this case and we got:

$SE = \frac{ME}{z_{\alpha/2}} =\frac{200}{0.994}=201.207$

And then for the new confidence interval we need to calculate the new $z_{\alpha}$ . For the first case the confidence was 95% so then $\alpha =1-0.95 =0.05$ and $\alpha/2 =0.025$ we can find a critical value on the normal standard distribution that accumulates 0.025 of the area on each tail and this value is $z=\pm 1.96$ , because P(z<-1.96) = 0.025 and P(Z>1.96) = 0.025. So then the new margin of error would be: