Suppose that a jewelry store tracked the amount of emeralds they sold each week to more accurately estimate how many emeralds to keep in stock. At the end of one year, the store used the weekly sales information to construct a 95% confidence interval for the mean number of emeralds sold per week. The confidence interval they calculated was ( 2.13 , 3.37 ) . Use the confidence interval to find the point estimate for the mean and the margin of error. Give your answers precise to two decimal places.

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[/tex] represent the sample mean

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

s represent the sample standard deviation

n 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)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

[tex]\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75[/tex]

And for the margin of error we have the following estimation:

[tex] ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62[/tex]